RD Sharma solution class 7 chapter 15 Properties of triangles Exercise 15.2

Exercise 15.2

Page-15.12

Question 1:

Two angles of a triangle are of measures 105° and 30°. Find the measure of the third angle.

Answer 1:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{third} \mathrm{angle} \mathrm{be} \mathrm{x}. \\ \mathrm{Sum} \mathrm{of} \mathrm{all} \mathrm{the} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} =180° \\ \therefore 105° + 30° + \mathrm{x} = 180° \\ \mathrm{or}, \mathrm{x} = 180° - 135° \\ \therefore \mathrm{x}=45° \\ \mathrm{The} \mathrm{third} \mathrm{angle} \mathrm{is} 45°. \end{gathered}$$

Question 2:

One of the angles of a triangle is 130°, and the other two angles are equal. What is the measure of each of these equal angles?

Answer 2:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{other} \mathrm{two} \mathrm{angles} \mathrm{be} \mathrm{x}. \\ \mathrm{Sum} \mathrm{of} \mathrm{all} \mathrm{the} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} =180° \\ \mathrm{i}.\mathrm{e}. 130° + \mathrm{x} + \mathrm{x} = 180° \\ \Rightarrow 2\mathrm{x} = 180° - 130° \\ \Rightarrow 2\mathrm{x} = 50° \\ \Rightarrow \mathrm{x} = \frac{50°}{2} \\ \Rightarrow \mathrm{x} = 25° \\ \mathrm{Therefore}, \mathrm{the} \mathrm{other} \mathrm{two} \mathrm{angles} \mathrm{are} 25° \mathrm{each}. \end{gathered}$$

Question 3:

The three angles of a triangle are equal to one another. What is the measure of each of the angles?

Answer 3:

$$\begin{gathered} \mathrm{Let} \mathrm{each} \mathrm{angle} \mathrm{of} \mathrm{the} \mathrm{triangle} \mathrm{be} \mathrm{x}. \\ \mathrm{Sum} \mathrm{of} \mathrm{all} \mathrm{the} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} =180° \\ \mathrm{i}.\mathrm{e}. \mathrm{x}+\mathrm{x}+\mathrm{x}=180° \\ \Rightarrow 3\mathrm{x}=180° \\ \Rightarrow \mathrm{x}=\frac{180°}{3} \\ \Rightarrow \mathrm{x}=60° \\ \mathrm{Therefore}, \mathrm{each} \mathrm{angle} \mathrm{is} \mathrm{equal} \mathrm{to} 60°. \end{gathered}$$

Question 4:

If the angles of a triangle are in the ratio 1 : 2 : 3, determine three angles.

Answer 4:

$$\begin{gathered} \mathrm{If} \mathrm{the} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{are} \mathrm{in} \mathrm{ratio} 1:2:3, \mathrm{then} \mathrm{let} \mathrm{us} \mathrm{take} \mathrm{the} \mathrm{first} \mathrm{angle} \mathrm{to} \mathrm{be} \mathrm{x}. \\ \mathrm{Which} \mathrm{means} \mathrm{that} \mathrm{the} \mathrm{second} \mathrm{angle} \mathrm{will} \mathrm{be} 2\mathrm{x} \mathrm{and} \mathrm{the} \mathrm{third} \mathrm{angle} \mathrm{will} \mathrm{be} 3\mathrm{x}. \\ \mathrm{Sum} \mathrm{of} \mathrm{all} \mathrm{the} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} = 180° \\ \therefore \mathrm{x} + 2\mathrm{x} + 3\mathrm{x} = 180° \\ \Rightarrow 6\mathrm{x} = 180° \\ \Rightarrow \mathrm{x} = \frac{180°}{6} \\ \Rightarrow \mathrm{x} = 30° \\ \Rightarrow 2\mathrm{x} = 2 \times 30° = 60° \\ \Rightarrow 3\mathrm{x} = 3 \times 30° = 90° \\ \mathrm{Therefore}, \mathrm{the} \mathrm{first} \mathrm{angle} \mathrm{is} \mathrm{equal} \mathrm{to} 30°, \mathrm{the} \mathrm{second} \mathrm{angle} \mathrm{is} \mathrm{equal} \mathrm{to} 60°, \mathrm{and} \mathrm{the} \mathrm{third} \mathrm{angle} \mathrm{is} \mathrm{equal} \mathrm{to} 90°. \end{gathered}$$

Question 5:

The angles of a triangle are (x − 40)°, (x − 20)° and ${\left(\frac{1}{2}x-10\right)}^{°}$. Find the value of x.

Answer 5:

$$\begin{gathered} \mathrm{Sum} \mathrm{of} \mathrm{all} \mathrm{the} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} = 180° \\ \Rightarrow (\mathrm{x} - 40)° + (\mathrm{x} - 20)° + (\frac{\mathrm{x}}{2} - 10)° = 180° \\ \Rightarrow \mathrm{x} + \mathrm{x} + \frac{\mathrm{x}}{2} - 40° - 20° - 10° = 180° \\ \Rightarrow \mathrm{x} + \mathrm{x} + \frac{\mathrm{x}}{2} - 70° = 180° \\ \Rightarrow \frac{5\mathrm{x}}{2} = 180° + 70° \\ \Rightarrow \frac{5\mathrm{x}}{2} = 250° \\ \Rightarrow \mathrm{x} = \frac{2}{5} \times 250° \\ \Rightarrow \mathrm{x} = 100° \\ \mathrm{Hence}, \mathrm{we} \mathrm{can} \mathrm{conclude} \mathrm{that} \mathrm{x} \mathrm{is} \mathrm{equal} \mathrm{to} 100°. \end{gathered}$$

Question 6:

The angles of a triangle are arranged in ascending order of magnitude. If the difference between two consecutive angles is 10°, find the three angles.

Answer 6:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{first} \mathrm{angle} \mathrm{of} \mathrm{the} \mathrm{triangle} \mathrm{be} \mathrm{x}. \\ \mathrm{Therefore}, \mathrm{we} \mathrm{can} \mathrm{say} \mathrm{that} \mathrm{the} \mathrm{second} \mathrm{angle} \mathrm{of} \mathrm{the} \mathrm{triangle} \mathrm{will} \mathrm{be} (\mathrm{x} + 10°) \mathrm{and} \mathrm{the} \mathrm{third} \mathrm{angle} \mathrm{of} \mathrm{the} \mathrm{triangle} \mathrm{will} \mathrm{be} (\mathrm{x} + 10° + 10°). \\ \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{the} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{is} \mathrm{equal} \mathrm{to} 180°. \\ \therefore \mathrm{x} + (\mathrm{x} + 10°) + (\mathrm{x} + 10° + 10°) = 180° \\ \Rightarrow \mathrm{x} + \mathrm{x} + \mathrm{x} + 10° + 10° + 10° = 180° \\ \Rightarrow 3\mathrm{x}+30°=180° \\ \Rightarrow 3\mathrm{x}=180°-30° \\ \Rightarrow 3\mathrm{x}=150° \\ \Rightarrow \mathrm{x} =\frac{150°}{3} \\ \Rightarrow \mathrm{x} = 50° \\ \mathrm{Now}, (\mathrm{x} + 10°) = 50° + 10° \\ \Rightarrow (\mathrm{x} + 10°) = 60° \\ \mathrm{And}, (\mathrm{x} + 10° + 10°) = 50° + 10° + 10° \\ \Rightarrow (\mathrm{x} + 10° + 10°) = 70° \\ \mathrm{Hence}, \mathrm{we} \mathrm{can} \mathrm{say} \mathrm{that} \mathrm{the} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{the} \mathrm{triangle} \mathrm{are} 50°, 60°\mathrm{and} 70°. \end{gathered}$$

Question 7:

Two angles of a triangle are equal and the third angle is greater than each of those angles by 30°. Determine all the angles of the triangle.

Answer 7:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{two} \mathrm{equal} \mathrm{angles} \mathrm{of} \mathrm{the} \mathrm{triangle} \mathrm{be} \mathrm{x}. \\ \mathrm{Hence}, \mathrm{the} \mathrm{third} \mathrm{angle} \mathrm{of} \mathrm{the} \mathrm{triangle} \mathrm{will} \mathrm{be} (\mathrm{x}+30°). \\ \mathrm{Sum} \mathrm{of} \mathrm{all} \mathrm{the} \mathrm{three} \mathrm{angle} \mathrm{of} \mathrm{a} \mathrm{triangle}=180° \\ \therefore \mathrm{x}+\mathrm{x}+(\mathrm{x}+30°)=180° \\ \Rightarrow \mathrm{x}+\mathrm{x}+\mathrm{x}+30°=180° \\ \Rightarrow 3\mathrm{x}+30°=180° \\ \Rightarrow 3\mathrm{x}=180°-30° \\ \Rightarrow 3\mathrm{x}=150° \\ \Rightarrow \mathrm{x}=\frac{150°}{3} \\ \mathbf{\Rightarrow }\mathbf{x}\mathbf{=}\mathbf{50}\mathbf{°} \\ (\mathrm{x}+30°)=50°+30° \\ \mathbf{\Rightarrow }\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{30}\mathbf{°}\mathbf{)}\mathbf{=}\mathbf{80}\mathbf{°} \\ \mathrm{Hence}, \mathrm{we} \mathrm{can} \mathrm{conclude} \mathrm{that} \mathrm{the} \mathrm{angles} \mathrm{of} \mathrm{the} \mathrm{triangle} \mathrm{are} 50°, 50° \mathrm{and} 80°. \end{gathered}$$

Question 8:

If one angle of a triangle is equal to the sum of the other two, show that the triangle is a right triangle.

Answer 8:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{the} \mathrm{triangle} \mathrm{be} \angle \mathrm{a}, \angle \mathrm{b} \mathrm{and} \angle \mathrm{c}. \\ \mathrm{Given}: \angle \mathrm{a}=\angle \mathrm{b}+\angle \mathrm{c} \\ \mathrm{Also}, \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{the} \mathrm{three} \mathrm{angle} \mathrm{of} \mathrm{a} \mathrm{triangle} = 180° \\ \mathrm{Or}, \angle \mathrm{a}+\angle \mathrm{b}+\angle \mathrm{c}=180° \\ \Rightarrow \angle \mathrm{a}+\angle \mathrm{a}=180° (\because \angle \mathrm{a}=\angle \mathrm{b}+\angle \mathrm{c}) \\ \Rightarrow 2\angle \mathrm{a}=180° \\ \Rightarrow \angle \mathrm{a}=\frac{180°}{2} \\ \Rightarrow \angle \mathrm{a}=90° \\ \mathrm{Hence}, \mathrm{we} \mathrm{can} \mathrm{conclude} \mathrm{that} \mathrm{the} \mathrm{given} \mathrm{triangle} \mathrm{is} \mathrm{a} \mathrm{right} \mathrm{angle} \mathrm{triangle}. \end{gathered}$$

Question 9:

If each angle of a triangle is less than the sum of the other two, show that the triangle is acute angled.

Answer 9:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{the} \mathrm{triangle} \mathrm{be} \angle \mathrm{a}, \angle \mathrm{b} \mathrm{and} \angle \mathrm{c}. \\ \mathrm{We} \mathrm{know}: \angle \mathrm{a}<\angle \mathrm{b}+\angle \mathrm{c} .....\left(\mathrm{i}\right) \left(\mathrm{Given}\right) \\ \mathrm{Which} \mathrm{means}: \angle \mathrm{b}<\angle \mathrm{a}+\angle \mathrm{c} \\ \mathrm{Or}, \angle \mathrm{c}<\angle \mathrm{a}+\angle \mathrm{b} \\ \mathrm{We} \mathrm{also} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{the} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{is} \mathrm{equal} \mathrm{to} 180°. \\ \mathrm{Which} \mathrm{means}: \angle \mathrm{a}+\angle \mathrm{b}+\angle \mathrm{c}=180° \\ \mathrm{Or}, \angle \mathrm{b}+\angle \mathrm{c}=180°-\angle \mathrm{a} \\ \mathrm{Putting} \mathrm{the} \mathrm{value} \mathrm{of} \angle \mathrm{b}+\angle \mathrm{c} \mathrm{in} \mathrm{equation} \left(\mathrm{i}\right): \\ \angle \mathrm{a}<180°-\angle \mathrm{a} \\ \Rightarrow 2\angle \mathrm{a}<180° \\ \Rightarrow \angle \mathrm{a}<90° \\ \mathrm{Similarly}: \\ \angle \mathrm{b}<90° \\ \angle \mathrm{c}<90° \\ \mathrm{Hence}, \mathrm{we} \mathrm{can} \mathrm{conclude} \mathrm{that} \mathrm{the} \mathrm{given} \mathrm{triangle} \mathrm{is} \mathrm{an} \mathrm{acute} \mathrm{triangle}. \end{gathered}$$

Question 10:

In each of the following, the measures of three angles are given. State in which cases, the angles can possibly be those of a triangle:
(i) 63°, 37°, 80°
(ii) 45°, 61°, 73°
(iii) 59°, 72°, 61°
(iv) 45°, 45°, 90°
(v) 30°, 20°, 125°

Answer 10:

$$\begin{gathered} \left(\mathrm{i}\right) \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{the} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{is} \mathrm{equal} \mathrm{to} 180°. \\ \mathrm{Now}, \mathrm{let} \mathrm{us} \mathrm{find} \mathrm{the} \mathrm{sum} \mathrm{of} 63°, 37° \mathrm{and} 80°. \\ 63°+37°+80°=180° \\ \mathrm{The} \mathrm{sum} \mathrm{of} 63°, 37° \mathrm{and} 80° \mathrm{is} \mathrm{equal} \mathrm{to} 180°. \\ \mathrm{Hence}, \mathrm{we} \mathrm{can} \mathrm{say} \mathrm{that} \mathrm{the} \mathrm{given} \mathrm{angles} \mathrm{can} \mathrm{be} \mathrm{those} \mathrm{of} \mathrm{a} \mathrm{triangle}. \end{gathered}$$
$$\begin{gathered} \left(\mathrm{ii}\right) \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{the} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{is} \mathrm{equal} \mathrm{to} 180°. \\ \mathrm{Now}, \mathrm{let} \mathrm{us} \mathrm{find} \mathrm{the} \mathrm{sum} \mathrm{of} 45°, 61° \mathrm{and} 73°. \\ 45°+61°+73°=179° \\ \mathrm{The} \mathrm{sum} \mathrm{of} 45°, 61° \mathrm{and} 73° \mathrm{is} \mathbf{not} \mathrm{equal} \mathrm{to} 180°. \\ \mathrm{Hence}, \mathrm{we} \mathrm{can} \mathrm{say} \mathrm{that} \mathrm{the} \mathrm{given} \mathrm{angles} \mathrm{cannot} \mathrm{be} \mathrm{those} \mathrm{of} \mathrm{a} \mathrm{triangle}. \end{gathered}$$
$$\begin{gathered} \left(\mathrm{iii}\right) \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{the} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{is} \mathrm{equal} \mathrm{to} 180°. \\ \mathrm{Now}, \mathrm{let} \mathrm{us} \mathrm{find} \mathrm{the} \mathrm{sum} \mathrm{of} 59°, 72° \mathrm{and} 61°. \\ 59°+72°+61°=192° \\ \mathrm{The} \mathrm{sum} \mathrm{of} 59°, 72° \mathrm{and} 61° \mathrm{is} \mathbf{not} \mathrm{equal} \mathrm{to} 180°. \\ \mathrm{Hence}, \mathrm{we} \mathrm{can} \mathrm{say} \mathrm{that} \mathrm{the} \mathrm{given} \mathrm{angles} \mathrm{cannot} \mathrm{be} \mathrm{those} \mathrm{of} \mathrm{a} \mathrm{triangle}. \end{gathered}$$
$$\begin{gathered} \left(\mathrm{iv}\right) \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{the} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{is} \mathrm{equal} \mathrm{to} 180°. \\ \mathrm{Now}, \mathrm{let} \mathrm{us} \mathrm{find} \mathrm{the} \mathrm{sum} \mathrm{of} 45°, 45° \mathrm{and} 90°. \\ 45°+45°+90°=180° \\ \mathrm{The} \mathrm{sum} \mathrm{of} 45°, 45° \mathrm{and} 90° \mathrm{is} \mathrm{equal} \mathrm{to} 180°. \\ \mathrm{Hence}, \mathrm{we} \mathrm{can} \mathrm{say} \mathrm{that} \mathrm{the} \mathrm{given} \mathrm{angles} \mathrm{can} \mathrm{be} \mathrm{those} \mathrm{of} \mathrm{a} \mathrm{triangle}. \end{gathered}$$
$$\begin{gathered} \left(\mathrm{v}\right) \mathrm{We} \mathrm{know} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{the} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{traingle} \mathrm{is} \mathrm{equal} \mathrm{to} 180°. \\ \mathrm{Now}, \mathrm{let} \mathrm{us} \mathrm{find} \mathrm{the} \mathrm{sum} \mathrm{of} 30°, 20° \mathrm{and} 125°. \\ 30°+20°+125°=175° \\ \mathrm{The} \mathrm{sum} \mathrm{of} 30°, 20° \mathrm{and} 125° \mathrm{is} \mathbf{not} \mathrm{equal} \mathrm{to} 180°. \\ \mathrm{Hence}, \mathrm{we} \mathrm{can} \mathrm{say} \mathrm{that} \mathrm{the} \mathrm{given} \mathrm{angles} \mathrm{cannot} \mathrm{be} \mathrm{those} \mathrm{of} \mathrm{a} \mathrm{triangle}. \end{gathered}$$
Therefore, we can conclude that in (i) and (iv), the angles can be those of a triangle.

Question 11:

The angles of a triangle are in the ratio 3 : 4 : 5. Find the smallest angle.

Answer 11:

$$\begin{gathered} \mathrm{If} \mathrm{the} \mathrm{angles} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{triangle} \mathrm{are} \mathrm{in} \mathrm{the} \mathrm{ratio} 3:4:5, \mathrm{then} \mathrm{let} \mathrm{us} \mathrm{take} \mathrm{the} \mathrm{smallest} \mathrm{angle} \mathrm{as} 3\mathrm{x}. \\ \mathrm{This} \mathrm{means} \mathrm{that} \mathrm{the} \mathrm{second} \mathrm{angle} \mathrm{will} \mathrm{be} 4\mathrm{x} \mathrm{and} \mathrm{the} \mathrm{third} \mathrm{angle} \mathrm{will} \mathrm{be} 5\mathrm{x}. \\ \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{the} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{is} \mathrm{equal} \mathrm{to} 180°. \\ \therefore 3\mathrm{x}+4\mathrm{x}+5\mathrm{x}=180° \\ \Rightarrow 12\mathrm{x}=180° \\ \Rightarrow \mathrm{x}=\frac{180°}{12} \\ \Rightarrow \mathrm{x}=15° \\ \mathrm{Now}, 3\mathrm{x}=3\times 15°=45° \\ \mathrm{Therefore}, \mathrm{we} \mathrm{can} \mathrm{conclude} \mathrm{that} \mathrm{the} \mathrm{smallest} \mathrm{angle} \mathrm{is} 45°. \end{gathered}$$

Question 12:

Two acute angles of a right triangle are equal. Find the two angles.

Answer 12:

$$\begin{gathered} \mathrm{Let} \mathrm{each} \mathrm{of} \mathrm{the} \mathrm{two} \mathrm{acute} \mathrm{angles} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{triangle} \mathrm{be} \mathrm{x}. \\ \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{third} \mathrm{angle} \mathrm{is} 90°. \left(\mathrm{Given}\right) \\ \mathrm{We} \mathrm{also} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{the} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{is} \mathrm{equal} \mathrm{to} 180°. \\ \mathrm{Which} \mathrm{means}: \mathrm{x}+\mathrm{x}+90°=180° \\ \Rightarrow 2\mathrm{x}=180°-90° \\ \Rightarrow \mathrm{x}=\frac{90°}{2} \\ \Rightarrow \mathrm{x}=45° \\ \mathrm{Hence}, \mathrm{we} \mathrm{can} \mathrm{conclude} \mathrm{that} \mathrm{each} \mathrm{of} \mathrm{the} \mathrm{two} \mathrm{acute} \mathrm{angles} \mathrm{is} \mathrm{equal} \mathrm{to} 45°. \end{gathered}$$

Question 13:

One angle of a triangle is greater than the sum of the other two. What can you say about the measure of this angle? What type of a triangle is this?

Answer 13:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{triangle} \mathrm{be} \angle \mathrm{a}, \angle \mathrm{b} \mathrm{and} \angle \mathrm{c}. \\ \mathrm{We} \mathrm{know}: \angle \mathrm{a}>\angle \mathrm{b}+\angle \mathrm{c} .....\left(\mathrm{i}\right) \left(\mathrm{Given}\right) \\ \mathrm{We} \mathrm{also} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{the} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{is} \mathrm{equal} \mathrm{to} 180°. \\ \therefore \angle \mathrm{a}+\angle \mathrm{b}+\angle \mathrm{c}=180° \\ \Rightarrow \angle \mathrm{b}+\angle \mathrm{c}=180°-\angle \mathrm{a} \\ \mathrm{Putting} \mathrm{the} \mathrm{value} \mathrm{of} \angle \mathrm{b}+\angle \mathrm{c} \mathrm{from} \mathrm{equation} \left(\mathrm{i}\right): \\ \angle \mathrm{a}>180°-\angle \mathrm{a} \\ \Rightarrow 2\angle \mathrm{a}>180° \\ \Rightarrow \angle \mathrm{a}>90° \\ \mathrm{Thus}, \mathrm{the} \mathrm{angle} \mathrm{is} \mathrm{more} \mathrm{than} 90°. \\ \mathrm{Hence}, \mathrm{we} \mathrm{can} \mathrm{conclude} \mathrm{by} \mathrm{saying} \mathrm{that} \mathrm{the} \mathrm{given} \mathrm{triangle} \mathrm{is} \mathrm{an} \mathrm{obtuse} \mathrm{triangle}. \end{gathered}$$

Page-15.13

Question 14:

In the six cornered figure, AC, AD and AE are joined. Find ∠FAB + ∠ABC + ∠BCD + ∠CDE + ∠DEF + ∠EFA.

Answer 14:

$$\begin{gathered} \mathrm{We} \mathrm{have} \mathrm{to} \mathrm{find} \angle \mathrm{FAB}+\angle \mathrm{ABC}+\angle \mathrm{BCD}+\angle \mathrm{CDE}+\angle \mathrm{DEF}+\angle \mathrm{EFA} ......\left(\mathrm{i}\right) \\ \mathrm{From} \mathrm{the} \mathrm{figure}, \mathrm{we} \mathrm{have}: \\ \angle \mathrm{FAB}=\angle \mathrm{FAE}+\angle \mathrm{EAD}+\angle \mathrm{DAC}+\angle \mathrm{CAB} \\ \angle \mathrm{BCD}=\angle \mathrm{ACB}+\angle \mathrm{ACD} \\ \angle \mathrm{CDE}=\angle \mathrm{ADC}+\angle \mathrm{ADE} \\ \angle \mathrm{DEF}=\angle \mathrm{AED}+\angle \mathrm{AEF} \\ \mathrm{Putting} \mathrm{the} \mathrm{values} \mathrm{of} \angle \mathrm{FAB},\angle \mathrm{BCD},\angle \mathrm{CDE},\angle \mathrm{DEF} \mathrm{in} \mathrm{equation} \left(\mathrm{i}\right): \\ (\angle \mathrm{FAE}+\angle \mathrm{EAD}+\angle \mathrm{DAC}+\angle \mathrm{CAB}) + \angle \mathrm{ABC}+(\angle \mathrm{ACB}+\angle \mathrm{ACD}) + (\angle \mathrm{ADC}+\angle \mathrm{ADE}) + (\angle \mathrm{AED}+\angle \mathrm{AEF}) + \angle \mathrm{EFA} \\ \Rightarrow (\angle \mathrm{ABC}+\angle \mathrm{ACB}+\angle \mathrm{CAB}) + (\angle \mathrm{FAE}+\angle \mathrm{AEF}+\angle \mathrm{EFA}) + (\angle \mathrm{FAE}+\angle \mathrm{AEF}+\angle \mathrm{EFA}) + (\angle \mathrm{ADC}+\angle \mathrm{ACD}+\angle \mathrm{DAC}) .....\left(\mathrm{ii}\right) \\ \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{the} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{is} \mathrm{equal} \mathrm{to} 180°. \\ \mathrm{Hence} \mathrm{we} \mathrm{can} \mathrm{say} \mathrm{the} \mathrm{following}: \\ \angle \mathrm{ABC}+\angle \mathrm{ACB}+\angle \mathrm{CAB} =180°(\mathrm{angles} \mathrm{of} △\mathrm{ABC}) \\ \angle \mathrm{FAE}+\angle \mathrm{AEF}+\angle \mathrm{EFA} =180°(\mathrm{angles} \mathrm{of} △\mathrm{AFE}) \\ \angle \mathrm{AED}+\angle \mathrm{ADE}+\angle \mathrm{EAD} =180°(\mathrm{angles} \mathrm{of} △\mathrm{AED}) \\ \angle \mathrm{ADC}+\angle \mathrm{ACD}+\angle \mathrm{DAC} =180°(\mathrm{angles} \mathrm{of} △\mathrm{ADC}) \\ \mathrm{Putting} \mathrm{these} \mathrm{values} \mathrm{in} \mathrm{equation} \left(\mathrm{ii}\right): \\ 180°+180°+180°+180° \\ \mathrm{Hence}, \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{angles} \mathrm{is} 720° \end{gathered}$$

Question 15:

Find x, y, z (whichever is required) in the figures given below:

Answer 15:

$$\begin{gathered} \left(\mathrm{i}\right) \mathrm{In} △\mathrm{ABC} \mathrm{and} △\mathrm{ADE}, \mathrm{we} \mathrm{have}: \\ \angle \mathrm{ADE}=\angle \mathrm{ABC} (\mathrm{corresponding} \mathrm{angles}) \\ \mathbf{\Rightarrow }\mathbf{x}\mathbf{°}\mathbf{=}\mathbf{40}\mathbf{°} \\ \angle \mathrm{AED}=\angle \mathrm{ACB} (\mathrm{corresponding} \mathrm{angles}) \\ \mathbf{\Rightarrow }\mathbf{y}\mathbf{°}\mathbf{=}\mathbf{30}\mathbf{°} \\ \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{the} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{is} \mathrm{equal} \mathrm{to} 180°. \\ \therefore \mathrm{x}°+\mathrm{y}°+\mathrm{z}°=180° (\mathrm{Angles} \mathrm{of} △\mathrm{ADE}) \\ \mathrm{Which} \mathrm{means}: 40°+30°+\mathrm{z}°=180° \\ \Rightarrow \mathrm{z}°=180°-70° \\ \mathbf{\Rightarrow }\mathbf{z}\mathbf{°}\mathbf{=}\mathbf{110}\mathbf{°} \\ \mathrm{Therefore}, \mathrm{we} \mathrm{can} \mathrm{conclude} \mathrm{that} \mathrm{the} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{triangle} \mathrm{are} 40°, 30° \mathrm{and} 110°. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right) \mathrm{We} \mathrm{can} \mathrm{see} \mathrm{that} \mathrm{in} △\mathrm{ADC}, \angle \mathrm{ADC} \mathrm{is} \mathrm{equal} \mathrm{to} 90°. (△\mathrm{ADC} \mathrm{is} \mathrm{a} \mathrm{right} \mathrm{triangle}) \\ \mathrm{We} \mathrm{also} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{the} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{is} \mathrm{equal} \mathrm{to} 180°. \\ \mathrm{Which} \mathrm{means}: 45°+90°+\mathrm{y}°=180° (\mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{angles} \mathrm{of} △\mathrm{ADC}) \\ \Rightarrow 135°+\mathrm{y}°=180° \\ \Rightarrow \mathrm{y}°=180°-135° \\ \mathbf{\Rightarrow }\mathbf{y}\mathbf{°}\mathbf{=}\mathbf{45}\mathbf{°} \\ \mathrm{We} \mathrm{can} \mathrm{also} \mathrm{say} \mathrm{that} \mathrm{in} △\mathrm{ABC}, \angle \mathrm{ABC}+\angle \mathrm{ACB}+\angle \mathrm{BAC} \mathrm{is} \mathrm{equal} \mathrm{to} 180°. (\mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{angles} \mathrm{of} △\mathrm{ABC}) \\ \Rightarrow 40°+\mathrm{y}°+(\mathrm{x}°+45°)=180° \\ \Rightarrow 40°+45°+\mathrm{x}°+45°=180° (\because \mathrm{y}°=45°) \\ \Rightarrow \mathrm{x}°=180°-130° \\ \mathbf{\Rightarrow }\mathbf{x}\mathbf{°}\mathbf{=}\mathbf{50}\mathbf{°} \\ \mathrm{Therefore}, \mathrm{we} \mathrm{can} \mathrm{say} \mathrm{that} \mathrm{the} \mathrm{required} \mathrm{angles} \mathrm{are} 45° \mathrm{and} 50°. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right) \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{the} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{is} \mathrm{equal} \mathrm{to} 180°. \\ \mathrm{Therefore}, \mathrm{for} △\mathrm{ABD}: \\ \angle \mathrm{ABD}+\angle \mathrm{ADB}+\angle \mathrm{BAD}=180° (\mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{angles} \mathrm{of} △\mathrm{ABD}) \\ \Rightarrow 50°+\mathrm{x}°+50°=180° \\ \Rightarrow 100°+\mathrm{x}°=180° \\ \Rightarrow \mathrm{x}°=180°-100° \\ \mathbf{\Rightarrow }\mathbf{x}\mathbf{°}\mathbf{=}\mathbf{80}\mathbf{°} \\ \mathrm{For} △\mathrm{ABC}: \\ \angle \mathrm{ABC}+\angle \mathrm{ACB}+\angle \mathrm{BAC}=180° (\mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{angles} \mathrm{of} △\mathrm{ABC}) \\ \Rightarrow 50°+\mathrm{z}°+(50°+30°)=180° \\ \Rightarrow 50°+\mathrm{z}°+50°+30°=180° \\ \Rightarrow \mathrm{z}°=180°-130° \\ \mathbf{\Rightarrow }\mathbf{z}\mathbf{°}\mathbf{=}\mathbf{50}\mathbf{°} \\ \mathrm{Using} \mathrm{the} \mathrm{same} \mathrm{argument} \mathrm{for} △\mathrm{ADC}: \\ \angle \mathrm{ADC}+\angle \mathrm{ACD}+\angle \mathrm{DAC}=180° (\mathrm{Sum} \mathrm{of} \mathrm{angles} \mathrm{of} △\mathrm{ADC}) \\ \Rightarrow \mathrm{y}°+\mathrm{z}°+30°=180° \\ \Rightarrow \mathrm{y}°+50°+30°=180° (\because \mathrm{z}° =50°) \\ \Rightarrow \mathrm{y}°=180°-80° \\ \mathbf{\Rightarrow }\mathbf{y}\mathbf{°}\mathbf{=}\mathbf{100}\mathbf{°} \\ \mathrm{Therefore}, \mathrm{we} \mathrm{can} \mathrm{conclude} \mathrm{that} \mathrm{the} \mathrm{required} \mathrm{angles} \mathrm{are} 80°, 50° \mathrm{and} 100°. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iv}\right) \mathrm{In} △\mathrm{ABC} \mathrm{and} △\mathrm{ADE}: \\ \angle \mathrm{ADE}=\angle \mathrm{ABC} (\mathrm{Corresponding} \mathrm{angles}) \\ \mathbf{\Rightarrow }\mathbf{y}\mathbf{°}\mathbf{=}\mathbf{50}\mathbf{°} \\ \mathrm{Also}, \angle \mathrm{AED}=\angle \mathrm{ACB} (\mathrm{Corresponding} \mathrm{angles}) \\ \mathbf{\Rightarrow }\mathbf{z}\mathbf{°}\mathbf{=}\mathbf{40}\mathbf{°} \\ \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{the} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{is} \mathrm{equal} \mathrm{to} 180°. \\ \mathrm{Which} \mathrm{means}: \mathrm{x}°+50°+40°=180° (\mathrm{Angles} \mathrm{of} △\mathrm{ADE}) \\ \mathrm{x}°=180°-90° \\ \mathbf{\Rightarrow }\mathbf{x}\mathbf{°}\mathbf{=}\mathbf{90}\mathbf{°} \\ \mathrm{Therefore}, \mathrm{we} \mathrm{can} \mathrm{conclude} \mathrm{that} \mathrm{the} \mathrm{required} \mathrm{angles} \mathrm{are} 50°, 40° \mathrm{and} 90°. \end{gathered}$$

Question 16:

If one angle of a triangle is 60° and the other two angles are in the ratio 1 : 2, find the angles.

Answer 16:

$$\begin{gathered} \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{one} \mathrm{of} \mathrm{the} \mathrm{angles} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{triangle} \mathrm{is} 60°. \left(\mathrm{Given}\right) \\ \mathrm{We} \mathrm{also} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{other} \mathrm{two} \mathrm{angles} \mathrm{of} \mathrm{the} \mathrm{triangle} \mathrm{are} \mathrm{in} \mathrm{the} \mathrm{ratio} 1:2. \\ \mathrm{Let} \mathrm{one} \mathrm{of} \mathrm{the} \mathrm{other} \mathrm{two} \mathrm{angles} \mathrm{be} \mathrm{x}. \\ \mathrm{Therefore}, \mathrm{the} \mathrm{second} \mathrm{one} \mathrm{will} \mathrm{be} 2\mathrm{x}. \\ \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{the} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{is} \mathrm{equal} \mathrm{to} 180°. \\ \Rightarrow 60°+\mathrm{x}°+2\mathrm{x}°=180° \\ \Rightarrow 3\mathrm{x}°=180°-60° \\ \Rightarrow 3\mathrm{x}°=120° \\ \Rightarrow \mathrm{x}°=\frac{120°}{3} \\ \Rightarrow \mathrm{x}°=40° \\ 2\mathrm{x}°=2\times 40 \\ \Rightarrow 2\mathrm{x}°=80° \\ \mathrm{Hence}, \mathrm{we} \mathrm{can} \mathrm{conclude} \mathrm{that} \mathrm{the} \mathrm{required} \mathrm{angles} \mathrm{are} 40° \mathrm{and} 80°. \end{gathered}$$

Question 17:

If one angle of a triangle is 100° and the other two angles are in the ratio 2 : 3, find the angles.

Answer 17:

$$\begin{gathered} \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{one} \mathrm{of} \mathrm{the} \mathrm{angles} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{triangle} \mathrm{is} 100°. \\ \mathrm{We} \mathrm{also} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{other} \mathrm{two} \mathrm{angles} \mathrm{are} \mathrm{in} \mathrm{the} \mathrm{ratio} 2:3. \\ \mathrm{Let} \mathrm{one} \mathrm{of} \mathrm{the} \mathrm{other} \mathrm{two} \mathrm{angles} \mathrm{be} 2\mathrm{x}. \\ \mathrm{Therefore}, \mathrm{the} \mathrm{second} \mathrm{angle} \mathrm{will} \mathrm{be} 3\mathrm{x}. \\ \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{is} 180°. \\ \Rightarrow 100°+2\mathrm{x}°+3\mathrm{x}°=180° \\ 5\mathrm{x}°=180°-100° \\ 5\mathrm{x}°=80° \\ \mathrm{x}°=\frac{80°}{5} \\ \Rightarrow \mathrm{x}°=16° \\ 2\mathrm{x}°=2\times 16 \\ \Rightarrow 2\mathrm{x}°=32° \\ 3\mathrm{x}°=3\times 16 \\ \Rightarrow 3\mathrm{x}°=48° \\ \mathrm{Thus}, \mathrm{the} \mathrm{required} \mathrm{angles} \mathrm{are} 32° \mathrm{and} 48°. \end{gathered}$$

Question 18:

In a ∆ABC, if 3∠A = 4 ∠B = 6 ∠C, calculate the angles.

Answer 18:

$$\begin{gathered} \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{for} \mathrm{the} \mathrm{given} \mathrm{triangle}, 3\angle \mathrm{A} \mathrm{is} \mathrm{equal} \mathrm{to} 6\angle \mathrm{C}. \\ \Rightarrow \angle \mathrm{A}=2\angle \mathrm{C} ...\left(\mathrm{i}\right) \\ \mathrm{We} \mathrm{also} \mathrm{know} \mathrm{that} \mathrm{for} \mathrm{the} \mathrm{same} \mathrm{triangle}, 4\angle \mathrm{B} \mathrm{is} \mathrm{equal} \mathrm{to} 6\angle \mathrm{C}. \\ \Rightarrow \angle \mathrm{B}=\frac{6}{4}\angle \mathrm{C} \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }...\left(\mathrm{ii}\right) \\ \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{is} 180°. \\ \mathrm{Therefore}, \mathrm{we} \mathrm{can} \mathrm{say} \mathrm{that}: \\ \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180° (\mathrm{Angles} \mathrm{of} △\mathrm{ABC}) ...\left(\mathrm{iii}\right) \\ \mathrm{On} \mathrm{putting} \mathrm{the} \mathrm{values} \mathrm{of}\angle \mathrm{A} \mathrm{and} \angle \mathrm{B} \mathrm{in} \mathrm{equation} \left(\mathrm{iii}\right), \mathrm{we} \mathrm{get}: \\ 2\angle \mathrm{C}+\frac{6}{4}\angle \mathrm{C}+\angle \mathrm{C}=180° \\ \frac{18}{4}\angle \mathrm{C}=180° \\ \Rightarrow \angle \mathrm{C}=40° \\ \mathrm{From} \mathrm{equation} \left(\mathrm{i}\right), \mathrm{we} \mathrm{have}: \\ \angle \mathrm{A}=2\angle \mathrm{C}=2\times 40 \\ \Rightarrow \angle \mathrm{A}=80° \\ \mathrm{From} \mathrm{equation} \left(\mathrm{ii}\right), \mathrm{we} \mathrm{have}: \\ \angle \mathrm{B}=\frac{6}{4}\angle \mathrm{C}=\frac{6}{4}\times 40° \\ \Rightarrow \angle \mathrm{B}=60° \\ \angle \mathrm{A}=80°, \angle \mathrm{B}=60°, \angle \mathrm{C}=40° \\ \mathrm{Therefore}, \mathrm{the} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{triangle} \mathrm{are} 80°, 60° \mathrm{and} 40°. \end{gathered}$$

Question 19:

Is it possible to have a triangle, in which
(i) two of the angles are right?
(ii) two of the angles are obtuse?
(iii) two of the angles are acute?
(iv) each angle is less than 60°?
(v) each angle is greater than 60°?
(vi) each angle is equal to 60°?
Give reasons in support of your answer in each case.

Answer 19:

(i) No, because if there are two right angles in a triangle, then the third angle of the triangle must be zero, which is not possible.
(ii) No, because as we know that the sum of all three angles of a triangle is always 180$°$. If there are two obtuse angles, then their sum will be more than 180$°$, which is not possible in case of a triangle.
(iii) Yes, in right triangles and acute triangles, it is possible to have two acute angles.
(iv) No, because if each angle is less than 60$°$, then the sum of all three angles will be less than 180$°$, which is not possible in case of a triangle.
Proof :
$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{the} \mathrm{triangle} \mathrm{be} \angle \mathrm{A}, \angle \mathrm{B} \mathrm{and} \angle \mathrm{C}. \\ \mathrm{As} \mathrm{per} \mathrm{the} \mathrm{given} \mathrm{information}, \\ \angle \mathrm{A}<60° ...\left(\mathrm{i}\right) \\ \angle \mathrm{B}<60° ...\left(\mathrm{ii}\right) \\ \angle \mathrm{C}<60° ...\left(\mathrm{iii}\right) \\ \mathrm{On} \mathrm{adding} \left(\mathrm{i}\right), \left(\mathrm{ii}\right) \mathrm{and} \left(\mathrm{iii}\right), \mathrm{we} \mathrm{get}: \\ \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}<60°+60°+60° \\ \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}<180° \\ \mathrm{We} \mathrm{can} \mathrm{see} \mathrm{that} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{three} \mathrm{angles} \mathrm{is} \mathrm{less} \mathrm{than} 180°, \mathrm{which} \mathrm{is} \mathrm{not} \mathrm{possible} \mathrm{for} \mathrm{a} \mathrm{triangle}. \\ \mathrm{Hence}, \mathrm{we} \mathrm{can} \mathrm{say} \mathrm{that} \mathrm{it} \mathrm{is} \mathrm{not} \mathrm{possible} \mathrm{for} \mathrm{each} \mathrm{angle} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{to} \mathrm{be} \mathrm{less} \mathrm{than} 60°. \end{gathered}$$

(v) No, because if each angle is greater than 60$°$, then the sum of all three angles will be greater than 180$°$, which is not possible.
Proof :
$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{triangle} \mathrm{be} \angle \mathrm{A}, \angle \mathrm{B} \mathrm{and} \angle \mathrm{C}. \\ \mathrm{As} \mathrm{per} \mathrm{the} \mathrm{given} \mathrm{information}, \\ \angle \mathrm{A}>60° ...\left(\mathrm{i}\right) \\ \angle \mathrm{B}>60° ...\left(\mathrm{ii}\right) \\ \angle \mathrm{C}>60° ...\left(\mathrm{iii}\right) \\ \mathrm{On} \mathrm{adding} \left(\mathrm{i}\right), \left(\mathrm{ii}\right) \mathrm{and} \left(\mathrm{iii}\right), \mathrm{we} \mathrm{get}: \\ \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}>60°+60°+60° \\ \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}>180° \\ \mathrm{We} \mathrm{can} \mathrm{see} \mathrm{that} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{triangle} \mathrm{are} \mathrm{greater} \mathrm{than} 180°, \mathrm{which} \mathrm{is} \mathrm{not} \mathrm{possible} \mathrm{for} \mathrm{a} \mathrm{triangle}. \\ \mathrm{Hence}, \mathrm{we} \mathrm{can} \mathrm{say} \mathrm{that} \mathrm{it} \mathrm{is} \mathrm{not} \mathrm{possible} \mathrm{for} \mathrm{each} \mathrm{angle} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{to} \mathrm{be} \mathrm{greater} \mathrm{than} 60°. \end{gathered}$$

(vi) Yes, if each angle of the triangle is equal to 60$°$, then the sum of all three angles will be 180$°$, which is possible in case of a triangle.
Proof :
$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{triangle} \mathrm{be} \angle \mathrm{A}, \angle \mathrm{B} \mathrm{and} \angle \mathrm{C}. \\ \mathrm{As} \mathrm{per} \mathrm{the} \mathrm{given} \mathrm{information}, \\ \angle \mathrm{A}= 60° ...\left(\mathrm{i}\right) \\ \angle \mathrm{B}= 60° ...\left(\mathrm{ii}\right) \\ \angle \mathrm{C}= 60° ...\left(\mathrm{iii}\right) \\ \mathrm{On} \mathrm{adding} \left(\mathrm{i}\right), \left(\mathrm{ii}\right) \mathrm{and} \left(\mathrm{iii}\right), \mathrm{we} \mathrm{get}: \\ \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=60°+60°+60° \\ \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180° \\ \mathrm{We} \mathrm{can} \mathrm{see} \mathrm{that} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{triangle} \mathrm{is} \mathrm{equal} \mathrm{to} 180°, \mathrm{which} \mathrm{is} \mathrm{possible} \mathrm{in} \mathrm{case} \mathrm{of} \mathrm{a} \mathrm{triangle}. \\ \mathrm{Hence}, \mathrm{we} \mathrm{can} \mathrm{say} \mathrm{that} \mathrm{it} \mathrm{is} \mathrm{possible} \mathrm{for} \mathrm{each} \mathrm{angle} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{to} \mathrm{be} \mathrm{equal} \mathrm{to} 60°. \end{gathered}$$

Question 20:

In ∆ABC, ∠A = 100°, AD bisects ∠A and AD ⊥ BC. Find ∠B.

Answer 20:

$$\begin{gathered} \mathrm{Consider}△\mathrm{ABD}. \\ \angle \mathrm{BAD}=\frac{100°}{2} (\because \mathrm{AD} \mathrm{bisects} \angle \mathrm{A}) \\ \Rightarrow \angle \mathrm{BAD}=50° \\ \angle \mathrm{ADB}=90° (\because \mathrm{AD}\perp \mathrm{BC}) \\ \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{is} 180°. \\ \mathrm{Thus}, \\ \angle \mathrm{ABD}+\angle \mathrm{BAD}+\angle \mathrm{ADB}=180° (\mathrm{Sum} \mathrm{of} \mathrm{angles} \mathrm{of} △\mathrm{ABD}) \\ \mathrm{Or}, \\ \angle \mathrm{ABD}+50°+90°=180° \\ \angle \mathrm{ABD}=180°-140° \\ \angle \mathrm{ABD}=40° \end{gathered}$$

Question 21:

In ∆ABC, ∠A = 50°, ∠B = 70° and bisector of ∠C meets AB in D. Find the angles of the triangles ADC and BDC.

Answer 21:

$$\begin{gathered} \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{is} \mathrm{equal} \mathrm{to} 180°. \\ \mathrm{Therefore}, \mathrm{for} \mathrm{the} \mathrm{given} △\mathrm{ABC}, \mathrm{we} \mathrm{can} \mathrm{say} \mathrm{that}: \\ \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180° (\mathrm{Sum} \mathrm{of} \mathrm{angles} \mathrm{of} △\mathrm{ABC}) \\ \Rightarrow 50°+70°+\angle \mathrm{C}=180° \\ \angle \mathrm{C}=180°-120° \\ \angle \mathrm{C}=60° \\ \angle \mathrm{ACD}=\angle \mathrm{BCD}=\frac{\angle \mathrm{C}}{2}(\mathrm{CD} \mathrm{bisects} \angle \mathrm{C} \mathrm{and} \mathrm{meets} \mathrm{AB}\hspace{0.17em}\mathrm{in}\hspace{0.17em}\mathrm{D}.) \\ \Rightarrow \angle \mathrm{ACD}=\angle \mathrm{BCD}=\frac{60°}{2}=30° \\ \mathrm{Using} \mathrm{the} \mathrm{same} \mathrm{logic} \mathrm{for} \mathrm{the} \mathrm{given} △\mathrm{ACD}, \mathrm{we} \mathrm{can} \mathrm{say} \mathrm{that}: \\ \angle \mathrm{DAC}+\angle \mathrm{ACD}+\angle \mathrm{ADC}=180° \\ \Rightarrow 50°+30°+\angle \mathrm{ADC}=180° \\ \angle \mathrm{ADC}=180°-80° \\ \angle \mathrm{ADC}=100° \\ \mathrm{If} \mathrm{we} \mathrm{use} \mathrm{the} \mathrm{same} \mathrm{logic} \mathrm{for} \mathrm{the} \mathrm{given} △\mathrm{BCD}, \mathrm{we} \mathrm{can} \mathrm{say} \mathrm{that}: \\ \angle \mathrm{DBC}+\angle \mathrm{BCD}+\angle \mathrm{BDC}=180° \\ \Rightarrow 70°+30°+\angle \mathrm{BDC}=180° \\ \angle \mathrm{BDC}=180°-100° \\ \angle \mathrm{BDC}=80° \\ \mathrm{Thus}, \\ \mathrm{For} ∆\mathrm{ADC}: \angle \mathrm{A} = 50°, \angle \mathrm{D} =100°, \angle \mathrm{C} = 30° \\ \mathrm{For} ∆\mathrm{BDC}: \angle \mathrm{B} = 70°, \angle \mathrm{D}= 80° , \angle \mathrm{C}= 30° \end{gathered}$$

Question 22:

In ∆ABC, ∠A = 60°, ∠B = 80° and the bisectors of ∠B and ∠C meet at O. Find
(i) ∠C
(ii) ∠BOC.

Answer 22:

$$\begin{gathered} \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{is} 180°. \\ \mathrm{Hence}, \mathrm{for} △\mathrm{ABC},\mathbf{ }\mathrm{we} \mathrm{can} \mathrm{say} \mathrm{that}: \\ \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180° (\mathrm{Sum} \mathrm{of} \mathrm{angles} \mathrm{of} △\mathrm{ABC}) \\ \Rightarrow 60°+80°+\angle \mathrm{C}=180° \\ \angle \mathrm{C}=180°-140° \\ \angle \mathrm{C}=40° \\ \mathrm{For} △\mathrm{OBC}: \\ \angle \mathrm{OBC}=\frac{\angle \mathrm{B}}{2}=\frac{80°}{2} (\mathrm{OB}\hspace{0.17em}\mathrm{bisects} \angle \mathrm{B}.) \\ \Rightarrow \angle \mathrm{OBC}=40° \\ \angle \mathrm{OCB}=\frac{\angle \mathrm{C}}{2}=\frac{40°}{2} (\mathrm{OC} \mathrm{bisects} \angle \mathrm{C}.) \\ \Rightarrow \angle \mathrm{OCB}=20° \\ \mathrm{If} \mathrm{we} \mathrm{apply} \mathrm{the} \mathrm{above} \mathrm{logic} \mathrm{to} \mathrm{this} \mathrm{triangle}, \mathrm{we} \mathrm{can} \mathrm{say} \mathrm{that}: \\ \angle \mathrm{OCB}+\angle \mathrm{OBC}+\angle \mathrm{BOC}=180° (\mathrm{Sum} \mathrm{of} \mathrm{angles} \mathrm{of} △\mathrm{OBC}) \\ 20°+40°+\angle \mathrm{BOC}=180° \\ \angle \mathrm{BOC}=180°-60° \\ \angle \mathrm{BOC}=120° \end{gathered}$$

Page-15.14

Question 23:

The  bisectors of the acute angles of a right triangle meet at O. Find the angle at O between the two bisectors.

Answer 23:

$$\begin{gathered} \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{is} 180°. \\ \mathrm{Hence}, \mathrm{for} △\mathrm{ABC}, \mathrm{we} \mathrm{can} \mathrm{say} \mathrm{that}: \\ \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180° \\ \Rightarrow \angle \mathrm{A}+90°+\angle \mathrm{C}=180° \\ \angle \mathrm{A}+\angle \mathrm{C}=180°-90° \\ \angle \mathrm{A}+\angle \mathrm{C}=90° \\ \mathrm{For}△\mathrm{OAC}: \\ \angle \mathrm{OAC}=\frac{\angle \mathrm{A}}{2} (\mathrm{OA} \mathrm{bisects} \angle \mathrm{A}.) \\ \angle \mathrm{OCA}=\frac{\angle \mathrm{C}}{2} (\mathrm{OC} \mathrm{bisects} \angle \mathrm{C}.) \\ \mathrm{On} \mathrm{applying} \mathrm{the} \mathrm{above} \mathrm{logic} \mathrm{to} △\mathrm{OAC}, \mathrm{we} \mathrm{get}: \\ \angle \mathrm{AOC}+\angle \mathrm{OAC}+\angle \mathrm{OCA}=180° (\mathrm{Sum} \mathrm{of} \mathrm{angles} \mathrm{of} △\mathrm{AOC}) \\ \Rightarrow \angle \mathrm{AOC}+\frac{\angle \mathrm{A}}{2} +\frac{\angle \mathrm{C}}{2} =180° \\ \angle \mathrm{AOC}+\frac{\angle \mathrm{A}+\angle \mathrm{C}}{2}=180° \\ \angle \mathrm{AOC}+\frac{90°}{2}=180° (\because \angle \mathrm{A}+\angle \mathrm{C}=90° ) \\ \angle \mathrm{AOC}=180°-45° \\ \angle AOC=135° \end{gathered}$$

Question 24:

In ∆ABC, ∠A = 50° and BC is produced to a point D. The bisectors of ∠ABC and ∠ACD meet at E. Find ∠E.

Answer 24:

$$\begin{gathered} \mathrm{In} \mathrm{the} \mathrm{given} \mathrm{triangle}, \angle \mathrm{ACD}=\angle \mathrm{A}+\angle \mathrm{B}. (\mathrm{Exterior} \mathrm{angle} \mathrm{is} \mathrm{equal} \mathrm{to} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{two} \mathrm{opposite} \mathrm{interior} \mathrm{angles}.) \\ \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{is} 180°. \\ \mathrm{Therefore}, \mathrm{for} \mathrm{the} \mathrm{given} \mathrm{triangle}, \mathrm{we} \mathrm{can} \mathrm{say} \mathrm{that}: \\ \angle \mathrm{ABC}+\angle \mathrm{BCA}+\angle \mathrm{CAB}=180°. (\mathrm{Sum} \mathrm{of} \mathrm{all} \mathrm{angles} \mathrm{of} △\mathrm{ABC}) \\ \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{BCA}=180° \\ \angle \mathrm{BCA}=180°-(\angle \mathrm{A}+\angle \mathrm{B}) \\ \angle \mathrm{ECA}=\frac{\angle \mathrm{ACD}}{2} (\because \mathrm{EC} \mathrm{bisects} \angle \mathrm{ACD}) \\ \angle \mathrm{ECA}=\frac{\angle \mathrm{A}+\angle \mathrm{B} }{2} (\because \angle \mathrm{ACD}=\angle \mathrm{A}+\angle \mathrm{B}) \\ \angle \mathrm{EBC}=\frac{\angle \mathrm{ABC}}{2} =\frac{\angle \mathrm{B}}{2} (\because \mathrm{EB} \mathrm{bisects} \angle \mathrm{ABC}) \\ \angle \mathrm{ECB}=\angle \mathrm{ECA}+\angle \mathrm{BCA} \\ \Rightarrow \angle \mathrm{ECB}=\frac{\angle \mathrm{A}+\angle \mathrm{B} }{2}+180°-(\angle \mathrm{A}+\angle \mathrm{B}) \\ \mathrm{If} \mathrm{we} \mathrm{use} \mathrm{the} \mathrm{same} \mathrm{logic} \mathrm{for}△\mathrm{EBC}, \mathrm{we} \mathrm{can} \mathrm{say} \mathrm{that}: \\ \angle \mathrm{EBC}+\angle \mathrm{ECB}+\angle \mathrm{BEC}=180° (\mathrm{Sum} \mathrm{of} \mathrm{all} \mathrm{angles} \mathrm{of} △\mathrm{EBC}) \\ \frac{\angle \mathrm{B}}{2}+\frac{\angle \mathrm{A}+\angle \mathrm{B} }{2}+180°-(\angle \mathrm{A}+\angle \mathrm{B})+\angle \mathrm{BEC}=180° \\ \angle \mathrm{BEC}=\angle \mathrm{A}+\angle \mathrm{B}-\left(\frac{\angle \mathrm{A}+\angle \mathrm{B} }{2}\right)-\frac{\angle \mathrm{B}}{2} \\ \angle \mathrm{BEC}=\frac{\angle \mathrm{A}}{2} \\ \Rightarrow \angle \mathrm{BEC}=\frac{50°}{2}=25° \end{gathered}$$

Question 25:

In ∆ABC, ∠B = 60°, ∠C = 40°, AL ⊥ BC and AD bisects ∠A such that L and D lie on side BC. Find ∠LAD.

Answer 25:

$$\begin{gathered} \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{is} 180°. \\ \mathrm{Therefore}, \mathrm{for} △\mathrm{ABC}, \mathrm{we} \mathrm{can} \mathrm{say} \mathrm{that}: \\ \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180° \\ \mathrm{Or}, \\ \angle \mathrm{A}+60°+40°=180° \\ \Rightarrow \angle \mathrm{A}=80° \\ \angle \mathrm{DAC}=\frac{\angle \mathrm{A}}{2} (\because \mathrm{AD}\hspace{0.17em}\mathrm{bisects} \angle \mathrm{A}) \\ \Rightarrow \angle \mathrm{DAC}=\frac{80°}{2}=40° \\ \mathrm{If} \mathrm{we} \mathrm{use} \mathrm{the} \mathrm{above} \mathrm{logic} \mathrm{on} △\mathrm{ADC}, \mathrm{we} \mathrm{can} \mathrm{say} \mathrm{that}: \\ \angle \mathrm{ADC}+\angle \mathrm{DCA}+\angle \mathrm{DAC}=180°. (\mathrm{Sum} \mathrm{of} \mathrm{all} \mathrm{the} \mathrm{angles} \mathrm{of} △\mathrm{ADC}) \\ \angle \mathrm{ADC}+40°+40°=180° \\ \angle \mathrm{ADC}=180°-80°=100° \\ \angle \mathrm{ADC}=\angle \mathrm{ALD}+\angle \mathrm{LAD} (\mathrm{Exterior} \mathrm{angle} \mathrm{is} \mathrm{equal} \mathrm{to} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{two} \mathrm{Interior} \mathrm{opposite} \mathrm{angles}.) \\ 100°=90°+\angle \mathrm{LAD} (\because \mathrm{AL}\perp \mathrm{BC}) \\ \angle \mathrm{LAD} =10° \mathbf{ } \end{gathered}$$

Question 26:

Line segments AB and CD intersect at O such that AC || DB. If ∠CAB = 35° and ∠CDB = 55°, find ∠BOD.

Answer 26:

$$\begin{gathered} \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{AC}\parallel \mathrm{BD} \mathrm{and} \mathrm{AB} \mathrm{cuts} \mathrm{AC} \mathrm{and} \mathrm{BD} \mathrm{at} \mathrm{A} \mathrm{and} \mathrm{B}, \mathrm{respectively}. \\ \therefore \angle \mathrm{CAB}=\angle \mathrm{DBA} (\mathrm{Alternate} \mathrm{interior} \mathrm{angles}) \\ \Rightarrow \angle \mathrm{DBA}=35° \\ \mathrm{We} \mathrm{also} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{is} 180°. \\ \mathrm{Hence}, \mathrm{for} △\mathrm{OBD}, \mathrm{we} \mathrm{can} \mathrm{say} \mathrm{that}: \\ \angle \mathrm{DBO}+\angle \mathrm{ODB}+\angle \mathrm{BOD}=180° \\ \Rightarrow 35°+55°+\angle \mathrm{BOD}=180° (\because \angle \mathrm{DBO}=\angle \mathrm{DBA} \mathrm{and}\angle \mathrm{ODB}=\angle \mathrm{CDB}) \\ \angle \mathrm{BOD}=180°-90° \\ =\angle \mathrm{BOD}=90° \end{gathered}$$

Question 27:

In Fig., ∆ABC is right angled at A. Q and R are points on line BC and P is a point such that QP || AC and RP || AB. Find ∠P.

Answer 27:

$$\begin{gathered} \mathrm{In} \mathrm{the} \mathrm{given} \mathrm{triangle}, \mathrm{AC}\parallel \mathrm{QP} \mathrm{and} \mathrm{BR} \mathrm{cuts} \mathrm{AC} \mathrm{and} \mathrm{QP} \mathrm{at} \mathrm{C} \mathrm{and} \mathrm{Q}, \mathrm{respectively}. \\ \therefore \angle \mathrm{QCA}=\angle \mathrm{CQP} (\mathrm{Alternate} \mathrm{interior} \mathrm{angles}) \\ \mathrm{Because} \mathrm{RP}\parallel \mathrm{AB} \mathrm{and} \mathrm{BR} \mathrm{cuts} \mathrm{AB} a\mathrm{nd} \mathrm{RP} \mathrm{at} \mathrm{B} \mathrm{and} \mathrm{R}, \mathrm{respectively}, \angle \mathrm{ABC}=\angle \mathrm{PRQ} (\mathrm{alternate} \mathrm{interior} \mathrm{angles}). \\ \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{is} 180°. \\ \mathrm{Hence}, \mathrm{for} △\mathrm{ABC}, \mathrm{we} \mathrm{can} \mathrm{say} \mathrm{that}: \\ \angle \mathrm{ABC}+\angle \mathrm{ACB}+\angle \mathrm{BAC}=180° \\ \Rightarrow \angle \mathrm{ABC}+\angle \mathrm{ACB}+90°=180° (\mathrm{Right} \mathrm{angled} \mathrm{at} \mathrm{A}) \\ \Rightarrow \angle \mathrm{ABC}+\angle \mathrm{ACB}=90° \\ \mathrm{Using} \mathrm{the} \mathrm{same} \mathrm{logic} \mathrm{for} △\mathrm{PQR}, \mathrm{we} \mathrm{can} \mathrm{say} \mathrm{that}: \\ \angle \mathrm{PQR}+\angle \mathrm{PRQ}+\angle \mathrm{QPR}=180° \\ \Rightarrow \angle \mathrm{ABC}+\angle \mathrm{ACB}+\angle \mathrm{QPR}=180° (\because \angle \mathrm{ABC}=\angle \mathrm{PRQ} \mathrm{and} \angle \mathrm{QCA}=\angle \mathrm{CQP} ) \\ \mathrm{Or}, \\ 90°+\angle \mathrm{QPR}=180° (\because \angle \mathrm{ABC}+\angle \mathrm{ACB}=90°) \\ \mathbf{\angle }\mathbf{QPR}\mathbf{=}\mathbf{90}\mathbf{°} \end{gathered}$$