RS Aggarwal 2019,2020 solution class 7 chapter 15 Property of Triangles Exercise 15B

Exercise 15B

Page-186

Question 1:

In the figure given alongside, find the measure of ∠ACD.

Answer 1:

We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.

$$\begin{gathered} \angle ACD= \angle CAB+\angle CBA \\ \angle ACD=75°+45°=120° \end{gathered}$$

Question 2:

In the figure given alongside, find the values of x and y.

Answer 2:

We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.

$$\begin{gathered} \therefore \angle BAC +\angle ABC =\angle ACD \\ x +68 = 130 \\ x = 62 \end{gathered}$$
Sum of the angles in any triangle is 180^o.

$$\begin{gathered} \therefore \angle BAC +\angle ABC+\angle ACB =180° \\ 62 + 68 + y = 180 \\ y = 50 \end{gathered}$$

Question 3:

In the figure given alongside, find the values of x and y.

Answer 3:

We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.

$$\begin{gathered} \therefore \angle BAC + \angle CBA = \angle ACD \\ 32 + x = 65 \\ x = 33 \end{gathered}$$

Also, sum of the angles in any triangle is 180$°$.

$$\begin{gathered} \therefore \angle BAC+\angle CBA +\angle ACB= 180° \\ 32 +33+ y = 180 \\ y = 115 \end{gathered}$$

∴ x= 33
    y =115

Question 4:

An exterior angle of a triangle measures 110° and its interior opposite angles are in the ratio 2 : 3. Find the angles of the triangle.

Answer 4:

Suppose the interior opposite angles are (2x)° and (3x)°.
We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.

∴ 3x +2x= 110
x = 22

The interior opposite angles are $(2\times 22)° \mathrm{and} (3\times 22)°, \mathrm{i}.\mathrm{e}.$ 44° and 66°.
Suppose the third angle of the triangle is y°.
Now, sum of the angles in any triangle is 180°.

∴ 44 + 66 + y = 180
y = 70

Hence, the angles of the triangle are 44°, 66° and 70°.

Question 5:

An exterior angle of a triangle is 100° and its interior opposite angles are equal to each other. Find the measure of each angle of the triangle.

Answer 5:

Suppose the interior opposite angles of an exterior angle 100^o are x^o and x^o.
We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.

∴ x + x = 100
2x= 100
x= 50

Also, sum of the angles of any triangle is  180°.
Let the measure of the third angle be y°.

∴ x + x + y = 180
50  + 50 + y= 180
y  = 80

Hence, the angles are of the measures 50°, 50° and 80°.

Question 6:

In the figure given alongside, find:

(i) ∠ACD
(ii) ∠AED

Answer 6:

We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.

In $△$ABC:
$$\begin{gathered} \angle ACD=\angle BAC+\angle ABC=25°+45° \\ \angle ACD=70° \\ \left(\mathrm{ii}\right) In △ECD: \\ \angle AED=\angle ECD+\angle EDC=70°+40° \\ =>\angle AED=110° \end{gathered}$$

Question 7:

In the figure given alongside, find:

(i) ∠ACD
(ii) ∠ADC
(iii) ∠DAE

Answer 7:

Sum of the angles of a triangle is 180$°$.

$$\begin{gathered} In△ABC: \\ \angle BAC+\angle CBA+\angle ACB=180° \\ \angle BAC=180°-(40°+100°) \\ =>\angle BAC=40° \end{gathered}$$ 


We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.

 $$\begin{gathered} \angle ACD=\angle BAC+\angle CBA=40°+40°=80° \\ \left(i\right) \angle ACD=80° \\ \left(ii\right) \mathrm{In} △ ACD: \\ \angle CAD+\angle ACD+\angle ADC=180° \\ =>\angle ADC=180°-(50°+80°) \\ =>\angle ADC=50° \\ \therefore \angle ADC= 50° \\ \left(iii\right) \angle DAB+\angle DAE=180° (s\mathrm{ince} BE \mathrm{is} \mathrm{a} \mathrm{straight} \mathrm{line}) \\ \angle DAE=180°-(\angle DAC+\angle CAB) \\ \angle DAE=180°-(50°+40°) \\ \angle DAE=90° \end{gathered}$$

Question 8:

In the figure given alongside, x : y = 2 : 3 and ∠ACD = 130°.
Find the values of x, y and z.

Answer 8:

$$\begin{gathered} \frac{x}{y}=\frac{2}{3} \\ \Rightarrow 3x = 2y \\ \Rightarrow x=\frac{2}{3}y \end{gathered}$$

We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.


∴ $\angle A + \angle B =\angle ACD$

x$°$+ y$°$ = 130$°$

$$\begin{gathered} \Rightarrow \frac{2y}{3}+y =130 \\ \Rightarrow 5y =130\times 3 \\ \Rightarrow 5y= 390 \\ \Rightarrow y = 78 \\ \Rightarrow x =\frac{2}{3}\times 78 \\ \Rightarrow x=52 \end{gathered}$$

Also, sum of the angles in any triangle is 180$°$
∴  x+ y + z = 180
z= 180- 78 - 52
z= 50