RD Sharma 2020 solution class 9 chapter 11 Triangles and Its Angles Exercise 11.1

Exercise 11.1

Page-11.10

Question 1:

In a Δ ABC, if ∠A = 55°, ∠B = 40°, find ∠C.

Answer 1:

$$\begin{gathered} \angle A+\angle B+\angle C=180° \left[\mathrm{The} \mathrm{sum} \mathrm{of} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{is} 180°.\right] \\ \Rightarrow 55°+40°+\angle C=180° \\ \Rightarrow 95°+\angle C=180° \\ \Rightarrow \angle C=180°-95° \\ \Rightarrow \angle C=85° \end{gathered}$$
 

Question 2:

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

Answer 2:

Let the angles of the given triangle be of xº, 2xº and 3xº. Then,

$$\begin{gathered} \therefore x+2x+3x=180 \left[\mathrm{The} \mathrm{sum} \mathrm{of} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{is} 180°\right] \\ \Rightarrow 6x=180 \\ \Rightarrow x=30 \end{gathered}$$

Hence, the angles of the triangle are 30º, 60º and 90º.

Question 3:

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

Answer 3:

Given angles are

$$\begin{gathered} \therefore \left(x-40\right)+\left(x-20\right)+\left(\frac{1}{2}x-10\right)=180 \\ \Rightarrow \frac{5}{2}x=180+70 \\ \Rightarrow \frac{5}{2}x=250 \\ \Rightarrow x=\frac{250\times 2}{5} \\ \Rightarrow x=100 \end{gathered}$$

Hence, the value of x is 100°.

Question 4:

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 4:

Let the two equal angles are x°, then the third angle will be (x + 30)°.

$$\begin{gathered} \therefore x+x+\left(x+30\right)=180 \left[\mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{is} 180°\right] \\ \Rightarrow 3x+30=180 \\ \Rightarrow 3x=150 \\ \Rightarrow x=50 \end{gathered}$$

Therefore, the angles of the given triangle are 50°, 50° and 80°.

Question 5:

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

Answer 5:

Let ABC be a triangle such that

$$\begin{gathered} \angle A=\angle B+\angle C \left[\mathrm{Since}, \mathrm{one} \mathrm{angle} \mathrm{is} \mathrm{sum} \mathrm{of} \mathrm{the} \mathrm{other} \mathrm{two}\right] \\ \therefore \angle A+\angle B+\angle C=180° \left[\mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{is} 180°\right] \\ \Rightarrow \angle A+\angle A=180° \\ \Rightarrow 2\angle A=180° \\ \Rightarrow \angle A=90° \end{gathered}$$
 

Hence, the given triangle is a right angled triangle.

Question 6:

Can a triangle have:

(i) Two right angles?

(ii) Two obtuse angles?

(iii) Two acute angles?

(iv) All angles more than 60°?

(v) All angles less than 60°?

(vi) All angles equal to 60°?

Answer 6:

(i) Let a triangle ABC has two angles equal to .  We know that sum of the three angles of a triangle is 180°.

Hence, if two angles are equal to , then the third one will be equal to zero which implies that A, B, C is collinear, or we can say ABC is not a triangle

A triangle can’t have two right angles.

(ii) Let a triangle ABC has  two obtuse angles

This implies that sum of only two angles will be equal to more than 180° which contradicts the theorem sum of all angles in a triangle is always equals 180°.

Therefore, a  triangle can’t have two obtuse angles.

(iii) Let a triangle ABC has two acute angles.

This implies that sum of two angles will be less than . Hence third angle will be the difference of 180° and sum of both acute angles

Therefore,  a triangle can have two acute angles.

(iv) Let a triangle ABC having angles are more than 60°.

This implies that the sum of three angles will be more than 180° which contradicts the theorem sum of all angles in a triangle is always equals 180°.

Therefore, a triangle can’t have all angles more than .

(v) Let a triangle ABC having anglesare less than 60°.

This implies that the sum of three angles will be less than 180° which contradicts the theorem sum of all angles in a triangle is always equals 180°.

Therefore,  a triangle can’t have all angles less than 60°.

(vi) Let a triangle ABC having angles all equal to 60°.

This implies that the sum of three angles will be equal to 180° which satisfies the theorem sum of all angles in a triangle is always equals 180°.

Therefore, a triangle can have all angles equal to 60°.

Question 7:

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 7:

Let the angles of a triangle are          [Since, the difference between two consceutive angles is 10°]         

$$\begin{gathered} \therefore x+\left(x+10\right)+\left(x+20\right)=180 \left[\mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{is} 180°\right] \\ \Rightarrow 3x+30=180 \\ \Rightarrow 3x=150 \\ \Rightarrow x=50 \end{gathered}$$

Therefore, the angles of the given triangle are 50°, (50 + 10)° and (50 + 20)° i.e. 50°, 60° and 70°.

 

Question 8:

ABC is a triangle in which ∠A = 72°, the internal bisectors of angles B and C meet in O. Find the magnitude of ∠BOC.

Answer 8:

 

Since OB and OC are the angle bisector of $\angle B \mathrm{and} \angle C$

$$\begin{gathered} \angle A+\angle B+\angle C=180° \\ \Rightarrow 72°+2\angle OBC+2\angle OCB=180° \left[\mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{is} 180°\right] \\ \Rightarrow 2\left(\angle OBC+\angle OCB\right)=108° \\ \Rightarrow \angle OBC+\angle OCB=54° \\ \Rightarrow 180°-\angle BOC=54° \left[\mathrm{Since}, \angle OBC+\angle OCB+\angle BOC=180°\right] \\ \Rightarrow \angle BOC=126° \end{gathered}$$

Hence magnitude of $\angle BOC \mathrm{is} 126°.$

Question 9:

The bisectors of base angles of a triangle cannot enclose a right angle in any case.

Answer 9:

Let ABC be a triangle and  BO and CO be the bisectors of the base anglerespectively.

We know that if the bisectors of angles ∠ABC and ∠ACB of a triangle ABC meet at a point O, then

$\angle BOC=90°+\frac{1}{2}\angle A$

From the above relation it is very clear that if is equals 90° then must be equal to zero.

Now, if possible let is equals zero but on other hand it represents that  A, B, C will be collinear, that is they do not form a triangle.

It leads to a contradiction.

Hence, the bisectors of base angles of a triangle cannot enclose a right angle in any case.

Question 10:

If the bisectors of the base angles of a triangle enclose an angle of 135°, prove that the triangle is a right triangle.

Answer 10:

Let ABC be a triangle and Let BO and CO be the bisectors of the base anglerespectively.

We know that if the bisectors of angles ∠ABC and ∠ACB of a triangle ABC meet at a point O, then

$\angle BOC=90°+\frac{1}{2}\angle A$

$$\begin{gathered} \therefore 135°=90°+\frac{1}{2}\angle A \\ \Rightarrow 45°=\frac{1}{2}\angle A \\ \Rightarrow \angle A=90° \end{gathered}$$

Hence the triangle is a right angled triangle.

Question 11:

In a Δ ABC, ∠ABC = ∠ACB and the bisectors of ∠ABC and ∠ACB intersect at O such that ∠BOC = 120°. Show that ∠A =∠B =∠C = 60°.

Answer 11:

Let ABC be a triangle and BO and CO be the bisectors of the base anglerespectively.

We know that if the bisectors of angles ∠ABC and ∠ACB of a triangle ABC meet at a point O, then

$\angle BOC=90°+\frac{1}{2}\angle A$

$$\begin{gathered} \therefore 120°=90°+\frac{1}{2}\angle A \\ \Rightarrow 30°=\frac{1}{2}\angle A \\ \Rightarrow \angle A=60° \end{gathered}$$

are equal as it is given that .

$$\begin{gathered} \angle A+\angle B+\angle C=180° \left[\mathrm{Sum} \mathrm{of} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{is} 180°\right] \\ \Rightarrow 60°+2\angle B=180° \left[\because \angle ABC=\angle ACB\right] \\ \Rightarrow \angle B=60° \end{gathered}$$

Hence, .

Question 12:

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

Answer 12:

Let a triangle ABC having angles.

It is given that the sum of two angles are less than third one.

We know that the sum of all angles of a triangle equal to 180°.

Similarly we can prove that

Since,  all angles are less than 90°.
Hence,  triangle is acute angled.