In $ \Delta X Y Z, X Y=X Z $. A straight line cuts $ X Z $ at $ P, Y Z $ at $ Q $ and $ X Y $ produced at $ R $. If $ Y Q=Y R $ and $ Q P=Q Z $, find the angles of $ \Delta X Y Z $.

Academic Mathematics NCERT Class 9

Given :

In Triangle XYZ ,

XY = XZ , YQ = YR ; QP = QZ.

To Find :

$$\displaystyle \angle XYZ\ ,\ \angle XZY\ ,\ \angle YXZ\ \ $$

Solution :

In triangle XYZ ,

XY = XZ

So, let's take

$\displaystyle \ \angle XYZ\ =\ \angle XZY\ =\ a$

In triangle YRQ ,

YQ = YR

So, let's take

$\displaystyle \ \angle YQR\ =\ \angle YRQ\ =\ b$

In triangle PQZ ,

PQ = QZ

So, let's take

$\displaystyle \ \angle PZQ\ =\ \angle QPZ\ =\ a$

For triangle YRQ, 'a' is Exterior angle, and 'b' , 'b' are interior angles.

By Exterior angle property,

Sum of two angles is equal to Exterior angle opposite to them.

$a=b+b$

$a=2b$............................... ( i)

$∠YQR =∠PQZ=b$ (Vertically opposite angles)

In triangle PQZ,

Sum of all angles of triangle is equal to 180°

$a+a+b=180°$

$2a+b=180°$ ...............................( ii)

Substitute (i) in (ii)

$2a+b=180°$

$2(2b)+b=180°$

$4b+b=180°$

$5b=180°$

$b=\frac{180°}{5}$

$b=36°$

Substitute b = 36° in (i)

$a=2b$

$a=2\times36°$

$a=72°$

$∠XYZ\ =∠XZY=a$

So,

∠XYZ = 72° and

∠XZY = 72°

$\angle XYZ\ \ +\ \angle XZY\ +\ \angle YZX\ =\ 180°$

72° +72° + ∠YZX = 180°

144° + ∠YZX =180°

∠YZX = 180° $-$144°

∠YZX = 36°

So, the angles of triangle XYZ are,

$\angle XYZ =72°$

$\angle XZY =72°$

$\angle YZX =36°$.

Simply Easy Learning

Updated on: 10-Oct-2022

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