Question

For ΔPQR and ΔXYZ, the correspondence PQR ⇔YZX is a similarity. m∠P= 2m∠Q and m∠X=120. Find m∠Y.

Answer 1

According to the problem given ,

∆PQR ~ ∆YZX

Let m<Q = x = m<Z

m<P = 2x = m<Y

m<X = 120°

Corresponding parts in similar triangles

are equal .

In ∆ YZX,

<Y + <Z + <X = 180°

2x + x + 120° = 180°

3x = 180 - 120

3x = 60

x = 60/3

x = 20°

Therefore ,

<Y = 2x = 2 × 20° = 40°

I hope this helps you.

: )

Answer 2

Given, ΔPQR and ΔXYZ, the correspondence PQR ⇔YZX is a similarity.

m∠P = 2m∠Q and m∠X = 120⁰

If the correspondence PQR⇔YZX is a similarity then

m∠P = m∠Y ........ (i)

m∠Q = m∠Z ........ (ii)

m∠R = m∠X ........ (iii)

from equation (iii), m∠R = m∠X = 120°

in ΔPQR , given m∠P = 2m∠Q .....................(iv)

now, m∠P + m∠Q + m∠R = 180°

⇒ 2m∠Q + m∠Q + 120° = 180° [ from equation (iv)

3m∠Q = 60° ⇒ m∠Q = 20°

hence, m∠P = 2m∠Q = 2 × 20° = 40°

now from equation (i), m∠Y = 40°