Question

If PQ ⊥ PS, PQ II SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y. In Fig. 6.43.

Answer 1

now here
in triangle qsr
by exterior angle theorem
angle sqr + angle Q S R =65 before
28 + X = 65
x =65 - 28
x=37
Y=53

therefore ,

Answer 2

Given: PQ is perpendicular to PS, PQ parallel SR, ∠SQR = 28° and, ∠QRT = 65°

To Find: Values of x and y

Proof : according to the question,

x + ∠SQR = ∠QRT (Alt. angles are equal)

x + 28° = 65°

x = 37°

Now,

In Δ PQS,

As we know that sum of int. angles of a triangle = 180°

=>∠ PQS + ∠ PSQ + ∠ QPS = 180°

Therefore,

=> y + 37° + 90° = 180°

=> y= 180° - 127°

=> y = 53°

So x=37° and y= 53°