In Fig., PQ = PS. The value of x is (a) 35° (b) 45° (c) 55° (d) 70

1.	In Fig., PQ = PS. The value of x is (a) 35° (b) 45° (c) 55° (d) 70°
Answer» (b) 45^o From the given figure, In triangle PQS, ∠PSQ + ∠QPS = 110^o … [from exterior angle property of a triangle] We know that, sum of all angle of the triangle is equal to 180^o. So, ∠PSQ + ∠QPS + ∠PQS = 180^o 110^o + ∠PQS = 180^o – 110^o ∠PQS = 70^o Now, consider the triangle PRS, ∠PSQ = x + 25^o … [from the exterior angle property of a triangle] x = 70^o – 25^o x = 45^o

In Fig., PQ = PS. The value of x is (a) 35° (b) 45° (c) 55° (d) 70°

Answer»

(b) 45^o

From the given figure,

In triangle PQS, ∠PSQ + ∠QPS = 110^o … [from exterior angle property of a triangle]

We know that, sum of all angle of the triangle is equal to 180^o.

So, ∠PSQ + ∠QPS + ∠PQS = 180^o

110^o + ∠PQS = 180^o – 110^o

∠PQS = 70^o

Now, consider the triangle PRS,

∠PSQ = x + 25^o … [from the exterior angle property of a triangle]

x = 70^o – 25^o

x = 45^o