Question

In ΔPQR, PQ = PR, ∠P = (2x + 20)° and ∠R = (x + 10)°. Find the value of x. Assign a special name to the triangle according to the angles.

Answer 1

Given:

In ΔPQR, PQ = PR, so triangle PQR is isosceles with PQ = PR.

∠P = (2x + 20)°

∠R = (x + 10)°

Since PQ = PR, the angles opposite to these equal sides are equal,

So, ∠Q = ∠R = (x + 10)°

The sum of the internal angles of a triangle is 180°,

Therefore, ∠P + ∠Q + ∠R = 180°.

Substitute the values:

(2x + 20) + (x + 10) + (x + 10) = 180° 

2x + 20 + x + 10 + x + 10 = 180° 

4x + 40 = 180° 

4x = 180 – 40

4x = 140

x = 35°

Now find the angles:

∠P = 2(35) + 20

= 70 + 20

= 90°

∠Q = ∠R

= 35 + 10

= 45°

Since ∠P = 90°, triangle PQR is a right triangle.

x = 35°

Triangle PQR is an isosceles right-angled triangle right △.