Question

In ΔPQR, PQ² + QR² = PR² and PR = 2QR. Write the measures of the angles of the triangle.

Answer 1

Given:

△PQR,

PQ² + QR² = PR²  ...[i]

PR = 2 QR   ...[ii]

Measures of the angles of triangle △PQR.

PQ² + QR² = PR²

∠Q = 90°

Step 1:

QR = x then

PR = 2x

PQ² + x² = (2x)² = 4x²    ...[using Pythagoras Theorem to find]

PQ² = 4x² − x² = 3x²

PQ² = 3x

PQ = `sqrt3​x`

QR = x

PR = 2x

Step 2:

Use tan⁡(∠P) = `(QR)/(PQ)`

= `x/(sqrt3x) = 1/sqrt3`

∠P = 30°

= ∠P + ∠Q + ∠R 

= 30° + 90° + ∠R = 180°

= ∠R = 60°

The measures of the angles of the triangle are: ∠P = 30°, ∠Q = 90°, ∠R = 60°.