Question

In Δ PQR, ∠P= 120° , PS⊥QR at Sand PQ+QS=SR , then the measure of ∠Q.

Answer 1

The measure of $\angle Q = 40^\circ$

Step-by-step explanation:

let  PQ = x

QS = z

taking a point T on the line SR such that ST = QS = z

in triangle PQT

PS ⊥ QT and QS = ST

Therefore , PQT is an isoceles triangle

now ,

PQ = PT = TR = x

since , PTR is an isoceles triangle

$\angle R = \angle TPR = \theta$

(angle opposite to equal sides are equal)

now, let the angle PTQ = $2\theta$

therefore

$\angle QPT = 180- (2\theta +2\theta )\\\angle QPT = 180- (4\theta )$

then ,

$\angle QPR = \angle QPT +\angle RPT \\120 = 180- 4\theta +\theta \\120 -180 = -3\theta\\-60 = -3\theta \\\theta = 20^\circ$

now ,

$\angle PQT = 2\theta \\\angle PQT = 2\times 20 = 40^\circ$

hence ,

The measure of $\angle Q = 40^\circ$

#Learn more:

Pqrs is a quadrilateral, show that pq+qr+sr+ps>2(pr+qs)

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