Question

Find the value of x in each case: ​

Answer 1

Answer:

x = 72°

Step-by-step explanation:

Given a triangle PQR in which point T lies on the PR such that QT is perpendicular to PR and point S lies on line QR. ∠SPR = 30°, ∠TQR = 48°, ∠QTR = 90° and ∠PSQ = x.

We need to find out the value of x.

In ∆TQR

We know that sum of all angles of the triangle is 180°. So,

→ ∠QTR + ∠TQR + ∠TRQ = 180°

Substitute the values,

→ 90° + 48° + ∠TRQ = 180°

→ 138° + ∠TRQ = 180°

→ ∠TRQ = 180° - 138°

→ ∠TRQ = 42°

In ∆PSR

Sum of interior angles is equal to exterior angle. So,

→ x = ∠SPR + ∠PRS

[ ∠PRS = ∠TRQ = 42° ]

Substitute the values,

→ x = 30° + 42°

→ x = 72°

Therefore, the value of x is 72°.

Answer 2

Answer:

t

Step-by-step explanation:

the correct value of x is 72°