Question

In the given figure, ST is parallel to QR. Find the measure of the following angles.

(a) ∠TSR

(b) ∠ PST

(c) ∠ SQR

(d) ∠ SPT

Answer 1

Explanation:

ANSWER

In triangle PQR,

∠RPU−∠PQR+∠PQR ...Exterior angle property

⟹2∠PQR=140

o

PQ=PR

⟹∠PQR=

2

140

o

⟹∠PQR=70

o

ST parallel to QR and QS is a transversal.

∴∠PQR=∠PST=70

o

...Corresponding angles

Now, in triangle QSR

∴∠SQR−∠RSQ=70

o

Here, PQ is a straight line.

⟹∠PST+∠TSR+∠RSQ=180

o

From equation 1 and 2

⟹∠TSR=180

o

−70

o

−70

o

⟹TSR=40

o