In the given figure, ST is parallel to QR. Find the measure of the following angles.
(a) ∠TSR
(b) ∠ PST
(c) ∠ SQR
(d) ∠ SPT
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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
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