Answer:
Angle PSQ is 61°
Step-by-step explanation:
In the triangle ΔPQR, we have;
m∠QPS = 65°
m∠PRQ = 43°
Line PSR is a straight line
m∠PQS : m∠SQR = 3:1
Therefore;
∠PQR = 180° - (∠QPS + ∠PRQ), by the angle sum theorem of a triangle
∴ ∠PQR = 180° - (65° + 43°) = 72°
∠PQR = 72°
∠PQR = m∠PQS + m∠SQR by angle addition postulate
m∠PQS : m∠SQR = 3:1
∴ m∠PQS = 3 × m∠SQR
∴ ∠PQR = m∠PQS + m∠SQR = 3 × m∠SQR + m∠SQR = 4 × m∠SQR
∠PQR = 4 × m∠SQR
∴ 4 × m∠SQR = ∠PQR = 72°
m∠SQR = 72°/4 = 18°
m∠PQS = 3 × m∠SQR = 3 × 18° = 54°
m∠PQS = 54°
∠PSQ = 180° - (∠QPS + ∠PQS) by the angle sum theorem
∴ ∠PSQ = 180° - (65° + 54°) = 61°
Angle ∠PSQ = 61°.
