In figure sides QP and RQ of ∆PQR are produced to points S and T respectively.
If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.
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∠PQT+∠PQR=180°. (Straight angle made on line TR at point Q)
110° +∠PQR=180°
∠PQR = 180° − 110° = 70°
In △PQR
∠PSR=∠PQR+∠PRQ (Exterior angle of triangle is eqal to sum of two interior angles)
135° = 70° +∠PRQ
∠PRQ = 135° −70° = 65°
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Answer:
∠PQT+∠PQR=180°. (Straight angle made on line TR at point Q)
110° +∠PQR=180°
∠PQR = 180° − 110° = 70°
In △PQR
∠PSR=∠PQR+∠PRQ (Exterior angle of triangle is eqal to sum of two interior angles)
135° = 70° +∠PRQ
∠PRQ = 135° −70° = 65°
Step-by-step explanation:
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