Lim Wei Lin's answer to Joan Cml's Primary 6 Maths Geometry Singapore question.
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∠STR
= 180° - 116° (∠ On a str. Line)
= 64°
∠QSR
= 180° - 87° - 65° (∠ Sum of triangle)
= 29°
∠QTR = 116° (vert. Opp ∠)
∠PRQ
= 189° - 116° - 29° (∠ Sum of triangle)
= 35°
∠QPR = 35° (isos. Triangle)
∠PQR
= 180° -35° - 35° (∠ Sum of triangle)
= 110°
= 180° - 116° (∠ On a str. Line)
= 64°
∠QSR
= 180° - 87° - 65° (∠ Sum of triangle)
= 29°
∠QTR = 116° (vert. Opp ∠)
∠PRQ
= 189° - 116° - 29° (∠ Sum of triangle)
= 35°
∠QPR = 35° (isos. Triangle)
∠PQR
= 180° -35° - 35° (∠ Sum of triangle)
= 110°
Date Posted:
3 years ago
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