In the figure, TKLR is a trapezium and triangles TRP and PTN are isosceles triangles. SH, SL and HL are straight lines. TR = TP = PN. Find
- ∠a
- ∠b
(a)
∠KTN = 180° - ∠a (Interior angles, TK//SL)
∠PNT = 180° - ∠a (Angles on a straight line)
∠PTN = 180° - ∠a (Isosceles triangle)
52° + 180° - ∠a + 180° - ∠a + ∠b + 16° = 180° (Angles on a straight line, HS)
52° + 180° + 180° + 16° - ∠a - ∠a + ∠b = 180°
428° - 2∠a + ∠b = 180°
2∠a - ∠b = 428° - 180°
2∠a - ∠b = 248°
∠b = 2∠a - 248° --- (1)
∠TPR
= ∠TRP
= 2 x (180° - ∠a)
= 360° - 2∠a (Exterior angle of a triangle)
∠b = 180° - (360° - 2∠a) - (360° - 2∠a)
∠b = 180° - 360° + 2∠a - 360° + 2∠a
∠b = 180° - 360° - 360° + 2∠a + 2∠a
∠b = 4∠a - 540° (Angles sum of triangle)
∠b = 4∠a - 540° --- (2)
(2) = (1)
4∠a - 540° = 2∠a - 248°
4∠a - 2∠a= 540° - 248°
2∠a = 292°
∠a
= 292° ÷ 2
= 146°
(b)
From (1)
∠b
= 2∠a - 248°
= 292° - 248°
= 44°
Answer(s): (a) 146°; (b) 44°