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
- ∠y
- ∠z
(a)
∠KTN = 180° - ∠y (Interior angles, TK//SL)
∠PNT = 180° - ∠y (Angles on a straight line)
∠PTN = 180° - ∠y (Isosceles triangle)
49° + 180° - ∠y + 180° - ∠y + ∠z + 21° = 180° (Angles on a straight line, HS)
49° + 180° + 180° + 21° - ∠y - ∠y + ∠z = 180°
430° - 2∠y + ∠z = 180°
2∠y - ∠z = 430° - 180°
2∠y - ∠z = 250°
∠z = 2∠y - 250° --- (1)
∠TPR
= ∠TRP
= 2 x (180° - ∠y)
= 360° - 2∠y (Exterior angle of a triangle)
∠z = 180° - (360° - 2∠y) - (360° - 2∠y)
∠z = 180° - 360° + 2∠y - 360° + 2∠y
∠z = 180° - 360° - 360° + 2∠y + 2∠y
∠z = 4∠y - 540° (Angles sum of triangle)
∠z = 4∠y - 540° --- (2)
(2) = (1)
4∠y - 540° = 2∠y - 250°
4∠y - 2∠y= 540° - 250°
2∠y = 290°
∠y
= 290° ÷ 2
= 145°
(b)
From (1)
∠z
= 2∠y - 250°
= 290° - 250°
= 40°
Answer(s): (a) 145°; (b) 40°