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)
50° + 180° - ∠y + 180° - ∠y + ∠z + 18° = 180° (Angles on a straight line, HS)
50° + 180° + 180° + 18° - ∠y - ∠y + ∠z = 180°
428° - 2∠y + ∠z = 180°
2∠y - ∠z = 428° - 180°
2∠y - ∠z = 248°
∠z = 2∠y - 248° --- (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 - 248°
4∠y - 2∠y= 540° - 248°
2∠y = 292°
∠y
= 292° ÷ 2
= 146°
(b)
From (1)
∠z
= 2∠y - 248°
= 292° - 248°
= 44°
Answer(s): (a) 146°; (b) 44°