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
- ∠i
- ∠j
(a)
∠KTN = 180° - ∠i (Interior angles, TK//SL)
∠PNT = 180° - ∠i (Angles on a straight line)
∠PTN = 180° - ∠i (Isosceles triangle)
53° + 180° - ∠i + 180° - ∠i + ∠j + 15° = 180° (Angles on a straight line, HS)
53° + 180° + 180° + 15° - ∠i - ∠i + ∠j = 180°
428° - 2∠i + ∠j = 180°
2∠i - ∠j = 428° - 180°
2∠i - ∠j = 248°
∠j = 2∠i - 248° --- (1)
∠TPR
= ∠TRP
= 2 x (180° - ∠i)
= 360° - 2∠i (Exterior angle of a triangle)
∠j = 180° - (360° - 2∠i) - (360° - 2∠i)
∠j = 180° - 360° + 2∠i - 360° + 2∠i
∠j = 180° - 360° - 360° + 2∠i + 2∠i
∠j = 4∠i - 540° (Angles sum of triangle)
∠j = 4∠i - 540° --- (2)
(2) = (1)
4∠i - 540° = 2∠i - 248°
4∠i - 2∠i= 540° - 248°
2∠i = 292°
∠i
= 292° ÷ 2
= 146°
(b)
From (1)
∠j
= 2∠i - 248°
= 292° - 248°
= 44°
Answer(s): (a) 146°; (b) 44°