In the figure, KCDG is a trapezium and triangles KGF and FKE are isosceles triangles. HB, HD and BD are straight lines. KG = KF = FE. Find
- ∠j
- ∠k
(a)
∠CKE = 180° - ∠j (Interior angles, KC//HD)
∠FEK = 180° - ∠j (Angles on a straight line)
∠FKE = 180° - ∠j (Isosceles triangle)
49° + 180° - ∠j + 180° - ∠j + ∠k + 21° = 180° (Angles on a straight line, BH)
49° + 180° + 180° + 21° - ∠j - ∠j + ∠k = 180°
430° - 2∠j + ∠k = 180°
2∠j - ∠k = 430° - 180°
2∠j - ∠k = 250°
∠k = 2∠j - 250° --- (1)
∠KFG
= ∠KGF
= 2 x (180° - ∠j)
= 360° - 2∠j (Exterior angle of a triangle)
∠k = 180° - (360° - 2∠j) - (360° - 2∠j)
∠k = 180° - 360° + 2∠j - 360° + 2∠j
∠k = 180° - 360° - 360° + 2∠j + 2∠j
∠k = 4∠j - 540° (Angles sum of triangle)
∠k = 4∠j - 540° --- (2)
(2) = (1)
4∠j - 540° = 2∠j - 250°
4∠j - 2∠j= 540° - 250°
2∠j = 290°
∠j
= 290° ÷ 2
= 145°
(b)
From (1)
∠k
= 2∠j - 250°
= 290° - 250°
= 40°
Answer(s): (a) 145°; (b) 40°