In the diagram below, ABCis a triangle. AGJB, AHC and BKC are straight lines. The straight line JK is perpendicular to BC.
It is given that BG= 40cm, GA = 33 cm, AH = 30 cm, GAH = 85o and JBK= 45o.
(a) Calculate the length, in cm of
(b) The area of triangle GAH is twice the area of triangle JBK. Calculate the length, in cm,
(c) Sketch triangle which has a different shape from triangle ABC such that, A’B’ = AB,
A’C’ = AC and ∠A’B’C’ = ∠ABC.
Using cosine rule,
GH2 = AG2 + AH2 – 2 (AG)(AH) ∠GAH
GH2= 332+ 302 – 2 (33)(30) cos 85o
GH2 = 1089 + 900 – 172.57
GH2 = 1816.43
GH = 42.62 cm