In the diagram below, ABC is 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 ÐA’B’C’ = ÐABC.
Using cosine rule,
GH2 = AG2 + AH2 – 2 (AG)(AH) ÐGAH
GH2= 332 + 302 – 2 (33)(30) cos 85o
GH2 = 1089 + 900 – 172.57
GH2 = 1816.43
GH = 42.62 cm