Solution of Triangles Long Questions (Question 4)


Question 4:
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
                          i.      GH
                        ii.      HC
(b) The area of triangle GAH is twice the area of triangle JBK. Calculate the length, in cm, 
      of BK.
(c) Sketch triangle  which has a different shape from triangle ABC such that, A’B’ = AB
     A’C’ = AC and ÐA’B’C’ = ÐABC.

Solution:
(a)(i)
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

(a)(ii)
ACD= 180 45 85 = 50 Using sine rule, AC sin 45 = 73 sin 50 AC= 73×sin 45 sin 50 AC=67.38 cm HC=67.3830=37.38 cm
(b)
Area of Δ GAH= 1 2 ( 33 )( 30 )sin 85 =493.12  cm 2 Let length of BK=JK=x × Area of Δ JBK = Area of Δ GAH 2×[ 1 2 ( x )( x ) ]=493.12                      x 2 =493.12                     x=22.21 cm                 BK=22.21 cm
(c)