(Long Questions) – Question 2

Posted on April 22, 2020 by user

Question 2:

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 = 85^o and JBK= 45^o.

(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 ∠A’B’C’ = ∠ABC.

Solution:
(a)(i)

Using cosine rule,

GH² = AG² + AH² – 2 (AG)(AH) ∠GAH

GH²= 33²+ 30² – 2 (33)(30) cos 85^o

GH² = 1089 + 900 – 172.57

GH² = 1816.43

GH = 42.62 cm

(a)(ii)

$\begin{array}{l} ∠ A C D = 180^{\circ} - 45^{\circ} - 85^{\circ} = 50^{\circ} \\ Using sine rule, \\ \frac{A C}{\sin 45^{\circ}} = \frac{73}{\sin 50^{\circ}} \\ A C = \frac{73 \times \sin 45^{\circ}}{\sin 50^{\circ}} \\ A C = 67.38 cm \\ ∴ H C = 67.38 - 30 = 37.38 cm \end{array}$

(b)

$\begin{array}{l} Area of Δ G A H = \frac{1}{2} (33) (30) \sin 85^{\circ} = 493.12 {cm}^{2} \\ Let length of B K = J K = x \\ 2 \times Area of Δ J B K = Area of Δ G A H \\ 2 \times [\frac{1}{2} (x) (x)] = 493.12 \\ x^{2} = 493.12 \\ x = 22.21 cm \\ B K = 22.21 cm \end{array}$

(c)