# 9.4.5 Solution of Triangles, Long Questions (Question 9)

Question 9
Diagram 6 shows a quadrilateral KLMN.

Calculate
(a)KML, [2 marks]

(b) the length, in cm, of KM,   [3 marks]

(c) area, in cm2, of triangle KMN,   [3 marks]

(d) a triangle K’L’M’ has the same measurements as those given for triangle KLM, that is K’L’= 12.4 cm, L’M’= 9.5 cm and ∠L’K’M’= 43.2o, which is different in shape to triangle KLM.
(i) Sketch the triangle K’L’M’,
(ii) State the size of ∠K’M’L’. [2 marks]

Solution:
(a)
$\begin{array}{l}\frac{\mathrm{sin}\angle KML}{12.4}=\frac{\mathrm{sin}{43.2}^{o}}{9.5}\\ \mathrm{sin}\angle KML=\frac{\mathrm{sin}{43.2}^{o}}{9.5}×12.4\\ \mathrm{sin}\angle KML=0.8935\\ \angle KML={63.32}^{o}\end{array}$

(b)
∠KLM = 180o – 43.2o – 63.32o = 73.48o
KM2 = 9.52 + 12.42 – 2(9.5)(12.4) cos 73.48o
KM2 = 244.01 – 66.99
KM2 = 177.02
KM = 13.30 cm

(c)
KM2 = MN2+ KN2– 2(MN)(KN) cos ∠KNM
13.302 = 5.42 + 9.92– 2(5.4)(9.9) cos ∠KNM
176.89 = 127.17 – 106.92 cos ∠KNM
$\begin{array}{l}\mathrm{cos}\angle KNM=\frac{127.17-176.89}{106.92}\\ \angle KNM={117.71}^{o}\end{array}$

Area of triangle KMN
$\begin{array}{l}=\frac{1}{2}\left(5.4\right)\left(9.9\right)\mathrm{sin}{117.71}^{o}\\ =23.66\text{}c{m}^{2}\end{array}$

(d)(i)

(d)(ii)
∠K’M’L’ = 180o – 63.32o = 116.68o