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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)
sinKML12.4=sin43.2o9.5sinKML=sin43.2o9.5×12.4sinKML=0.8935KML=63.32o 

(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
cosKNM=127.17176.89106.92KNM=117.71o
 
Area of triangle KMN
=12(5.4)(9.9)sin117.71o=23.66cm2
  
(d)(i)


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

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