Solution of Triangles Long Questions (Question 3)


Question 3:
The diagram below shows a triangle ABC.

(a) Calculate the length, in cm, of AC.
(b) A quadrilateral ABCDis now formed so that AC is a diagonal, ACD = 45° and 
      AD = 14 cm. Calculate the two possible values of ÐADC.
(c) By using the acute ADC from (b), calculate
     (i) the length, in cm, of CD,
     (ii) the area, in cm2, of the quadrilateral ABCD

Solution:
(a)
Using cosine rule,
AC2 = AB2 + BC2 – 2 (AB)(BC) ABC
AC2 = 162 + 122 – 2 (16)(12) cos 70o
AC2 = 400 – 131.33
AC2 = 268.67
AC = 16.39 cm

(b)

Using sine rule, sin A D C 16.39 = sin 45 14 sin A D C = 16.39 × sin 45 14 sin A D C = 0.8278 A D C = 55.87  or  ( 180 55.87 ) A D C = 55.87  or 124 .13
(c)(i)
Acute angle of  A D C = 55.87 C A D = 180 45 55.87 = 79.13 C D sin 79.13 = 14 sin 45 C D = 14 × sin 79.13 sin 45 = 19.44  cm
(c)(ii)
Area of quadrilateral  A B C D = Area of  Δ   A B C + Area of  Δ   A C D = 1 2 ( 16 ) ( 12 ) sin 70 + 1 2 ( 16.39 ) ( 14 ) sin 79.13 = 90.21 + 112.67 = 202.88  cm 2

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