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 ABCD is now formed so that AC is a diagonal, ÐACD = 45° and 
      AD = 14 cm. Calculate the two possible values of ÐADC.
(c) By using the acute ÐADC 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) ÐABC
AC2 = 162 + 122 – 2 (16)(12) cos 70o
AC2 = 400 – 131.33
AC2 = 268.67
AC = 16.39 cm

(b)

Using sine rule, sinADC 16.39 = sin 45 14 sinADC= 16.39×sin 45 14 sinADC=0.8278 ADC= 55.87  or ( 180 55.87 ) ADC= 55.87  or 124 .13
(c)(i)
Acute angle of ADC= 55.87 CAD= 180 45 55.87 = 79.13 CD sin 79.13 = 14 sin 45 CD= 14×sin 79.13 sin 45 =19.44 cm
(c)(ii)
Area of quadrilateral ABCD = Area of Δ ABC+Area of Δ ACD = 1 2 ( 16 )( 12 )sin 70 + 1 2 ( 16.39 )( 14 )sin 79.13 =90.21+112.67 =202.88  cm 2