Solution of Triangles Long Questions (Question 5 & 6)


Question 5:
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, ACD = 45° and 
  AD = 14 cm. Calculate the two possible values of  ADC.
(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


Question 6:
Diagram below shows trapezium ABCD.
(a) Calculate
(i) ∠BAC.
(ii) the length, in cm, of AD.
(b) The straight line AB is extended to B’ such that BC = B’C.
(i) Sketch the trapezium AB’CD.
(ii) Calculate the area, in cm2, of ∆BB’C.  

Solution:
(a)(i)
5 2 = 4 2 + 7 2 2( 4 )( 7 )cosBAC 25=16+4956cosBAC 56cosBAC=40 cosBAC= 40 56  BAC= cos 1 40 56    = 44 o 25


(a)(ii)
AD sinDCA = 7 sin 115 o AD sin 44 o 25 = 7 sin 115 o ( DCA=BAC )   AD= 7 sin 115 o ×sin 44 o 25   AD=5.406 cm


(b)(i)




(b)(ii)
sinABC 7 = sin 44 o 25 5 sinABC= sin 44 o 25 5 ×7    = 78 o 28 ABC= 180 o 78 o 28 ABC= 101 o 32( obtuse angle ) CBB= 180 o 101 o 32= 78 o 28 BCB= 180 o 78 o 28 78 o 28= 23 o 4 Area of BBC= 1 2 ×5×5× 23 o 4  =4.898  cm 2

Leave a Reply

Your email address will not be published. Required fields are marked *