Solution of Triangles, SPM Paper 2 (Long Question)


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Your email address will not be published.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 *