SPM 2021 Add Maths Paper 2, Question 14Solution of Triangles

Solution by scale drawing is not accepted. Diagram 6 shows a quadrilateral ABCD. Point E is the intersection point of the straight lines AC and BD. It is given that AB = 2.8 cm, BC = 3 cm, DC = 3.6 cm, the area of triangle ABC is equal to the area of triangle BCD, AE = 4 cm, angle ABC = 143° and angle BCD is an obtuse angle. (a) Calculate (i) angle BCD, (ii) the length, in cm, of CE, (iii) the length, in cm, of BE.

From part (a), triangle ABC has AC = 5.501 cm, BC = 3 cm, angle BAC = 19.16° and angle ABC = 143°. (b)(i) Sketch a triangle A'B'C' which has a different shape from triangle ABC such that A'C' = AC, B'C' = BC and angle B'A'C' = angle BAC. (b)(ii) Hence, state the size of angle A'B'C'.