SPM 2019 Add Maths Paper 2, Question 13Solution of Triangles

Solved on video in English and Bahasa Melayu · 10 marks · Form 4, Solution of Triangles · 7:29 video

The question

Solution by scale drawing is not accepted. Diagram 6 shows a quadrilateral ABCD such that AC and BD are straight lines. It is given that the area of triangle ABC = 6 cm² and angle ABC is obtuse. (a) Find (i) angle ABC, (ii) the length, in cm, of AC, (iii) angle BAC. (b) Given BD = 7.3 cm and angle BCD = 90 degrees, calculate the area, in cm², of triangle ACD.

[10 marks]

SPM 2019 Add Maths Paper 2, Question 13: the question as drawn in Eduly's video7:29

Step 1

Area of triangle ABC = (1/2)(AB)(BC)sin(ABC) = 6: (1/2)(4)(3.5)sin(ABC) = 6, so sin(ABC) = 12/14 = 0.8571.

Why this step

We know both sides and the area, so the angle between them is the only unknown left in the area formula.

9 more steps, each with the reason for it, in the full solution

Final answer

∠ABC = 121°, AC = 6.532 cm, ∠BAC = 27.34°, area of triangle ACD ≈ 17.81 cm²

Where students lose marks

Watch out: ABC is obtuse, so use 180° − 59°, and at C subtract angle BCA from 90° (don't add).

Watch every step of this question solved on video (7:29), with the reason behind each step and a box that checks your own answer.

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