SPM 2017 Add Maths Paper 2, Question 15Solution of Triangles

Solved on video in English and Bahasa Melayu · 10 marks · Form 4, Solution of Triangles · 6:55 video

The question

Solution by scale drawing is not accepted. Diagram 8 shows a quadrilateral ABCD on a horizontal plane, with AD = 10 m, angle DBC = 64°, BC = 20.5 m and DC = 22 m. VBDA is a pyramid such that AB = 12 m and V is 5 m vertically above A. Find (a) angle BDC, (b) the length, in m, of BD, (c) the area, in m², of inclined plane BVD.

[10 marks]

SPM 2017 Add Maths Paper 2, Question 15: the question as drawn in Eduly's video6:55

Step 1

In triangle BDC, use the sine rule: sin(∠BDC)/BC = sin(∠DBC)/DC => sin(∠BDC)/20.5 = sin64°/22

Why this step

Pair each side with the angle opposite it: BC faces ∠BDC, DC faces ∠DBC. Mixing these pairs up is a common way to lose marks here.

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

Final answer

(a) ∠BDC = 56°53'; (b) BD = 21.007 m; (c) Area = 62.65 m²

Where students lose marks

Watch out: in the sine rule, pair each side with its opposite angle. V directly above A makes VAB and VAD right-angled at A.

Watch every step of this question solved on video (6:55), with the reason behind each step and a box that checks your own answer.

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