SPM 2020 Add Maths Paper 2, Question 13Solution of Triangles

Solved on video in English · Form 4, Solution of Triangles · 7:16 video

The question

(a)(i)

(Solution by scale drawing is NOT accepted.) Diagram 7 shows a triangle ABC marked on horizontal ground and a lamp post PVB. VB is a slant part with length 9 m. PV is vertical, directly above the line AB. The angle of depression of point A from point V is 28 degrees, angle VBA = 80 degrees, angle BAC = 35 degrees and AC = 20 m. (a)(i) Calculate the length, in m, of AB.

(a)(ii)

(a)(ii) Calculate the length, in m, of BC.

(b)(i)

(b)(i) Sketch a triangle A'B'C' which has a different shape from triangle ABC such that B'C' = BC, A'B' = AB and angle B'A'C' = angle BAC. (This is the ambiguous case.)

(b)(ii)

(b)(ii) Calculate angle A'B'C'. Hence, find the area, in m², of triangle A'B'C'.

SPM 2020 Add Maths Paper 2, Question 13: the question as drawn in Eduly's video7:16

Part (a)(i)

Step 1

In triangle AVB: the angle of depression of A from V equals the angle of elevation of V from A, so angle VAB = 28 degrees.

Why this step

The angle of depression is measured from a horizontal line through V. The ground is horizontal too, so alternate angles put the same 28° at A, inside our triangle.

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

Final answer

AB = 18.23 m

Where students lose marks

Watch out: don't assume ∠PVB = 90°. Use ∠VAB = 28° to get ∠AVB = 180° − 28° − 80° = 72°, not 62°.

Part (a)(ii)

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

Final answer

BC = 11.62 m

Where students lose marks

Watch out: the cosine rule is BC² = AB² + AC² − 2(AB)(AC)cos 35°. Don't use + before the 2(AB)(AC)cos term.

Part (b)(i)

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

Final answer

Sketch (see video)

Where students lose marks

Watch out: the arc from B' cuts the 35° arm twice. Take the near crossing to get the obtuse triangle, not a copy of ABC.

Part (b)(ii)

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

Final answer

∠A'B'C' = 29.14°, area = 51.58 m²

Where students lose marks

Watch out: for the second triangle use the obtuse 115.86°, not the acute 64.14°. The acute angle gives ∠A'B'C' ≈ 80.85° and a wrong area.

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