MRSM Trial 2025 Add Maths Paper 2, Question 12Solution of Triangles

Solved on video in English · Form 4, Solution of Triangles · 7:50 video

The question

(a)(i)

Solution by scale drawing is not accepted. Diagram 8 shows an isosceles triangle ABC (AB = AC = 15 cm) and straight lines AD and BCD, with C on BD. It is given that BD = 20 cm and AD = 17.46 cm.

(a)(i) Calculate ∠ABD.

(a)(ii)

75°.

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

(a)(iii)

01 cm.

(a)(iii) Calculate the area, in cm², of triangle ABC by using Heron's formula.

(b)(i)

(b)(i) It is given that triangle A'C'D' has a different shape from triangle ACD such that A'C' = AC, C'D' = CD and ∠C'A'D' = ∠CAD. Sketch triangle A'C'D'.

(b)(ii)

Triangle A'C'D' has A'C' = AC, C'D' = CD and ∠C'A'D' = ∠CAD, but a different (obtuse) shape from triangle ACD.

(b)(ii) Calculate ∠A'D'C'.

MRSM Trial 2025 Add Maths Paper 2, Question 12: the question as drawn in Eduly's video7:50

Part (a)(i)

Step 1

In triangle ABD apply the cosine rule with AD opposite ∠ABD: AD² = AB² + BD² − 2·AB·BD·cos∠ABD, i.e. 17.46² = 15² + 20² − 2(15)(20)cos∠ABD.

Why this step

The side opposite the angle we want goes on the left. AD faces ∠ABD, and 20 is BD, not BC. Mixing these up is the usual slip.

1 more step, with the reason for it, in the full solution

Final answer

∠ABD = 57.75°

Where students lose marks

Watch out: use triangle ABD, where 20 cm is BD (not BC) and AD is the side opposite ∠ABD.

Part (a)(ii)

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

Final answer

BC = 16.01 cm

Where students lose marks

Watch out: use the isosceles base angles to get ∠BAC = 64.50°, then pair each side with its opposite angle in the sine rule.

Part (a)(iii)

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

Final answer

Area ≈ 101.55 cm²

Where students lose marks

Watch out: s is half the perimeter, and Heron uses s(s − a)(s − b)(s − c), one factor per side.

Part (b)(i)

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

Final answer

Sketch (see video)

Part (b)(ii)

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

Final answer

∠A'D'C' = 133.40°

Where students lose marks

Watch out: don't stop at the acute 46.60°. The second triangle has the obtuse angle, so use 180° − ∠ADC.

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