SPM 2022 Add Maths Paper 2, Question 12Solution of Triangles

Solved on video in English and Bahasa Melayu · 2 marks · Form 4, Solution of Triangles · 7:43 video

The question

(a)

Solution by scale drawing is not accepted. Diagram 6 shows quadrilateral JKMN with an area of L cm². It is given that KN = √p cm and ∠KMN = 110°. In triangle JKN, JN = 10 cm, JK = 12 cm and ∠NJK = 60°. In triangle KMN, ∠KNM = 30°.

(a) Find (i) the value of p, (ii) the value of L.

(b)

(b) Sketch triangle K'M'N' which has a different shape from triangle KMN, such that K'M' = KM, K'N' = KN and ∠K'N'M' = ∠KNM. Hence, state the value of ∠M'K'N'.

[2 marks]

SPM 2022 Add Maths Paper 2, Question 12: the question as drawn in Eduly's video7:43

Part (a)

Step 1

In triangle JKN use the cosine rule on KN with the included angle at J = 60°:

Why this step

In JKN we know two sides and the angle between them, but no angle opposite a known side, so the cosine rule is the one that works.

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

Final answer

p = 124, L ≈ 73.17 cm²

Where students lose marks

Watch out: KM must be accurate. Use the correct sine rule setup and keep 4 decimal places (5.9251), or L will be wrong.

Part (b)

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

Final answer

∠M'K'N' = 80°

Where students lose marks

Watch out: for the different triangle use ∠K'M'N' = 70°, not 110°. Recomputing with 110° gives 40° again, which is wrong.

Watch every step of this question solved on video (7:43), with the reason behind each step and a box that checks your own answer.

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