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Geometry · isosceles triangles and angles on a straight line
P6 · Geometry · PSLE Maths
Question
Siti pasted two triangles on a straight line during art class. B, C and D lie on a straight line, AB = AC and CE = DE. Given that ∠ABC = 38° and ∠CED = 64°, find the size of ∠ACE.
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Hint
First find the base angles in the two isosceles triangles, then use 180° at point C on the straight line.
Show the answer and step-by-step explanation
Answer: ∠ACE 84 °
- Read the diagram — B, C and D are on a straight line. On the left, AB = AC, so it is an isosceles triangle. On the right, CE = DE, so it is also an isosceles triangle.
- Left base angle — Since AB = AC, triangle ABC is isosceles. So its base angles are equal: ∠ACB = ∠ABC = 38°.
- Right base angles — Since CE = DE, the base angles ∠ECD and ∠CDE are equal. Using the angle sum of a triangle: the two base angles add up to 180° − 64° = 116°, so each is 58°.
- Use 180° — At point C, B, C and D form a straight line, so the three adjacent angles on the straight line add up to 180°. That is, 38° + x + 58° = 180°.
- Find x — So x = 180° − 38° − 58° = 84°. Answer: ∠ACE = 84°. 📌 Key idea: mark equal base angles in isosceles triangles first; use 180° in a triangle and 180° on a straight line.
Method: In an isosceles triangle, the base angles are equal; angles in a triangle add up to 180°; angles on a straight line add up to 180°.
Practice: 3 similar questions
- Mei Ling drew two triangles. B, C and D lie on a straight line, AB = AC and CE = DE. Given that ∠ABC = 42° and ∠CED = 70°, find ∠ACE.
Answer
83 ° — AB = AC, so ∠ACB = ∠ABC = 42°. CE = DE, so ∠ECD = ∠CDE. Each base angle = (180° − 70°) ÷ 2 = 55°. On the straight line BCD: 42° + ∠ACE + 55° = 180°. ∠ACE = 180° − 42° − 55° = 83°.
- Raju is designing a playground pattern. P, Q and R lie on a straight line, MP = MQ and NQ = NR. Given that ∠PMQ = 76° and ∠QNR = 50°, find ∠MQN.
Answer
63 ° — In isosceles triangle PMQ, MP = MQ, so base angle ∠PQM = (180° − 76°) ÷ 2 = 52°. In isosceles triangle QNR, NQ = NR, so base angle ∠NQR = (180° − 50°) ÷ 2 = 65°. P, Q and R lie on a straight line: 52° + ∠MQN + 65° = 180°. ∠MQN = 180° − 52° − 65° = 63°.
- Ali made a design using a straight ruler. B, C, D and F lie on a straight line, AB = AC and CE = DE. Given that ∠ABC = 47° and the exterior angle ∠EDF = 118°, find ∠ACE.
Answer
71 ° — AB = AC, so ∠ACB = ∠ABC = 47°. C, D and F lie on a straight line, so ∠CDE = 180° − 118° = 62°. CE = DE, so ∠ECD = ∠CDE = 62°. On the straight line at C: 47° + ∠ACE + 62° = 180°. ∠ACE = 180° − 47° − 62° = 71°.
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