A triangular pennant ABC is sewn for a school sports day. In triangle ABC, $\angle BAC = 50°$. The side AB is extended to a point D so that A, B and D lie on a straight line. The exterior angle $\angle CBD$ at vertex B is $110°$. Find the size of $\angle BCA$.
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Worked Solution
Step 1: Since A, B and D lie on a straight line, ∠ABC and ∠CBD are angles on a straight line. So ∠ABC + ∠CBD = 180°.\n
Step 2: ∠ABC = 180° − 110° = 70°.\n
Step 3: The three interior angles of triangle ABC sum to 180°: ∠BAC + ∠ABC + ∠BCA = 180°.\n
Step 4: 50° + 70° + ∠BCA = 180° → ∠BCA = 180° − 120° = 60°.\n
Step 5: (Alternative shortcut: the exterior-angle theorem gives ∠CBD = ∠BAC + ∠BCA, so 110 = 50 + ∠BCA, giving ∠BCA = 60° directly.)
Answer: 60°
Correct answer: 60°
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