Problem

For the triangle shown, \(A~=~40°\), \(B~=~65°\), and \(a~=~12\). Find \(\text{all}~\text{three}~\text{angles}\) and side lengths, then verify the angle sum and the order of opposite sides and angles.

Prompt triangle with A=40°, B=65°, and opposite side a=12; angle C and sides b and c are withheld.

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What this problem is really about

Complete the angle sum first so every opposite side-angle pair is available. Use the fully known pair to anchor one Law of Sines scale, then apply that same scale to each missing side without rounding early. Finish by checking both the angle sum and whether the side order matches the opposite-angle order.

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C=180°-40°-65°=75°. Then b=12sin(65°)/sin(40°)≈16.9 and c=12sin(75°)/sin(40°)≈18.0; the angles sum to 180° and c is opposite the largest angle.

Use C=75° but round sin(40°) to 0.6, sin(65°) to 0.9, and sin(75°) to 1.0 first, giving b=18.0 and c=20.0.

Use C=180°-40°+65°=205°; then b≈16.9 and c=12sin(205°)/sin(40°)≈-7.9, so the listed angles total 310°.

C=75°, but reverse both sine ratios: b=12sin(40°)/sin(65°)≈8.5 and c=12sin(40°)/sin(75°)≈8.0.

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For the triangle shown, \(A~=~40°\), \(B~=~65°\), and \(a~=~12\). Find \(\text{all}~\text{three}~\text{angles}\) and side lengths, then verify the angle sum and the order of opposite sides and angles.

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