The lengths of the sides of a triangle are 15m, 21m, and 28cm. What is the size of the angle facing the side 21m (correct to one decimal place)?

Note: the question has a unit typo (28cm should be 28m). The sides are 15, 21 and 28, all in the same unit, so we treat them that way.

Concept: When all three sides are known, use the cosine rule to find an angle:

cos A = (b² + c² − a²) / (2bc)

where A is the angle opposite side a.

Solution:

The angle faces the 21 m side, so a = 21, b = 15, c = 28.

• b² = 225
• c² = 784
• a² = 441

cos A = (225 + 784 − 441) / (2 × 15 × 28) cos A = 568 / 840 cos A = 0.67619

A = cos⁻¹(0.67619)

Checking: cos 47° = 0.6820 and cos 48° = 0.6691. The value 0.6762 lies between them.

(0.6820 − 0.6762) / (0.6820 − 0.6691) = 0.0058 / 0.0129 ≈ 0.45

So A ≈ 47.45°, which is about 47.5° (a more precise calculator value is 47.54°, rounding to 47.5°).

Answer: The angle is about 47.5°, which matches option B.

Tip: When you have three sides (SSS), go straight to the cosine rule. Always put the side opposite the wanted angle as "a", and subtract its square in the numerator. Also check that the answer is sensible: the largest angle faces the longest side (28), and 47.5° for the middle side is reasonable.