A triangular field \(ABC\) has \(BC = 45\) m, \(CA = 38\) m and \(AB = 52\) m, as shown. ABC45 m38 m52 m© EAGLE BEACON GLOBAL Work out the size of the large...
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A triangular field \(ABC\) has \(BC = 45\) m, \(CA = 38\) m and \(AB = 52\) m, as shown.
Work out the size of the largest angle of the field. Give your answer correct to \(1\) decimal place. (4)
Answer Details
This question uses the cosine rule to find an angle in a triangle when all three side lengths are known, applying the fact that the largest angle is always opposite the longest side.
The longest side is \(AB = 52\) m, so the largest angle is the one opposite it, angle \(C\). Using the cosine rule rearranged to find an angle:
\[\cos C = \dfrac{45^2 + 38^2 - 52^2}{2 \times 45 \times 38}\][2 marks]
\[\cos C = \dfrac{2025+1444-2704}{3420} = \dfrac{765}{3420} = 0.224 \text{ (3 s.f.)}\][1 mark]
\[C = \cos^{-1}(0.224) = 77.1^\circ \text{ (1 d.p.)}\][1 mark]
The largest angle of the field is \(77.1^\circ\). Checking which angle is required before choosing which side to isolate in the cosine rule (here \(AB\), so the formula is rearranged for angle \(C\)) avoids a common labelling error.
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