Question

If the ratio of three sides of a triangle is a:b:c = 7:8:9 then show that cosA:cosB:cosC = 14:11:16

Answer 1

Answer on Question #44425 – Math – Trigonometry

Problem

If the ratio of three sides of a triangle is a:b:c = 7:8:9 then show that $\cos A:\cos B:\cos C = 14:11:16$

Solution

Use cosine theorem for side a, b and c in same order

$$a^2 = b^2 + c^2 - b * c * \cos A$$ $$b^2 = a^2 + c^2 - a * c * \cos B$$ $$c^2 = b^2 + a^2 - b * a * \cos C$$

Putting values:

$$49 = 64 + 81 - 144 \cos A$$ $$-96 = -144 \cos A$$ $$\cos A = \frac{96}{144} = \frac{2}{3}$$

Do in same way with other equations we get

$$\cos B = \frac{66}{126} = \frac{11}{21}$$

And

$$\cos C = \frac{32}{112} = \frac{2}{7}$$

So, there is ratio

$$\cos A: \cos B: \cos C = \frac{2}{3}: \frac{11}{21}: \frac{2}{7}$$

Multiply it with 21 we get

$$\cos A: \cos B: \cos C = 14:11:6$$

P.S. There is a mistake in problem condition

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