Question

if the ration 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 #39216 – Math – Trigonometry

Question: If the ratio of three sides of a triangle is $a:b:c = 7:8:9$, then show that $\cos A : \cos B : \cos C = 14:11:6$.

Solution. Let us apply the law of cosines three times, using each angle once.

Start with $\cos A$ and thus the side $a$:

$$\begin{array}{l} a^{2} = b^{2} + c^{2} - 2bc \cos A \\ - 2bc \cos A = a^{2} - b^{2} - c^{2} \\ - 2 \cos A = \frac{a^{2}}{bc} - \frac{b^{2}}{bc} - \frac{c^{2}}{bc} \\ - 2 \cos A = \frac{a}{b} \cdot \frac{a}{c} - \frac{b}{c} - \frac{c}{b} \\ \end{array}$$

Now note that we are given each of the ratios $\frac{a}{b}, \frac{a}{c}, \frac{b}{c}$. Therefore,

$$\begin{array}{l} - 2 \cos A = \frac{7}{8} \cdot \frac{7}{9} - \frac{8}{9} - \frac{9}{8} \\ - 2 \cos A = \frac{7^{2} - 8^{2} - 9^{2}}{8 \cdot 9} = \frac{49 - 64 - 81}{72} \\ 2 \cos A = \frac{96}{72} \\ 2 \cos A = \frac{96}{72} \\ \cos A = \frac{2}{3}. \\ \end{array}$$

Similarly, for $\cos B$ we have

$$b^{2} = a^{2} + c^{2} - 2ac \cos B.$$

Making the same transformations as above, we arrive at

$$\begin{array}{l} - 2 \cos B = \frac{b}{a} \cdot \frac{b}{c} - \frac{a}{c} - \frac{c}{a} \\ - 2 \cos B = \frac{8}{7} \cdot \frac{8}{9} - \frac{7}{9} - \frac{9}{7} \\ \end{array}$$

and find

$$\cos B = \frac{11}{21}.$$

Finally, for $\cos C$

$$\begin{array}{l} c^{2} = a^{2} + b^{2} - 2ab \cos C \\ - 2 \cos C = \frac{c}{a} \cdot \frac{c}{b} - \frac{a}{b} - \frac{b}{a} \\ - 2 \cos C = \frac{9}{7} \cdot \frac{9}{8} - \frac{7}{8} - \frac{8}{7} \\ \end{array}$$

and

$$\cos C = \frac {2}{7}.$$

Now we only need to find the ratio of obtained cosines:

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

Bring all the fractions on the left side of the equality to a common denominator:

$$\cos A: \cos B: \cos C = \frac {14}{21}: \frac {11}{21}: \frac {6}{21},$$

or

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