Law of cosines (Need a creative solution)

[gusrohar]

I have encountered a problem which I need to find a creative solution for:

In the triangle ABC, a=BC, b=CA, c=AB and the angles A=alpha, B=beta and C=gamma.

Find c, given that the triangle is obtuse at C (or gamma) and a=4, b=3 and sin(gamma)=8/9

I understand that I need to use the law of cosines which gives me:

c^2=4^2 + 3^2 - 2*3*4 cos(gamma)

However the problem is I need to anser in integers and therefor I need a creative way to turn sin(gamma) into cos(gamma) in terms of integers if that makes sense. Essentially I cannot answer in decimals. And I have difficulty finding a solution.

Thanks!

 

[HallsofIvy]

Presumably you know that \(\displaystyle sin^2(\theta)+ cos^2(\theta)= 1\) for any \(\displaystyle \theta\). Since you are given \(\displaystyle sin(\gamma)= 8/9\), you know that \(\displaystyle \frac{64}{81}+ cos^2(\gamma)= 1\) so that \(\displaystyle cos^2(\gamma)= 1- \frac{64}{81}= \frac{17}{81}\) and then \(\displaystyle cos(\gamma)= \frac{\sqrt{17}}{9}\). You say "I need a creative way to turn sin(gamma) into cos(gamma) in terms of integers" but that's just not going to happen unless you are willing to accept a not very good approximation.

An equivalent way to do this is to recognize that if \(\displaystyle sin(\gamma)= 8/9\) then we can think of \(\displaystyle \gamma\) as an angle in a right triangle with "opposite side" of length 8 and hypotenuse of length 9. By the Pythagorean theorem, the "near side" has length \(\displaystyle \sqrt{9^2- 8^2}= \sqrt{17}\) giving the same answer as before.

If you are willing to accept an approximation, you could note that since 17 is pretty close to 16, \(\displaystyle \sqrt{17}\) is pretty close to 4 and use \(\displaystyle \frac{4}{9}\). Or, writing \(\displaystyle \sqrt{17}= 4.1231...\) a better approximation would be \(\displaystyle \frac{41231}{90000}\)