Trig assignment

[courteous]

»In a right triangle ABC $\displaystyle a+b+c=20$ and $\displaystyle \beta=15$. What are length for a,b,c (and sizes of other two angles, $\displaystyle \alpha$ and $\displaystyle \gamma$)?« I'm assuming that by agreement $\displaystyle \gamma=90$.

I'm just solving it by using 'law of sine' and thus making myself (long ) quadratic equation. I have a gut feeling that there are more (and shorter) ways to solve it.

 

[Denis]

So other angle = 75, and c = 20 - a - b ; got that?
Since a^2 + b^2 = (20 - a - b)^2, then get b in terms of a : b = (200 - 20a) / (20 - a) [1]
Then you have a/SIN(15) = b/SIN(75) ; so b = aSIN(75) /SIN(15) [2]

Solve using [1] and [2].
The quadratic ain't that bad; quit complaining :wink:

 

[courteous]

Thank you, Denis!

First I was left with sth1cos^2x + sth2cosx - sth2 = 0 (sth is something ), which led to a mess...

 

[Denis]

Easier still if you use TAN(15) = a/b