Question

If a, b and c are all acute angles in a triangle and sinA = m and sinB = n, find sin c in terms of m and n?

I do not understand how to do this one, please help!

Answer 1

`sin(c)=msqrt(1-n^2)+nsqrt(1-m^2)`

Explanation:

Since they are angles of a triangle, we have `a+b+c=pi`, so we can deduce that `c = pi-a-b = pi-(a+b)`

So, `sin(c) = sin(pi-(a+b))`

Since `sin(theta)=sin(pi-theta)`, we have that `sin(pi-(a+b))=sin(a+b)`

In turn,

`sin(a+b)=sin(a)cos(b)+sin(b)cos(a)` (1)

So far, we only know the sine values, but we don't know the cosines. Using the fundamental relation `sin^2+cos^2=1`, we can cosines from sines:

`sin^2(a)+cos^2(a)=1 \implies cos(a) = sqrt(1-sin^2(a))`

(we're taking the positive root because we know that alla angles are between `0` and `pi/2`, so both sine and cosine are positive).

So, we have:

• `sin(a) = m`
• `cos(a) = sqrt(1-m^2)`
• `sin(b) = n`
• `cos(b) = sqrt(1-n^2)`

Plug these values into the formula (1) to get the answer.