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?

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

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Answer 1 · 2018-05-03T12:24:04

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

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:

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