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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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!

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Answer 1 · 2018-05-03T12:24:04

$sin (c) = m \sqrt{1 - n^{2}} + n \sqrt{1 - m^{2}}$ $sin(c)=msqrt(1-n^2)+nsqrt(1-m^2)$

Explanation:

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

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

Since $sin (θ) = sin (π - θ)$ $sin(theta)=sin(pi-theta)$ , we have that $sin (π - (a + b)) = sin (a + b)$ $sin(pi-(a+b))=sin(a+b)$

In turn,

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

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

${sin}^{2} (a) + {cos}^{2} (a) = 1 \Rightarrow cos (a) = \sqrt{1 - {sin}^{2} (a)}$ $sin^2(a)+cos^2(a)=1 \implies cos(a) = sqrt(1-sin^2(a))$

(we're taking the positive root because we know that alla angles are between $0$ $0$ and $\frac{π}{2}$ $pi/2$ , so both sine and cosine are positive).

So, we have:

$sin (a) = m$ $sin(a) = m$
$cos (a) = \sqrt{1 - m^{2}}$ $cos(a) = sqrt(1-m^2)$
$sin (b) = n$ $sin(b) = n$
$cos (b) = \sqrt{1 - n^{2}}$ $cos(b) = sqrt(1-n^2)$

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

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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!

1 Answer

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