Find all solutions for each triangle.

A = 30

a = 4

b = 8

4 / 30 = 8 / SIN B

4 SIN B = 8 ( SIN30 )

SIN B = 8 (SIN 30) / 4 = 1

Did I do that right?

I know there is only one solution.

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- Aug 27th 2008, 05:58 PMtopaz192find the triangle
Find all solutions for each triangle.

A = 30

a = 4

b = 8

4 / 30 = 8 / SIN B

4 SIN B = 8 ( SIN30 )

SIN B = 8 (SIN 30) / 4 = 1

Did I do that right?

I know there is only one solution. - Aug 27th 2008, 06:12 PMShyam
Yes, you are right.

$\displaystyle

=\frac{4}{sin30}= \frac{8}{sinB}

$

$\displaystyle

sinB=1

$

$\displaystyle

B=90

$

Now, in right triangle, right angled at B,

$\displaystyle

c=\sqrt{8^2-4^2}

$

$\displaystyle

c=\sqrt{48}=\;4\sqrt{3}

$

$\displaystyle

C=180-(90+30)=60

$ - Aug 27th 2008, 06:16 PMtopaz192
where did 90 come from?

i get everything else. - Aug 27th 2008, 06:18 PMicemanfan
- Aug 27th 2008, 06:20 PMtopaz192
Ooooh Ok duh. lol thnks!