How do you solve the triangle ABC given A=30, b=4, c=6?

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Answer 1 · 2017-02-20T13:34:16

$a = 2 \sqrt{13 - 6 \sqrt{3}} ≅ 3.23$ $a=2sqrt(13-6sqrt3)~=3.23$

$hat B=sin^(-1)(1/(sqrt(13-6sqrt3)))~=38.26°$

$hat C~=180°-hatA-hatB~=111.74°$

You would apply the cosine theorem to find the third side a:

$a=sqrt(b^2+c^2-2bc cos hatA$

Then

$a=sqrt(4^2+6^2-2*4*6*cos30°)$

$=sqrt(16+36-cancel48^24*sqrt3/cancel2)$

$=sqrt(4(4+9-6sqrt(3))$

$=2sqrt(13-6sqrt3)~=3.23$

To find the angle $hat B$ , you would apply the sine theorem:

$a/(sin hat A)=b/(sin hatB)$

or

$sin hat B=b/a*sin hat A$

$=cancel4^2/(cancel2sqrt(13-6sqrt3))*sin 30°$

$cancel2/(sqrt(13-6sqrt3))*1/cancel2$

$=1/(sqrt(13-6sqrt3))$

then

$hat B=sin^(-1)(1/(sqrt(13-6sqrt3)))~=38.26°$

$hat C~=180°-hatA-hatB~=111.74°$

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