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=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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