How do you solve the triangle given α = 15.6 degrees, b = 10.25, and c = 5.5?

Answer 1

#a=5.17#
#B=147.7^o#
#C=16.67^o#

This is a side-angle-side triangle so we must use the laws of cosine. First we'll solve for side a #a^2=b^2+c^2-2ac cosA#
#a^2=10.25^2+5.5^2-(2*10.25*5.5*cos15.6)# which gives 26.716. Take the square root of that and we have #a=5.17#
From there we can use the laws of sine #sinA/a=sinB/b=sinC/c# or we can stick with the laws of cosine #cosB=(a^2+c^2-b^2)/(2ac)#
I stuck with cosine #cosB=(5.17^2+5.5^2-10.25^2)/(2*5.17*5.5)# which gives us -0.846. Then we take #cos^-1(-0.846)=147.73^o#
We know that triangles have #180^o# so #180-15.6-147.73=12.27^o#
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Answer from HIX Tutor

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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