How do you solve the triangle ABC given a=11, b=13, c=16?

1 Answer
Oct 6, 2017

Answer:

#A = 43.0491# (4 .d.p.)

#B = 53.7785# (4 .d.p.)

#C= 83 . 1724# (4 .d.p.)

Explanation:

Labelling triangle in the conventional way i.e. side a opposite angle A, and so on.

We need to use the cosine rule to start, since we know three sides, but no angles.

Cosine Rule:

#a^2+b^2-2bc (CosA)#

We will first find #cos(A)# and hence angle A

Rearrange the cosine formula so #cos(A)# is the subject:

#cos(A) = (a^2-b^2-c^2)/(-2bc)#

Putting known values into this:

#cos(A) = ((11)^2-(13)^2-(16)^2)/(-2(13)(16)) = (-304)/-416= 19/26#

arc-cosine:

#cos^-1(19/26)= 43.0491# (4 .d.p.)

We now have angle A. We can find a second angle using the cosine rule, or we could switch to the sine rule. The sine rule is less cumbersome to use, but we will stick to the cosine rule for now.

We will now find angle B.

We need to rearrange the cosine formula for #cos(B)# to be the subject. This can be achieved by just switching a and b.

#cos(B) = (b^2-a^2-c^2)/(-2ac)#

Putting in values:

#cos(B) = ((13)^2-(11)^2-(16)^2)/(-2(11)(16))=(-208)/(-352)=13/22#

arc-cosine:

#cos^-1(13/22) = 53.7785# (4 .d.p.)

Angle C #= 180 - (53.7785 + 43.0491)= 83 . 1724#

So:

#A = 43.0491# (4 .d.p.)

#B = 53.7785# (4 .d.p.)

#C= 83 . 1724# (4 .d.p.)