in triangleABC,AB=6CM,BC=5.2&<B =150 degre

Question

Find AC, find angle C

Answer 1

We use the cosine rule:

AC^2=AB^2+BC^2-2AB•BCcosB=36+27.04-2*6*5.2cos150=117.08 approx.

AC=√117.08=10.82cm approx.

Now we use the sine rule: sinA/BC=sinB/AC=sinC/AB. We need angle C and we have angle B and sides AB and AC, so sin150/10.82=sinC/6; sinC=6sin150/10.82=0.2773 approx. C=16.10 degrees approx. (sin150=0.5, cos150=-√3/2=-0.8660 approx).

 

Answer 2

Given AB=6CM,BC=5.2&<B =150 degre

Using cosine rule:

AC^2 = AB^2+BC^2-2AB•BCcosB
         = 36+27.04-2*6*5.2cos150
         = 117.08

AC = √117.08=10.82 cm

sinB/AC = sinC/AB

sin 150/10.82 = sinC/6

C = 16.10 degree

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