The second problem is that you can also use the cosine theorem to find the cosine value of angle B and deduce its sine value. Then sin(b+π/3) can become sinb* radical number 3/2+cosb* 1/2.

or

(1) Cosine Theorem: COSC = (A 2+B 2-C 2)/2AB

=(9+25-49)/(2×3×5)

=- 1/2

And 0

(2) Cosine theorem: COSB = (A 2+C 2-B 2)/2AC.

=(9+49-25)/(2×3×7)

= 1 1/ 14

And 0

So sin (b+π/3) = sinb *1/2+cosb * √ 3/2.

=5√3/ 14* 1/2+ 1 1/ 14*√3/2

=4√3/7

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