The answer to the question is,
12 sin(∠B) = 2, 12 sin(∠C) = √3+√35
Sin rule is,
sin(A)/BC = sin(B)/AC = sin(C)/AB
sin(B) = (sin(A)×AC)/BC
sin(C) =(sin(B)×AB)/AC
To solve this question we apply sin rule,
Now we take
12 sin(∠B) =12 (sin(∠A)×AC)/BC
12 sin(∠B) =12 (sin(π/6)×(x/3x)) where ∠A =π/6 and AC=x, BC =3x
12 sin(∠B) =12 ((1/2)×(1/3))
12 sin(∠B) =12/6 = 2
12 sin(∠B) = 2
now we find the value of 12 sin(∠C)
12 sin(∠C) = 12(sin B)×(AB/BC)
now put the value of 12 sin(∠B) = 2
12 sin(∠C) = 2×(AB/BC)
12 sin(∠C) = (2/AC)×[AC×(cos(π/6))+BC×(cos B)]
12 sin(∠C) = 2×[cos(π/6)+(BC/AC)×(cos B)]
cos (π/6) = √3/2 and BC/AC =3
then
12 sin(∠C) = 2×[(√3/2)+3×(cos B)]
cos B= √1-sin²B
12 sin(∠C) = 2×[(√3/2)+3×√(1-sin²B)]
12 sin(∠B) = 2
sin B = 1/6
12 sin(∠C) = 2×[(√3/2)+3×√(1-(1/6)²]
12 sin(∠C) = 2×[(√3/2)+3×√1-(1/36)]
12 sin(∠C) = 2×[(√3/2)+3×√(36-1)/36]
12 sin(∠C) = 2×[(√3/2)+3×√(35/36)]
12 sin(∠C) = 2×[(√3/2)+(3/6)×√35]
12 sin(∠C) = 2×[(√3/2)+(1/2)×√35]
12 sin(∠C) = 2×[(1/2)(√3+√35)]
12 sin(∠C) = √3+√35
Hence the answer is,
12 sin(∠B) = 2, 12 sin(∠C) = √3+√35
Learn more about triangles rules from:
brainly.com/question/27998693
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![\[12 \sin(\angle B), 12 \sin(\angle C)\]](https://tex.z-dn.net/?f=%5C%5B12%20%5Csin%28%5Cangle%20B%29%2C%2012%20%5Csin%28%5Cangle%20C%29%5C%5D)
in that order?