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Consider [tex]$\triangle ABC$ such that $BC = 3AC$ and $\angle A = \pi/6$[/tex]
What are [tex]\[12 \sin(\angle B), 12 \sin(\angle C)\][/tex] in that order?

Consider Textriangle ABC Such That BC 3AC And Angle A Pi6tex What Are Tex12 Sinangle B 12 Sinangle Ctex In That Order class=

Sagot :

varshamittal029

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:

https://brainly.com/question/27998693

#SPJ10

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