If a - b - c are the sides opposite to angles A - B - C of a triangle ABC- respectively and - A -3- b - c -3-1- 2- then the value of - B - C isA- -2B- -12C- -cAD- -4

Bc=√(3)+12 ⇒b cb+c=√(3)+1 2√(3)+1+2 ⇒b cb+c=√(3) 1√(3)+3 Now, tanB C2=b cb+cA2 ⇒tanB C2=√(3) 1√(3)+3·√(3) ⇒tanB C2=√(3) 1√(3)+1=2 √(3) ⇒B C2=π12 ⇒ B C=π6 ...

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Byju's Answer

Standard XII

Mathematics

Napier’s Analogy

If a , b , c ...

Question

If $a, b, c$ are the sides opposite to angles $A, B, C$ of a triangle $A B C,$ respectively and $∠ A = \frac{π}{3}, b : c = \sqrt{3} + 1 : 2,$ then the value of $∠ B - ∠ C$ is

\frac{π}{12}

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\frac{π}{6}

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\frac{π}{4}

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\frac{π}{2}

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Solution

The correct option is B $\frac{π}{6}$
$\frac{b}{c} = \frac{\sqrt{3} + 1}{2}$
$\Rightarrow \frac{b - c}{b + c} = \frac{\sqrt{3} + 1 - 2}{\sqrt{3} + 1 + 2}$
$\Rightarrow \frac{b - c}{b + c} = \frac{\sqrt{3} - 1}{\sqrt{3} + 3}$

Now, $tan \frac{B - C}{2} = \frac{b - c}{b + c} cot \frac{A}{2}$
$\Rightarrow tan \frac{B - C}{2} = \frac{\sqrt{3} - 1}{\sqrt{3} + 3} \cdot \sqrt{3}$
$\Rightarrow tan \frac{B - C}{2} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = 2 - \sqrt{3}$
$\Rightarrow \frac{B - C}{2} = \frac{π}{12}$
$\Rightarrow B - C = \frac{π}{6}$

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If a,b,c are the sides opposite to angles A,B,C of a triangle ABC, respectively and ∠A=π3, b:c=√3+1:2, then the value of ∠B−∠C is

The correct option is B π6bc=√3+12 ⇒b−cb+c=√3+1−2√3+1+2 ⇒b−cb+c=√3−1√3+3 Now, tanB−C2=b−cb+ccotA2 ⇒tanB−C2=√3−1√3+3⋅√3 ⇒tanB−C2=√3−1√3+1=2−√3 ⇒B−C2=π12 ⇒B−C=π6

If $a, b, c$ are the sides opposite to angles $A, B, C$ of a triangle $A B C,$ respectively and $∠ A = \frac{π}{3}, b : c = \sqrt{3} + 1 : 2,$ then the value of $∠ B - ∠ C$ is