In a triangle ABC- - A-60- and b- c-3-1- 2-then the value of - B- C-A- 45-B- 60-C- 15-D- 30-

The correct option is D 30°bc=3+12⇒b cb+c=3+1 23+1+2=3 1(3+1)3Using nbsp;Napier's nbsp;rule, nbsp;we nbsp;get tan nbsp;B C2=b cb+ccot nbsp;A2⇒3 1(3+ ...

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Byju's Answer

Standard XII

Mathematics

Napier’s Analogy

In a triangle...

Question

In a triangle ABC,
$∠ A = 60 ° a n d b : c = \sqrt{3} + 1 : 2, t h e n t h e v a l u e o f ∠ B - ∠ C =$ .

45 °

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60 °

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15 °

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30 °

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Solution

The correct option is D $30 °$
$\frac{b}{c} = \frac{\sqrt{3} + 1}{2} \Rightarrow \frac{b - c}{b + c} = \frac{\sqrt{3} + 1 - 2}{\sqrt{3} + 1 + 2} = \frac{\sqrt{3} - 1}{(\sqrt{3} + 1) \sqrt{3}} U sin g N a p i e r' s r u l e, w e g e t - tan \frac{B - C}{2} = \frac{b - c}{b + c} c o t \frac{A}{2} \Rightarrow (\frac{\sqrt{3} - 1}{(\sqrt{3} + 1) \sqrt{3}}) \times \sqrt{3} = 2 - \sqrt{3} \Rightarrow \frac{B - C}{2} = 15 ° H e n c e, B - C = 30 °$

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In a triangle ABC, ∠A=60° and b:c=√3+1:2,then the value of ∠B−∠C= .

The correct option is D 30°bc=√3+12⇒b−cb+c=√3+1−2√3+1+2=√3−1(√3+1)√3Using Napier's rule, we get−tan B−C2=b−cb+ccot A2⇒(√3−1(√3+1)√3)×√3=2−√3⇒B−C2=15°Hence, B−C=30°

In a triangle ABC,
$∠ A = 60 ° a n d b : c = \sqrt{3} + 1 : 2, t h e n t h e v a l u e o f ∠ B - ∠ C =$ .