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JEE Main 201510 AprilOnline
Exam
JEE Main
Year
2015
Date
10 April
Mode
Online
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

JEE Main 2015 · 10 April · Online · Mathematics · Trigonometric Ratios & Identities

In a ABC , (a)/(b) = 2 + 3 , and ∠C = 60 ° . Then the ordered pair...

In a δABC\delta ABC, ab=2+3\dfrac{a}{b} = 2 + \sqrt{3}, and ∠C=60°.∠C = 60 ° . Then the ordered pair (∠A,∠B)( ∠A, ∠B ) is equal to:

Options

  1. A(105°,15°)\left(105 ° , 15 °\right)
  2. B(15°,105°)\left(15 ° , 105 °\right)
  3. C(45°,75°)\left(45 ° , 75 °\right)
  4. D(75°,45°)( 75 ° , 45 ° )

Correct answer

A. (105°,15°)\left(105 ° , 15 °\right)

Step-by-step solution

ab>1⇒∠A>∠B\begin{array}{ll}\dfrac{a}{b} > 1 ⇒ ∠ A > ∠ B\end{array}

A+B=120°A + B = 120°

⇒sin⁡(120°−B)sinB=2+3⇒ \dfrac{\sin ( 120° - B )}{sinB} = 2 + \sqrt{3}

⇒32cotB+12=2+3\begin{array}{ll}\text{⇒}\dfrac{\sqrt{3}}{2}cotB + \dfrac{1}{2} = 2 + \sqrt{3}\end{array}

⇒cotB=23(32+3)\begin{array}{ll}⇒ cotB = \dfrac{2}{\sqrt{3}}\left(\dfrac{3}{2} + \sqrt{3}\right)\end{array}

⇒cotB=2+3⇒ cotB = 2 + \sqrt{3}

⇒B=15°\begin{array}{ll}⇒ B = 15°\end{array}

Hence, (∠A,∠B)=(105°,15°)( ∠ A, ∠ B ) = ( 105°, 15° )

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