In a ABC , if ∠ A = ∠ B = (1)/(2) ( ^ - 1 ( 6 + 1 2 3 ) + ^ - 1 ( 1 3 )) and length of the side opposite to ∠ C is c = 6 ⋅ 3 (1)/(4) , then the area of ABC is

Correct answer: 27 2∠ B = ^ - 1 ( 6 + 1 2 3 ) + ^ - 1 ( 1 3 ) = ^ - 1 ( 6 + 1 3 - 2 ) + ^ - 1 ( 1 2 ) = ^ - 1 ( 6 + 1 3 - 2 + 1 2 1 - 6 + 1 3 - 2 · 1 2 ) = ^ - 1 ( ( 6 + 1) 2 + ( 3 - 2 ) 2 ( 3 - 2 ) - ( 6 + 1) ) = ^ - 1 ( 1 2 + 2 + 3 - 2 6 - 2 - 6 - 1 ) = ^ - 1 ( 2 3 + 3 - 3 ) = ^ - 1 ( 3 3 - 3 ) = ^ - 1 (- 3 ) = (2 π)/(3) According to question ∠ A = ∠ B = (1)/(2) ( ^ - 1 ( 6 + 1 2 3 ) + ^ - 1 ( 1 3 )) ⇒ ∠ A = ∠ B = (1)/(2) ((2 π)/(3)) ⇒ ∠ A = ∠ B = (π)/(3) ∴ ∠ C = π - ∠ A - ∠ B = π - (2 π)/(3) = (π)/(3) Since Δ ABC is an equilateral Δ ∴ ar (Δ ABC) = 3 4 a ^ 2 = 3 4 c ^ 2 = 3 4 · (6 ⋅ 3 ^ 1 / 4 ) ^ 2 = 3 4 3 ⋅ 6 × 6 = 3 × 3 × 3 = 2 7

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Inverse Trigonometric Functions

NTA Abhyas 2020 · Mathematics · Inverse Trigonometric Functions

In a ABC , if ∠ A = ∠ B = (1)/(2) ( ^ - 1 ( 6 + 1 2 3 ) + ^ - 1 ( 1...

In a

\delta ABC

, if

∠ \text{A} = ∠ \text{B} = \dfrac{1}{2}

\left({\text\sin}^{- 1}

\left(\dfrac{\sqrt{6} + 1}{2 \sqrt{3}}\right) + {\text\sin}^{- 1} \left(\dfrac{1}{\sqrt{3}}\right)\right)

and length of the side opposite to

∠ \text{C}

\text{c} = 6 ⋅ 3 \dfrac{1}{4}

, then the area of

\delta ABC

Correct answer

Step-by-step solution

\text{2}∠ \text{B} = {\text\sin}^{- 1} \left(\dfrac{\sqrt{6} + 1}{2 \sqrt{3}}\right) + {\text\sin}^{- 1} \left(\dfrac{1}{\sqrt{3}}\right)

= {\text\tan}^{- 1} \left(\dfrac{\sqrt{6} + 1}{\sqrt{3} - \sqrt{2}}\right) + {\text\tan}^{- 1} \left(\dfrac{1}{\sqrt{2}}\right)

= {\text\tan}^{- 1} \left(\dfrac{\dfrac{\sqrt{6} + 1}{\sqrt{3} - \sqrt{2}} + \dfrac{1}{\sqrt{2}}}{1 - \dfrac{\sqrt{6} + 1}{\sqrt{3} - \sqrt{2}} \cdot \dfrac{1}{\sqrt{2}}}\right)

= {\text\tan}^{- 1} \left(\dfrac{\left(\sqrt{6} + 1\right) \sqrt{2} + \left(\sqrt{3} - \sqrt{2}\right)}{\sqrt{2} \left(\sqrt{3} - \sqrt{2}\right) - \left(\sqrt{6} + 1\right)}\right)

= {\text\tan}^{- 1} \left(\dfrac{\sqrt{1 2} + \sqrt{2} + \sqrt{3} - \sqrt{2}}{\sqrt{6} - 2 - \sqrt{6} - 1}\right)

= {\text\tan}^{- 1} \left(\dfrac{2 \sqrt{3} + \sqrt{3}}{- 3}\right)

= {\text\tan}^{- 1} \left(\dfrac{3 \sqrt{3}}{- 3}\right)

= {\text\tan}^{- 1} \left(- \sqrt{3}\right)

= \dfrac{2 \pi}{3}

According to question

∠ \text{A} = ∠ \text{B} = \dfrac{1}{2} \left({\text\sin}^{- 1} \left(\dfrac{\sqrt{6} + 1}{2 \sqrt{3}}\right) + {\text\sin}^{- 1} \left(\dfrac{1}{\sqrt{3}}\right)\right)

⇒ ∠ \text{A} = ∠ \text{B} = \dfrac{1}{2} \left(\dfrac{2 \pi}{3}\right)

⇒ ∠ \text{A} = ∠ \text{B} = \dfrac{\pi}{3}

∴ ∠ \text{C} = \pi - ∠ \text{A} - ∠ \text{B}

= \pi - \dfrac{2 \pi}{3}

= \dfrac{\pi}{3}

Since

\Delta \text{ABC}

is an equilateral

\Delta

∴ \text{ar} \left(\Delta \text{ABC}\right) = \dfrac{\sqrt{3}}{4} {\text{a}}^{2}

= \dfrac{\sqrt{3}}{4} {\text{c}}^{2}

= \dfrac{\sqrt{3}}{4} \cdot {\left(6 ⋅ {3}^{1 / 4}\right)}^{2}

= \dfrac{\sqrt{3}}{4} \sqrt{3} ⋅ 6 \times 6

= 3 \times 3 \times 3

= 2 7

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