If 0 < A < B < π, sin A + sin B = (3)/(2) and cos A + cos B = 1 2 , then A =

Correct answer: A. 1 5 ^ o (sin A + sin B) ^ 2 + (cos A + cos B) ^ 2 = (3)/(2) + (1)/(2) = (4)/(2) 2 + 2(sin Asin B + cos AcosB) = 2 (B - A) = 0 ⇒ B - A = (π)/(2) ⇒ B = (π)/(2) + A cos A + cos B = 1 2 ⇒ cos A + ((π)/(2) + A) = 1 2 ⇒ cos A - sin A = 1 2 ⇒ cos A ⋅ 1 2 - sin A 1 2 = (1)/(2) (4 5 ^ o - A) = (1)/(2) = 3 0 ^ o ⇒ 4 5 ^ o - A = 3 0 ^ o ⇒ A = 1 5 ^ o

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Trigonometric Equations

NTA Abhyas 2020 · Mathematics · Trigonometric Equations

If 0 < A < B < π, sin A + sin B = (3)/(2) and cos A + cos B = 1 2 ,...

0 < A < B < \pi, sin A + sin B = \sqrt{\dfrac{3}{2}}

and

cos A + cos B = \dfrac{1}{\sqrt{2}},

then

A =

Options

A $1{5}^{o}$
B $3{0}^{o}$
C $4{5}^{o}$
D $22{\dfrac{1}{2}}^{o}$

Correct answer

A. $1{5}^{o}$

Step-by-step solution

{\left(sin A + sin B\right)}^{2} + {\left(cos A + cos B\right)}^{2} = \dfrac{3}{2} + \dfrac{1}{2} = \dfrac{4}{2}

2 + 2\left(sin Asin B + cos AcosB\right) = 2

\cos \left(B - A\right) = 0 ⇒ B - A = \dfrac{\pi}{2} ⇒ B = \dfrac{\pi}{2} + A

cos A + cos B = \dfrac{1}{\sqrt{2}} ⇒ cos A + \cos \left(\dfrac{\pi}{2} + A\right) = \dfrac{1}{\sqrt{2}}

⇒ cos A - sin A = \dfrac{1}{\sqrt{2}} ⇒ cos A ⋅ \dfrac{1}{\sqrt{2}} - sin A\dfrac{1}{\sqrt{2}} = \dfrac{1}{2}

\sin \left(4{5}^{o} - A\right) = \dfrac{1}{2} = \sin 3{0}^{o} ⇒ 4{5}^{o} - A = 3{0}^{o}

⇒ A = 1{5}^{o}

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