If sin A -1-10 and sin B -1-5- where A and B are positive acute angles- then A - B -A- -B- - - 2C- - - 3D- - - 4

The correct option is D We know that sin(A + B) = sinA cosB + cosA sinB = nbsp; 1/√(10)√(1 1/5) + nbsp;1/√(5)√(1 1/10) = nbsp;1/√(10)√(4/5) + nbsp ...

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Byju's Answer

Standard XI

Mathematics

Trigonometric Ratios of Compound Angles

If sin A =1/√...

Question

If sinA = $\frac{1}{\sqrt{10}}$ and sinB = $\frac{1}{\sqrt{5}}$ , where A and B are positive acute angles, then A+B=

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Solution

The correct option is D

We know that sin(A + B) = sinA cosB + cosA sinB

= $\frac{1}{\sqrt{10}}$ $\sqrt{1 - \frac{1}{5}}$ + $\frac{1}{\sqrt{5}}$ $\sqrt{1 - \frac{1}{10}}$

= $\frac{1}{\sqrt{10}}$ $\sqrt{\frac{4}{5}}$ + $\frac{1}{\sqrt{5}}$ $\sqrt{\frac{9}{10}}$ = $\frac{1}{\sqrt{50}}$ (2 + 3) = $\frac{5}{\sqrt{50}}$ = $\frac{1}{\sqrt{2}}$

$\Rightarrow$ sin(A + B) = sin $\frac{π}{4}$

Hence, A + B = $\frac{π}{4}$ .

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Similar questions

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If sinA = 1√10 and sinB = 1√5, where A and B are positive acute angles, then A+B=

The correct option is D We know that sin(A + B) = sinA cosB + cosA sinB = 1√10√1−15 + 1√5√1−110 = 1√10√45 + 1√5√910 = 1√50(2 + 3) = 5√50 = 1√2 ⇒ sin(A + B) = sin π4 Hence, A + B = π4.

If sinA = $\frac{1}{\sqrt{10}}$ and sinB = $\frac{1}{\sqrt{5}}$ , where A and B are positive acute angles, then A+B=