Question

1. Given that sinA = 1/√2and cosB =1/√2 then value of (A + B) is​

Answer 1

Given:-

• SinA = $\sf{\dfrac{1}{\sqrt{2}}}$

• CosB = $\sf{\dfrac{1}{\sqrt{2}}}$

To Find:-

• The value of (A + B)

Solution:-

It is given that SinA = $\sf{\dfrac{1}{\sqrt{2}}}$

From the trigonometric table we know that:-

Sin45° = $\sf{\dfrac{1}{\sqrt{2}}}$

Hence We can write:-

$\sf{SinA = Sin45^\circ}$

=> A = 45°

Also we know that,

Cos45° = $\sf{\dfrac{1}{\sqrt{2}}}$

Hence we can write:-

$\sf{CosB = Cos45^\circ}$

=> B = 45°

Now,

We were asked to find the value of A + B

Putting the respective Values:-

= 45° + 45°

= 90°

Hence,

The value of A + B = 90°

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Important:-

We need to learn the trigonometric table gor solving these kind of sums.

The trigonometric table as follows:-

$Large{ \begin{tabular}{|c|c|c|c|c|c|} \cline{1-6} \theta & \sf 0^{\circ} & \sf 30^{\circ} & \sf 45^{\circ} & \sf 65^{\circ} & \sf 90^{\circ} \\ \cline{1-6} \sin & 0 & \dfrac{1}{2 } & \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} & 1 \\ \cline{1-6} \cos & 1 & \dfrac{ \sqrt{ 3 }}{2} } & \dfrac{1}{ \sqrt{2} } & \dfrac{ 1 }{ 2 } & 0 \\ \cline{1-6} \tan & 0 & \dfrac{1}{ \sqrt{3} } & 1 & \sqrt{3} & \infty \\ \cline{1-6} \cot & \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } &0 \\ \cline{1 - 6} \sec & 1 & \dfrac{2}{ \sqrt{3}} & \sqrt{2} & 2 & \infty \\ \cline{1-6} \csc & \infty & 2 & \sqrt{2 } & \dfrac{ 2 }{ \sqrt{ 3 } } & 1 \\ \cline{1 - 6}\end{tabular}}$

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How Did I solve?

→ Firstly we know that Sin45° = $\sf{\dfrac{1}{\sqrt{2}}}$. So, on comparing SinA with Sin45° we got A = 45°. Similarly, we know that Cos45° = $\sf{\dfrac{1}{\sqrt{2}}}$. So, on comparing CosB with Cos45° we got the value of B as 45°. And as we were asked the value of A + B we added the values and got A + B = 90°

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Answer 2

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