# The value of $\sin ({22^ \circ }30')$ is

A. $\sqrt {\dfrac{{\sqrt 2 + 1}}{{2\sqrt 2 }}} $

B. $\sqrt {\sqrt 2 - 1} $

C. $\sqrt {\dfrac{{\sqrt 2 - 1}}{{2\sqrt 2 }}} $

D. None of these

Answer

Verified

384k+ views

Hint: In this type of question, since the angle is expressed both in degree and minutes, we will make use of the half angle formula of sin and solve it.

Complete step-by-step answer:

We have to find the value of $\sin ({22^ \circ }30')$

We know that 60 minutes = 1 degree

So, 30’=1/2 degree

So, we can express the given equation as sin $22{\dfrac{1}{2}^0}$ =sin${22.5^0}$

Making use of the half angle formula for sin, we can write $\sin \left( {\dfrac{\theta }{2}} \right) = \pm \sqrt {\dfrac{{1 - \cos \theta }}{2}} $

Here, this is nothing but equal to $\dfrac{\theta }{2} = {22.5^0}$ $ \Rightarrow \theta = {45^0}$

Here, we need only the positive value since ${22.5^0}$ lies in the first quadrant where sin is positive

$ \Rightarrow \sin ({22.5^0}) = \sqrt {\dfrac{{1 - \cos {{45}^0}}}{2}} $

$

{\text{ }}\sqrt {\dfrac{{1 - \dfrac{1}{{\sqrt 2 }}}}{2}} \\

= \sqrt {\dfrac{{\sqrt 2 - 1}}{{2\sqrt 2 }}} \\

\Rightarrow \sin ({22^0}33') = \sqrt {\dfrac{{\sqrt 2 - 1}}{{2\sqrt 2 }}} \\

$

So, option C is the correct answer for this question

Note: Whenever solving these types of problems first convert the value of angle given in minutes to degrees and solve it or we can also convert it into radians and solve it further.

Complete step-by-step answer:

We have to find the value of $\sin ({22^ \circ }30')$

We know that 60 minutes = 1 degree

So, 30’=1/2 degree

So, we can express the given equation as sin $22{\dfrac{1}{2}^0}$ =sin${22.5^0}$

Making use of the half angle formula for sin, we can write $\sin \left( {\dfrac{\theta }{2}} \right) = \pm \sqrt {\dfrac{{1 - \cos \theta }}{2}} $

Here, this is nothing but equal to $\dfrac{\theta }{2} = {22.5^0}$ $ \Rightarrow \theta = {45^0}$

Here, we need only the positive value since ${22.5^0}$ lies in the first quadrant where sin is positive

$ \Rightarrow \sin ({22.5^0}) = \sqrt {\dfrac{{1 - \cos {{45}^0}}}{2}} $

$

{\text{ }}\sqrt {\dfrac{{1 - \dfrac{1}{{\sqrt 2 }}}}{2}} \\

= \sqrt {\dfrac{{\sqrt 2 - 1}}{{2\sqrt 2 }}} \\

\Rightarrow \sin ({22^0}33') = \sqrt {\dfrac{{\sqrt 2 - 1}}{{2\sqrt 2 }}} \\

$

So, option C is the correct answer for this question

Note: Whenever solving these types of problems first convert the value of angle given in minutes to degrees and solve it or we can also convert it into radians and solve it further.

Recently Updated Pages

Which of the following would not be a valid reason class 11 biology CBSE

What is meant by monosporic development of female class 11 biology CBSE

Draw labelled diagram of the following i Gram seed class 11 biology CBSE

Explain with the suitable examples the different types class 11 biology CBSE

How is pinnately compound leaf different from palmately class 11 biology CBSE

Match the following Column I Column I A Chlamydomonas class 11 biology CBSE

Trending doubts

Which of the following Chief Justice of India has acted class 10 social science CBSE

Green glands are excretory organs of A Crustaceans class 11 biology CBSE

What if photosynthesis does not occur in plants class 11 biology CBSE

What is 1 divided by 0 class 8 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference Between Plant Cell and Animal Cell

10 slogans on organ donation class 8 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

What is the past tense of read class 10 english CBSE