Question

Convert the following angles in to radian?

1) 40°48'
2) 75°30'


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Answer 1

$\large\underline{\sf{Solution-}}$

Given angle is

$\rm :\longmapsto\:40 \degree \: 48'$

We know,

In sexagesimal system of measuring of angle, in Trigonometry, we have

$\rm\implies \:\boxed{\tt{ 1\degree = 60' \: }}$

So, using this, we have

$\rm \:  =  \: \:\bigg[40 \: \dfrac{48}{60} \bigg]^{\degree }$

$\rm \:  =  \: \:\bigg[40 \: \dfrac{4}{5} \bigg]^{\degree }$

$\rm \:  =  \: \dfrac{204}{5}^{\degree }$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ 1\degree = {\bigg[\dfrac{\pi}{180} \bigg]}^{c} \: }}}$

So, using this, we get

$\rm \:  =  \: \dfrac{204}{5} \times \dfrac{\pi}{180}$

$\rm \:  =  \: \dfrac{51\pi}{225}$

Hence,

$\rm\implies \:\boxed{\tt{ \rm \: 40\degree 48' =  \: \dfrac{51\pi}{225} \: }}$

$\large\underline{\sf{Solution-2}}$

Given angle is

$\rm :\longmapsto\:75\degree 30'$

can be rewritten as

$\rm \:  =  \: \:\bigg[75 \: \dfrac{30}{60} \bigg]^{\degree }$

$\rm \:  =  \: \:\bigg[75 \: \dfrac{1}{2} \bigg]^{\degree }$

$\rm \:  =  \: \dfrac{151}{2}^{\degree }$

$\rm \:  =  \: \dfrac{151}{2} \times \dfrac{\pi}{180}$

$\rm \:  =  \: \dfrac{151\pi}{360}$

Hence,

$\rm\implies \:\boxed{\tt{ \rm \: 75\degree 30' =  \: \dfrac{151\pi}{360} \: }}$

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LEARN MORE

$\boxed{\tt{ 1 \: rt. \: angle = 90\degree = {\bigg[\dfrac{\pi}{2} \bigg]}^{c} = {100}^{g}}}$

$\boxed{\tt{ l \: = \: r \: \times \: \theta \: }}$

One radian is defined as an angle subtended at the centre of the circle by an arc whose length is equal to radius of the circle.

Answer 2

Question:-

Convert the following angles into radian?

1) 40°48'.

2) 75°30'.

Given:-

• 40°48'.

• 75°30'.

To Find:-

• Need to convert the following angles into radian.

Solution:-

1) 40°48'.

= 40°48'

$\sf \large = 40 \degree + ( \frac{48}{60} ) \degree$

$\sf \large = 40 \degree + ( \frac{4}{5} ) \degree$

$\sf \large = ( \frac{204 \degree}{5} \times \frac{\pi}{180} ) {}^{rad}$

$\sf \large = \frac{204}{180 \times 5} \times \pi$

$\sf \large = \frac{51}{45 \times 5} = \frac{51\pi}{225 \: radian}$

________________________________________

2) 75°30'.

= 75°30'

$\sf \large = 75 \degree + ( \frac{30}{60} ) \degree$

$\sf \large = (75 + \frac{1}{2} ) \degree$

$\sf \large (\frac{151}{2} ) \degree$

$\sf \large = ( \frac{151}{2} ) \times ( \frac{\pi}{180} )$

$\sf \large = \frac{151\pi}{360 \: radian}$

Answer:-

$\sf \large{ \boxed{ \sf \red{1) \: 40°48' = \frac{51\pi}{225 \: radian}. }}}$

$\sf \large { \boxed{ \sf \blue{2) \:75°30' = \frac{151\pi}{360 \: radian} . }}}$

Hope you have satisfied. ⚘