Question

Without calculator,steps for finding cos inverse(8/208)?

Answer 1

\[\arccos{\frac{8}{208}}=\arccos{\frac{1}{26}}=\frac{\pi}{2}-\arcsin{\frac{1}{26}}\approx \frac{\pi}{2}-\frac{1}{26}\]because for small \(\theta\), \(\sin\theta=\theta\Rightarrow\arcsin\theta=\theta \)

Answer 2

Answer had to come in degrees which is 88 @amilapsn ..
Please proceed again...

Answer 3

please do all in steps.... Till I get 88 degrees

Answer 4

Nice!
@arindameducationusc are you saying you don't know how to convert \(\frac{\pi}{2}-\frac{1}{26}\) from radians to degrees ?

Answer 5

\[\arccos{\frac{8}{208}}\approx\frac{\pi}{2}-\frac{1}{26}=\left(\left(\frac{\pi}{2}-\frac{1}{26}\right)\times \frac{180}{\pi}\right)^o\approx88^o\]

Answer 6

Are you ok @arindameducationusc ?

Answer 7

It should be (pi/2-1/26)*pi/180.... @amilapsn
@ganeshie8

Answer 8

No, I got it....my bad..... silly mistakes do happen! LOL...!!!!
I got it@amilapsn

Answer 9

whichever you're converting to, that must be on top :

radians to degrees : 180/pi

degrees to radians : pi/180

Answer 10

Thank you.
one question @ganeshie8
why did we use pi/2-arcsin(theota)?????

Answer 11

property?
Send me a link about this property?

Answer 12

familiar with this identity \(\arcsin(x)+\arccos(x)=\frac{\pi}{2}\) ?

Answer 13

o yes, got it...! thank you.....

Answer 14

np :) you also need to know the small angle approximation of \(\sin\theta\)

when \(\theta\) is small, we can replace \(\sin\theta \) by \(\theta\)