# Math Help - solve

1. ## solve

I need some help with these.

Solve the following

cos-1(x) = 60° 23' 45"
sin-1 (x) = 60° 23' 45"
tan-1 (x) = 60° 23' 45"

2. Your calculator should have a cos/sin/tan^-1 function. If not you can use google.

Remember, there are sixty seconds in a minute and sixty minutes in a degree, so when entering it into your calculator, you want to do ((23)(60) + 45)/(3600) Find that decimal, put it after the sixty and find the inverse cos/sin/tan from there.

3. Originally Posted by Wildman
I need some help with these.

Solve the following

[1] cos-1(x) = 60° 23' 45"
[2] sin-1 (x) = 60° 23' 45"
[3] tan-1 (x) = 60° 23' 45"
Hi Wildman,

I hope you intended these to be 3 separate exercises.

You are taking the inverse cos, inverse sin, and inverse tan of some ratio to arrive at $60^{\circ}23'45"$

Convert: $60^{\circ}23'45"\Longrightarrow 60+\frac{23}{60}+\frac{45}{3600}=\left(\frac{2899} {48}\right)^{\circ}$

Now just take the Cos, Sin, and Tan of that amount to find the value of x in each case.

[1] $\cos\left(\frac{2899}{48}\right)^{\circ}\approx .4940$

[2] $\sin\left(\frac{2899}{48}\right)^{\circ}\approx .8695$

[3] $\tan\left(\frac{2899}{48}\right)^{\circ}\approx 1.7600$