anonymous one year ago I need help, but I think I understand most of it.

1. anonymous

From a boat on the lake, the angle of elevation to the top of a cliff is 18°41'. If the base of the cliff is 991 feet from the boat, how high is the cliff (to the nearest foot)?

2. anonymous

So I already know I need to use tan

3. jim_thompson5910

you should have something like this drawn up |dw:1440632228679:dw|

4. anonymous

tan (h/991)

5. jim_thompson5910

it should be tan(18°41') = h/991

6. jim_thompson5910

7. anonymous

Oh okay

8. anonymous

Does 18'41 mean the same as 18.41?

9. jim_thompson5910

no

10. jim_thompson5910

the 41' means "41 arcminutes". Sometimes the book will drop the "arc" and just say "41 minutes"

11. jim_thompson5910

Rule: 1 degree = 60 arcminutes think of it as 60 minutes in an hour

12. jim_thompson5910

divide both sides of 1 degree = 60 arcminutes by 60 to get (1/60) degree = 1 arcminute then multiply both sides by 41 to get (41/60) degree = 41 arcminutes

13. jdoe0001

hmm actually, my calculator takes the minutes for the trig functions

14. jdoe0001

but you could convert it to decimals, and just use the flat degree

15. anonymous

Give me .6833->

16. jim_thompson5910

so that means 18°41' is equivalent to 18.6833° approximately

17. anonymous

$\tan (18.6833)=\frac{ x }{ 991 }$

18. anonymous

Then I just solve for x

19. jim_thompson5910

correct

20. anonymous

Thank you

21. jim_thompson5910

what value of x do you get

22. anonymous

335ft

23. jim_thompson5910

yep I'm getting 335.113085763609 which rounds to 335

24. anonymous

mhm Thank you for the help

25. jim_thompson5910

you're welcome