frogs29 2 years ago Find the value of x. Round to the nearest tenth. The diagram is not drawn to scale

1. frogs29

2. frogs29

@Opcode

3. frogs29

@hba

4. frogs29

@amistre64

5. amistre64

which trig function would you use with the information given?

6. frogs29

wouldnt you use cosine?

7. amistre64

start out by writing down the 3 basic trig functions for me ...

8. frogs29

well there are sine and cosine i know for sure

9. amistre64

good, and also tangent. how do we define them in terms of the sides of a triangle? using opposite, adjacent, and hypotenuse ...

10. frogs29

tangent and cotangent?

11. frogs29

in not sure to be honest

12. amistre64

then that is what you need to work on. sin = opp/hyp cos = adj/hyp tan = opp/adj given these 3 basic trig functions, which one uses the information given?

13. amistre64

you might wanna use words that you like :) i tend to use over, next.to, and slant becasue they sound less babbley

14. frogs29

So this triangle i would use tangent then?

15. amistre64

correct tan(a) = over/next.to , lets fill in whats given tan(15) = x/12 how do we move this about to get x all alone?

16. frogs29

i have a quetion first. where and how do you get 15?

17. amistre64

i have old eyes :) i read the 12 and 35 together and saw 15

18. frogs29

oh ok lol so it would be tan(35)=x/12?

19. amistre64

yes, then its just a little algebra to solve for x

20. frogs29

so then i would multiply 12 and 35 together?

21. amistre64

you are on the right track, but lets clarify that alittle 12 * tan(35) = x you need a table or calculator for the tan(35) part

22. frogs29

oh ok let me see

23. frogs29

so it would be 8.40249045851651735

24. frogs29

and rounding to the nearest tenth would be 8.4

25. amistre64

correct

26. frogs29

ok so that would be x then?

27. amistre64

lets see. 12*tan(35) = x 8.4 = x yep, im pretty sure that completes it :)

28. frogs29

ok thank you so much would you be able to help me with one more?

29. amistre64

prolly not, ive got a class starting soon

30. frogs29

oh ok well thank you for your help i appreciate it!

31. amistre64

good luck :)