Question

In △JKL, what is the length of line segment KL?
33radical 3

66

33radical 2

16.5

Answer 1

@NoelGreco can you help please

Answer 2

You have to angles, and a side. Now think of your trig ratios, especially SOH-CAH-TOA.

What should you use to solve for the missing side? @brianjr227

Answer 3

Tan? @genius12

Answer 4

Good. That works if you choose the 30 degrees angle. Then you can do tan(30) and as you know from "TOA", tangent is the ratio of the opposite side over the adjacent. We know that tan(30) = 1/sqrt(3) and the "adjacent" side is what we're missing hence we do:\[\bf \tan(30)=\frac{ 33 }{ KL } \implies KL=\frac{ 33 }{ \tan(30) }\]

Answer 5

you could also do tan(60) which would be KL/33 and then solve for KL from there..

@brianjr227

Answer 6

@genius12 not sure if i did this right but I think it's 33square root 2

Answer 7

@brianjr227 are you sure that it's \(\bf 33 \sqrt{2}\) ?

Answer 8

@genius12 not really :(

Answer 9

Ok. Firstly we need to figure what tan(30) is. You should really know your basic trig ratios like sin(30), sin(45), sin(60), cos(30), tan(30)...The special angles are the ones you should know. Here is a table from which you can learn the trig ratios:

http://library.thinkquest.org/20991/media/alg2_table.gif

It may also help to know where these come from.

Anyway, here we need to know what tan(30) is, which is:\[\bf \tan(30)=\frac{ \sqrt{3} }{ 3 }\]Now we plug this in which gives us:\[\bf KL =\frac{ 33 }{ \frac{ \sqrt{3} }{ 3 } } \implies KL =33 \times \frac{ 3 }{ \sqrt{3} }=33 \times \frac{ \sqrt{3} \cancel{\sqrt{3}} }{ \cancel{\sqrt{3}} }=33\sqrt{3}\]

Answer 10

@brianjr227

Answer 11

alright im going to take a look at the link next but it looks like i made a mistake. but i see it now. Wow I think I understand it better and thanks!!! @genius12

Answer 12

yw

btw fan? @brianjr227 :D

Answer 13

just saying, @brianjr227-nice dp