Question

find the measure of < x using trigonometric ratios

Answer 1

@ganeshie8 IM HORRIBLE AT THESE

Answer 2

we'll handle this the same way... SOH-CAH-TOA
go to the angle corner, and figure out given sides as opposite,adjacent and hyp

Answer 3

Use tangent here. Tanx = opposite/adjacent. What side is opposite angle x?
\[\Large \tan x = \frac{ \text{opposite} }{\text{adjacent} }\]

Answer 4

opposite of <x 8
adjacent to <x 5
hypotenuse unknown

Answer 5

Hypotenuse we don't need, use tanx = opp/adj.

Answer 6

i know i like to list what i know :DD

Answer 7

Okay :)

Answer 8

That's a good tactic, especially for subjects like physics :)

Answer 9

\[\tan(x)=\frac{ 8 }{ 5 }\]
\[(5)*\tan x=\frac{ 8 }{ 5 }*(5)\]

Answer 10

like that ?

Answer 11

nopes ! leave it as it is tanx = 8/5

Answer 12

yes ! just divide the right side, then take inverse tan both sides

Answer 13

i am used to having to to that i suppose

Answer 14

haha i bet you're !

Answer 15

No need for extra work here :P

\[\tan(x)=\frac{ 8 }{ 5 }\] so \[x = \tan^{-1} \frac{ 8 }{ 5 }\]which you need a calculator for.

Answer 16

x\[\approx 58\]

Answer 17

Looks about right!

Answer 18

right right !

Answer 19

gw !!

Answer 20

so when finding angles like ^^ use tan^-1

Answer 21

see, you have

tan x = 8/5

Answer 22

you want to find x, so to find x you have to isolate x

Answer 23

tan^-1 is the "opposite" of tan, so you use it to undo it. Just like a square root is the opposite of squared, a square root undoes the squard.

Answer 24

take tan^-1 both sides :-

tan x = 8/5

tan^-1 tan x = tan^-1 8/5

x = tan^-1 8/5

Answer 25

oh ok that makes sense :D

Answer 26

left side, tan^-1 and tan cancel out

Answer 27

TY soo much @agent0smith && @ganeshie8

Answer 28

np :)

Answer 29

If you have sinx = some number, then you'd use sin^-1(some number) to undo it. Same for cosx = some number. But make sure you isolate the sinx or cosx, so it's like sinx = ... or cosx = ...

ie you can't just do cos^-1 of 10cosx = 5, you'd have to divide by 10 first.