Question

In △JKL, solve for x.
66.73
74.89
15.44
38.16

Answer 1

Do you know soh cah toa?

Answer 2

no

Answer 3

\[\sin(\theta) = \frac{opposite}{hypotenuse} ~~~~~ \cos(\theta)= \frac {adjacent}{hypotenuse} ~~~~~ \tan(\theta) = \frac {opposite}{adjacent} ~ the~adjacent~and~opposite~angles~are~coming~from~the~reference~\angle~while~the~hypotenuse~is~just~the~regular~hypotenuse\]

Answer 4

oops o-O

Answer 5

ima cracker

Answer 6

So look from angle J, the 27, which of the 2 do you have out of adjacent, opposite and hypotenuse

Answer 7

34, and x

Answer 8

but like what are they adjacent opposite hypotenuse...?

Answer 9

the sine one

Answer 10

Yes so can you write out the equation using sin(theta) = opposite/hypotenuse ?

Answer 11

<X=sinx= opposite over hypotenuse = x over 34

Answer 12

i think thats right, im not sure

Answer 13

\[\sin27 = \frac {34}{x}\]

Answer 14

so i was right ?

Answer 15

Well, now find x

Answer 16

how would i find it ?

Answer 17

thank yuuu

Answer 18

yw~