Question

can someone help me with this problem?

Answer 1

whats the problem?

Answer 2

I'll draw it up

Answer 3

question says label triangle write ratio

Answer 4

Which right triangle trig ratio relates the adjacent leg to the hypotenuse?

Answer 5

the x is the opposite because its the opposite from the angle. the hypotenuse is the blank side and the adj is the 23?

Answer 6

The hypotenuse is the side opposite the right angle. It is the side you have marked to be x.

The adjacent side is 23.

The angle is 63.

So, which trig function?

Answer 7

Add 63 + 23 and put 86 over X and simplify i think

Answer 8

i hope this helped

Answer 9

ok so i understand it alittle then I dont :/

Answer 10

recall you SOH CAH TOA again


\(\bf {\color{brown}{ sin}}(\theta)=\cfrac{opposite}{hypotenuse}= \cfrac{opposite}{x}\implies x=\cfrac{opposite}{sin(\theta)}
\\ \quad \\

% cosine
{\color{blue}{ cos}}(\theta)=\cfrac{adjacent}{hypotenuse}= \cfrac{adjacent}{x}\implies x=\cfrac{adjacent}{cos(\theta)}
\\ \quad \\

% tangent
tan(\theta)=\cfrac{opposite}{adjacent}\implies tan(\theta)\cdot adjacent=opposite\)

Answer 11

also keep in mind that
the opposite side is always the one "facing off" the angle
and the other leg/side is the adjacent one

Answer 12

but how can I find X?

Answer 13

@jdoe0001

Answer 14

well. notice above, use either the sine or cosine functions
what's the opposite side? well, notice the tangentt function above

recall that you are given the angle, and the adjacent side

Answer 15

Simplify 86 over X and X='s whatever you divided by to get that answer i don't know if i'm correct

Answer 16

ohhhh ok ty

Answer 17

sorry about that

Answer 18

ok so like you see how like you did 23+63=86 I put 86/x but what can I divide? im kind of stuck?

Answer 19

@nalias

Answer 20

hmmm
recall that
adjacent side = 23
and the angle = 63 degrees
thus \(\bf {\color{brown}{ sin}}(\theta)=\cfrac{opposite}{hypotenuse}= \cfrac{opposite}{x}\implies x=\cfrac{opposite}{sin(63^o)}
\\ \quad \\

% cosine
{\color{blue}{ cos}}(\theta)=\cfrac{adjacent}{hypotenuse}= \cfrac{adjacent}{x}\implies x={\color{purple}{\cfrac{23}{cos(63^o)}}}
\\ \quad \\

% tangent
tan(\theta)=\cfrac{opposite}{adjacent}\implies {\color{brown}{ tan(63^o)\cdot 23}}=opposite\)

Answer 21

Thank you !

Answer 22

yw