Can anyone help me out? Steps would be useful but not needed. Its Trig-Missing Sides.

Answer:
10.3
Step-by-step explanation:
Recall SOH, CAH, TOA
In this case we have the hypotonouse is given (the hyptonouse is the side that is opposite from the right angle) and we need to solve for the opposite
therefore we will use SOH
we have
sin(69)=x/11
11*sin(69)=x
x=10.269
which rounds to
10.3
Answer:
[tex] \displaystyle \: x= 10.3[/tex]
Step-by-step explanation:
we are given hypotenuse and [tex]\theta[/tex]
we want to figure out x (opposite)
remember that,
[tex] \displaystyle \: \sin( \theta) = \frac{opp}{hypo} [/tex]
here our [tex]\theta [/tex] is 69°
and hypo is 11 and opp is x
so substitute:
[tex] \displaystyle \: \sin( 69 ^{ \circ} ) = \frac{ x}{11} [/tex]
cross multiplication:
[tex] \displaystyle \: x= 11 \sin( 69 ^{ \circ} ) [/tex]
by using calculator:
[tex] \displaystyle \: x= 10.3[/tex]