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

Can anyone help me out Steps would be useful but not needed Its TrigMissing Sides class=

Respuesta :

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

Nayefx

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]

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