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Can someone help me with this!!!!!

Can someone help me with this!!!!!-example-1
User Jrhicks
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1 Answer

5 votes

Answer:

30.
x=5\\y=5√(3)

31.
x=4\\y=4√(3)

32.
x=6\\y=6√(3)

33.
x=4.5\\y=(9√(3) )/(2)

34.
x=40\\y=40√(3)

35.
x=7.5\\y=(15√(3) )/(2)

Explanation:

Use SOHCAHTOA to find one of the missing sides, then use the pythagorean theorem (
x^(2)+ y^(2) =z^(2)) to find the other one


sin=(opposite)/(hypotenuse)


cos=(adjacent)/(hypotenuse)


tan=(opposite)/(adjacent)

30.

First find
x using Sin


sin(30)=(x)/(10)


10*sin(30)=x


x=5

Then use the pythagorean theorem to find
y


5^(2)+ y^(2) =10^(2)


25+y^(2) =100


y^(2) =75


\sqrt{y^(2)} =√(75)


y=5√(3)

32.

First use Sin to find
x


sin(30)=(x)/(8)


8*sin(30)=x


x=4

Then use the pythagorean theorem to find
y


4^(2)+ y^(2) =8^(2)


16+y^(2) =64


y^(2) =48


\sqrt{y^(2) } =√(48)


y=4√(3)

32.

First use Sin to find
x


sin(30)=(x)/(12)


12*sin(30)=x


x=6

Then use the pythagorean theorem to find
y


6^(2)+ y^(2) =12^(2)


36+y^(2) =144


y^(2) =108


\sqrt{y^(2) } =√(108)


y=6√(3)

33.

First use Sin to find
x


sin(30)=(x)/(9)


9*sin(30)=x


x=4.5

Then use the pythagorean theorem to find
y


4.5^(2)+ y^(2) =9^(2)


20.25+y^(2) =81


y^(2) =60.75


\sqrt{y^(2) } =√(60.75)


y=(9√(3) )/(2)

34.

First use Sin to find
x


sin(30)=(x)/(80)


80*sin(30)=x


x=40

Then use the pythagorean theorem to find
y


40^(2)+ y^(2) =80^(2)


1600+y^(2) =6400


y^(2) =4800


\sqrt{y^(2) } =√(4800)


y=40√(3)

35.

First use Sin to find
x


sin(30)=(x)/(15)


15*sin(30)=x


x=7.5

Then use the pythagorean theorem to find
y


7.5^(2)+ y^(2) =15^(2)


56.25+y^(2) =225


y^(2) =168.75


\sqrt{y^(2) }=√(168.75)


y=(15√(3) )/(2)

User Popo
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