Question

I need to know the improper fractions answers for:

y=
x=
b=
If I could get an answer that would be awesome.

Answer 1

Part 1)
`x=(15)/(2)\ units`

Part 2)
`z=(15√(3))/(2)\ units`

Part 3)
`y= (15√(3))/(4)\ units`

Part 4)
`b= (45)/(4)\ units`

Explanation:

step 1

Find the value of x

In the large right triangle

`cos(60^o)=(x)/(15)` ----> by CAH (adjacent side divided by the hypotenuse)

Remember that

`cos(60^o)=(1)/(2)`

substitute

`(1)/(2)=(x)/(15)`

solve for x

`x=(15)/(2)\ units` ---> improper fraction

step 2

Find the value of z

In the large right triangle

Applying the Pythagorean Theorem

`15^2=x^2+z^2`

substitute the value of x

`15^2=((15)/(2))^2+z^2`

solve for z

`z^2=15^2-((15)/(2))^2`

`z^2=225-(225)/(4)`

`z^2=(675)/(4)`

`z=(√(675))/(2)\ units`

simplify

`z=(15√(3))/(2)\ units`

step 3

Find the value of y

In the right triangle of the right

`sin(30^o)=(y)/(z)` ---> by SOH (opposite side divided by the hypotenuse)

substitute the given values of y and z

Remember that

`sin(30^o)=(1)/(2)`

so

`(1)/(2)=y:(15√(3))/(2)`

solve for y

`(1)/(2)= (2y)/(15√(3))`

`y= (15√(3))/(4)\ units`

step 4

Find the value of b

In the right triangle of the right

`cos(30^o)=(b)/(z)` ---> by CAH (adjacent side divided by the hypotenuse)

substitute the given values of y and z

Remember that

`cos(30^o)=(√(3))/(2)`

so

`(√(3))/(2)=b:(15√(3))/(2)`

solve for y

`(√(3))/(2)= (2b)/(15√(3))`

`b= (45)/(4)\ units`