Question

I NEED HELP PLZZZ!!! I AM AN IDIOT!

What is the value of x?


Enter your answer in the box.

x =

Answer 1

4

Explanation:

When explaining my solution I will assume that you know SOH CAH TOA. If you don't leave a comment below and I will explain it

We know the line KL and angle K which is enough to get the rest of the lenghts of this triangle

Using CAH we have

Cos30=x/8sqrt2

solve for x and get

8sqrt2Cos30=x

Put this into your calculator and get that line KJ= 9.7979

With this we can solve for the rest of the trinagle using a^2+b^2=c^2

So we have

9.7979^2+LJ^2=(8sqrt2)^2

Compute what you can and get x by itself to get

LJ^2=32

LJ=sqrt32 which is 5.6568

Now that we know the length of line LJ we can solve the second part of the triange.

Knowing that the hypotonous is equal to 5.6568 and angle j=45 we can use SOH

we have

sin45=x/5.6568

Solve this and get

5.6568sin45=4

x=4

Answer 2

x = 4

Explanation:

Using the sine ratio in right Δ JKL and exact value

sin30° =
`(1)/(2)` , then

sin30° =
`(opposite)/(hypotenuse)` =
`(JL)/(KL)` =
`(JL)/(8√(2) )` =
`(1)/(2)` ( cross- multiply )

2JL = 8
`√(2)` ( divide both sides by 2 )

JL = 4
`√(2)`

---------------------------------------------------------------

Using the sine ratio in Δ JLM and exact value

sin45° =
`(1)/(√(2) )` , then

sin45° =
`(ML)/(JL)` =
`(x)/(4√(2) )` =
`(1)/(√(2) )` ( cross- multiply )

x ×
`√(2)` = 4
`√(2)` ( divide both sides by
`√(2)` )

x = 4