Question

Find the missing part

Answer 1

I used Sin30.
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Answer 2

Here we have right triangles, which we can solve using trigonometric reasons, bceause we know the acute angles and the hypothenuse of the bigger right triangle.

Let's the sin function

`sin30=(x)/(15)\\ x=15(sin30)\\x=15((1)/(2) )\\x=(15)/(2)`

Then, we use the cosine function

`cos30=(z)/(15)\\ z=15((√(3) )/(2) )`

Now, we use the sin to find
`y` in the middle size right triangle

`sin30=(y)/(15(√(3) )/(2) )\\ y=(1)/(2) * 15(√(3) )/(2)\\ y=15(√(3) )/(4)`

Then, we use cosine in the smaller triangle

`cos60=(a)/((15)/(2) ) \\a=(15)/(2) * (1)/(2)\\ a=(15)/(4)`

Also, we know that

`a+b=15`, and
`a=(15)/(4)`, so

`b=15-a\\b=15-(15)/(4)\\ b=(60-15)/(4)\\ b=(45)/(4)`

Therefore, all the answers are

`x= (15)/(2)\\ y=15(√(3) )/(4)\\ z=15(√(3) )/(2)\\a=(15)/(4)\\ b=(45)/(4)`