Question

Find the missing part

Answer 1

I used Sin30.
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Answer 2

Answer:

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

$sin30=\frac{x}{15}\\ x=15(sin30)\\x=15(\frac{1}{2} )\\x=\frac{15}{2}$

Then, we use the cosine function

$cos30=\frac{z}{15}\\ z=15(\frac{\sqrt{3} }{2} )$

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

$sin30=\frac{y}{15\frac{\sqrt{3} }{2} }\\ y=\frac{1}{2} \times 15\frac{\sqrt{3} }{2}\\ y=15\frac{\sqrt{3} }{4}$

Then, we use cosine in the smaller triangle

$cos60=\frac{a}{\frac{15}{2} } \\a=\frac{15}{2} \times \frac{1}{2}\\ a=\frac{15}{4}$

Also, we know that

$a+b=15$, and $a=\frac{15}{4}$, so

$b=15-a\\b=15-\frac{15}{4}\\ b=\frac{60-15}{4}\\ b=\frac{45}{4}$

Therefore, all the answers are

$x= \frac{15}{2}\\ y=15\frac{\sqrt{3} }{4}\\ z=15\frac{\sqrt{3} }{2}\\a=\frac{15}{4}\\ b=\frac{45}{4}$