# Question b66d2

Mar 28, 2017

If we assume c is the unknown length of $s i \mathrm{de} C$

sina = 9/18 = 0.5 = 30deg;sin60 = .866;#

$s i \mathrm{de} C = .866 \cdot 18 = 15.6$in

#### Explanation:

The right triangle given has a hypotenuse of $18$in.

It has an opposite of $9$in.

From these we can find the sine of the angle we will call $a$ which is across the triangle from it, or opposite to it.

There is a famous formula: $\sin a = \frac{o p p}{h y p}$

Here, $\sin a = \frac{9}{18} = 0.5$

An angle with $\sin = 0.5$ is $30 \mathrm{de} g$, which will fit into our triangle because we can see the angle across from our opposite is acute.

Now we know the three angles of the triangle add up to $180 \mathrm{de} g$ so the remaining angle is $180 - 90 - 30 = 60 \mathrm{de} g$

We can find that $\sin 60 \mathrm{de} g = 0.866$

$\sin 60 \mathrm{de} g = 0.866 = \frac{o p p}{h y p} = \frac{C}{h y p} = \frac{C}{18}$

$C = .866 \cdot 18 = 15.6$in

So the unknown $s i \mathrm{de} C$ of the triangle is $0.866 \cdot 18 = 15.6$in