Math / Geometry

Question8. $\begin{array}{l} x= \\ y= \end{array}$ $\qquad$

Studdy Solution

STEP 1

1. We are given a right triangle.
2. The hypotenuse is $20$ .
3. One angle is $60^\circ$ .
4. We need to find the lengths of the sides labeled $x$ and $y$ .

STEP 2

1. Identify the type of triangle and the relationships between the angles and sides.
2. Use trigonometric functions to find the lengths of the sides $x$ and $y$ .

STEP 3

Identify the type of triangle. We have a right triangle with one angle $60^\circ$ . The other non-right angle must be:
$180^\circ = 90^\circ + 60^\circ + ( \text{third angle} )$ $180^\circ - 90^\circ - 60^\circ = ( \text{third angle} )$ $30^\circ = ( \text{third angle} )$
This is a 30-60-90 triangle, which is a special right triangle.

STEP 4

In a 30-60-90 triangle, the sides have a specific ratio: the side opposite the $30^\circ$ angle is $\frac{1}{2}$ of the hypotenuse, the side opposite the $60^\circ$ angle is $\frac{\sqrt{3}}{2}$ of the hypotenuse, and the hypotenuse is the longest side.

STEP 5

Use the trigonometric function sine to find $y$ , the side opposite the $60^\circ$ angle:
$\sin(60^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{y}{20}$ $\sin(60^\circ) = \frac{\sqrt{3}}{2}$
Solving for $y$ :
$\frac{\sqrt{3}}{2} = \frac{y}{20}$ $y = 20 \cdot \frac{\sqrt{3}}{2}$ $y = 10\sqrt{3}$

STEP 6

Use the trigonometric function cosine to find $x$ , the side adjacent to the $60^\circ$ angle:
$\cos(60^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{x}{20}$ $\cos(60^\circ) = \frac{1}{2}$
Solving for $x$ :
$\frac{1}{2} = \frac{x}{20}$ $x = 20 \cdot \frac{1}{2}$ $x = 10$
The lengths of the sides are: $x = 10$ $y = 10\sqrt{3}$

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