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3 years ago

6

The sector of a circle with a 12 in. radius has a central angle measure of 60°. What is the exact area of the sector in terms of

π ?

1 answer:

IRINA_888 [86]3 years ago

4 0

The sector of the circle formed by its two radius of length 12 inches has an area of 24 pi. This is obtained using the formula, Area is equal to the product of pi, radius squared and central angle all over 360.

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Artemon [7]

<h2>The missing measures of the angles are:</h2>

$m\angle 1= 60^{\circ}\\\\m\angle 2= 39^{\circ}\\\\m\angle 3= 21^{\circ}\\\\m\angle 4 = 39^{\circ}\\\\m\angle 5 = 21^{\circ}$

<u>Given</u>:

$m\angle Z = 138^{\circ}$

<em><u>Note the following:</u></em>

An equilateral triangle has all its three angles equal to each other. Each angle = $60^{\circ}$ .

This implies that, since $\triangle WXY$ is equilateral, therefore, $m\angle Y = m\angle XWY = m\angle XYW = 60^{\circ}$

Base angles of an isosceles triangle are congruent to each other.

This implies that, since $\triangle WZY$ is isosceles, therefore, $m\angle 3 = \angle 5$

Applying the above stated, let's find the measure of each angle:

Find $m\angle 1$

$m\angle 1 = 60^{\circ}$ (an angle in an equilateral triangle equals 60 degrees)

Find $m\angle 2$

$m\angle 2 = 60 - m \angle 3$

$m\angle 2 = 60 -\frac{1}{2}(180 - 138)$ $(Note: \frac{1}{2}(180 - 138) = 1 $ base $ angle $ of $ \triangle WZY)$

$m\angle 2 = 60 -21\\\\m\angle 2 = 39^{\circ}$

Find $m\angle 3$ and $m\angle 5$ (base angles of isosceles triangle WZY)

$m\angle 3 = \frac{1}{2}(180 - 138) (1 $ base $ angle $ of $ \triangle WZY)$

$m\angle 3 = \frac{1}{2}(42) \\\\m\angle3 = 21^{\circ}$

$m\angle3 = m\angle 5$ (base angles of isosceles triangle are congruent)

Therefore,

$m\angle5 = 21^{\circ}$

Find $m\angle 4$

$m\angle 4 = 60 - m\angle 5$

Substitute

$m\angle 4 = 60 - 21\\\\m\angle 4 = 39^{\circ}$

The missing measures of the angles are:

$m\angle 1= 60^{\circ}\\\\ m\angle 2= 39^{\circ}\\\\m\angle 3= 21^{\circ}\\\\m\angle 4 = 39^{\circ}\\\\m\angle 5 = 21^{\circ}$

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brainly.com/question/2944195

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