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Bas_tet [7]
2 years ago
12

AM and AN are tangent to circle C. If AM measures 2.5 cm, what is the measure of AN?

Mathematics
1 answer:
Vera_Pavlovna [14]2 years ago
3 0

The measure of AN form the given figure is 2. 5cm. Option D

<h3>How to determine the length</h3>

It is important to note that if from one external point, two tangents are drawn to a circle then they have equal tangent segments.

We have that,

AM ≅ AN

so, if AM  measures 2. 5cm

AN measures 2. 5cm

Thus, the measure of AN form the given figure is 2. 5cm. Option D

Learn more about tangent lines here:

brainly.com/question/9636512

#SPJ1

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========================================================

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Use the trig identity below and plug in the given value to get...

\sin\left(\frac{x}{2}\right) = \pm \sqrt{\frac{1-\cos(x)}{2}}\\\\\sin\left(\frac{2\theta}{2}\right) = \pm \sqrt{\frac{1-\cos(2\theta)}{2}}\\\\\sin(\theta) = \sqrt{\frac{1-\cos(2\theta)}{2}} \ \ \text{since } \ \sin(\theta) > 0\\\\\sin(\theta) = \sqrt{\frac{1-\frac{12}{13}}{2}}\\\\

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Edit:

To find cos(theta), we use the pythagorean identity below

\sin^2(\theta) + \cos^2(\theta) = 1\\\\\cos^2(\theta) = 1-\sin^2(\theta)\\\\\cos(\theta) = \sqrt{1-\sin^2(\theta)}\ \ \text{cosine is positive in Q1}\\\\\cos(\theta) = \sqrt{1-\left(\frac{1}{\sqrt{26}}\right)^2}\\\\\cos(\theta) = \sqrt{1-\frac{1}{26}}\\\\\cos(\theta) = \sqrt{\frac{25}{26}}\\\\\cos(\theta) = \frac{\sqrt{25}}{\sqrt{26}}\\\\\cos(\theta) = \frac{5}{\sqrt{26}}\\\\\cos(\theta) = \frac{5\sqrt{26}}{\sqrt{26}*\sqrt{26}}\\\\\cos(\theta) = \frac{5\sqrt{26}}{26}\\\\

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