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tatyana61 [14]
3 years ago
11

Find the value of a. A. 58 B. 130 C. 86 D. 65

Mathematics
1 answer:
tatiyna3 years ago
5 0

Answer:

C. \ \ \ 86°

Step-by-step explanation:

1. Approach

In order to solve this problem, one must first find a relationship between arc (a) and arc (c). This can be done using the congruent arcs cut congruent segments theorem. After doing so, one can then use the secants interior angle to find the precise measurement of arc (a).

2. Arc (a) and arc (c)

A secant is a line or line segment that intersects a circle in two places. The congruent segments cut congruent arcs theorem states that when two secants are congruent, meaning the part of the secant that is within the circle is congruent to another part of a secant that is within that same circle, the arcs surrounding the congruent secants are congruent. Applying this theorem to the given situation, one can state the following:

a = c

3. Finding the degree measure of arc (a),

The secants interior angle theorem states that when two secants intersect inside of a circle, the measure of any of the angles formed is equal to half of the sum of the arcs surrounding the angles. One can apply this here by stating the following:

86=\frac{a+c}{2}

Substitute,

86=\frac{a+c}{2}

86=\frac{a+a}{2}

Simplify,

86=\frac{a+a}{2}

86=\frac{2a}{2}

86=a

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True or false. Tan^2 x = 1 - cos2x/ 1 + cos 2x
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<u>ANSWER</u>

True

<u>EXPLANATION</u>

The given trigonometric equation is

\tan^{2} (x)  =  \frac{1 -  \cos(2x) }{1 +  \cos(2x) }

Recall the double angle identity:

\cos(2x)  =  \cos^{2} x -   \sin^{2}x

We apply this identity to obtain:

\tan^{2} (x)  =  \frac{1 - (\cos^{2} x -   \sin^{2}x) }{1 +  (\cos^{2} x -   \sin^{2}x) }

We maintain the LHS and simplify the RHS to see whether they are equal.

Expand the parenthesis

\tan^{2} (x)  =  \frac{1 - \cos^{2} x  +  \sin^{2}x }{1 +  \cos^{2} x -   \sin^{2}x}

\implies\tan^{2} (x)  =  \frac{1 - \cos^{2} x  +  \sin^{2}x }{1  -   \sin^{2}x  + \cos^{2} x }

Recall that:

1  -   \sin^{2}x  =  \cos^{2}x

1  -   \cos^{2}x  =  \sin^{2}x

We apply these identities to get:

\implies\tan^{2} (x)  =  \frac{\sin^{2}x +  \sin^{2}x }{\cos^{2} x + \cos^{2} x }

\implies\tan^{2} (x)  =  \frac{2\sin^{2}x }{ 2\cos^{2} x }

\implies\tan^{2} (x)  =  \frac{\sin^{2}x }{ \cos^{2} x }

\implies \tan^{2} (x)  =(  \frac{\sin x }{ \cos x })^{2}

Also

\frac{\sin x }{ \cos x } =  \tan(x)

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\implies \tan^{2} (x)  =\tan^{2} (x)

Therefore the correct answer is True

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