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3 years ago

Geometry, Triangle in a circle. Find X show work and explain please

Mathematics

1 answer:

nikklg [1K]3 years ago

7 0

Answer: $x=7$

Step-by-step explanation:

By the Intersecting Secants Theorem, we know that:

$(5)(x+5)=(6)(6+4)$

Having this, we can find the value of "x" by solving for "x":

Applying Distributive property:

$(5)(x+5)=(6)(24)\\5x+25=60$

Subtract 25 from both sides of the equation:

$5x+25-25=60-25\\5x=35$

And finally dviding both sides of the equation by 5, we get:

$\frac{5x}{5}=\frac{35}{5}\\x=7$

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tiny-mole [99]

Answer:

30°, 60°

Step-by-step explanation:

Let x represent the measure of the smaller angle in degrees. Then the larger unknown angle is 2x, and the sum of the angles in the triangle is 180°:

x + 2x + 90° = 180° . . . the sum of the angles in the triangle

3x = 90° . . . . . . . . . . . . subtract 90°, collect terms

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In mathematics, an equation is a statement that asserts the equality of two expressions, which are connected by the equals sign "=".

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3 years ago

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Natali [406]

Answer: 24

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3 years ago

Please Help! will give brainlest!

zzz [600]

Answer:

$\theta = \frac{\pi}{3} , \pi, \frac{5\pi}{ 3}$

Step-by-step explanation:

we want to solve the following trigonometric equation:

$\displaystyle - 2 { \sin}^{2} (\theta )+ \cos( \theta) + 1 = 0, \: \theta \in [0,2\pi)$

The first step of solving trigonometric equation is to rewrite the equation in terms of one trigonometric function . With Pythagorean theorem, we know that sin²x=1-cos²x . It will be helpful to rewrite the equation in terms of one trig functions. Therefore, substitute 1-cos² $\theta$ in the place of sin² $\theta$ :

$\displaystyle - 2 (1 - \cos ^{2} ) + \cos( \theta) + 1 = 0, \: \theta \in [0,2\pi)$

simplify:

$\displaystyle\implies - 2 (1 - \cos ^{2} (\theta) + \cos( \theta) + 1 = 0, \: \theta \in [0,2\pi)\\\implies- 2 + \cos ^{2} (\theta)+ \cos( \theta) + 1 = 0, \: \theta \in [0,2\pi)\\\implies\cos ^{2} (\theta)+ \cos( \theta) -1 = 0$

Consider cos² $\theta\implies$ x. Thus,

$2 {x}^{2} + x - 1 = 0$

solving the quadratic equation yields:

${x}^{} = \frac{1}{2} \\ x = - 1$

back-substitute:

$\begin{cases} \cos( \theta) = \dfrac{1}{2} \\ \cos( \theta) = - 1 \end{cases} \theta \in[0,2\pi)$

take inverse trig in both sides

$\implies \begin{cases} \theta = \dfrac{\pi}{3} + 2n\pi\\\theta= \frac{5\pi}{3} + 2n\pi \\ \theta = \pi + 2n\pi\end{cases} \theta \in[0,2\pi) \\\\\implies\theta= \frac{\pi}{3}+\dfrac{2n\pi}{3},\theta\in [0,2\pi)\\\text{when n=0}\\ \implies \theta = \frac{\pi}{3} \\ \text{when n=1}\\ \theta= \pi\\ \text{when n=2}\\\theta=\frac{5\pi}{ 3}$

In conclusion,

$\displaystyle\theta = \frac{\pi}{3} , \pi, \frac{5\pi}{ 3}$

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2 years ago