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2 years ago

What is the value of x? Show your work in arriving at your answer.

Mathematics

1 answer:

Trava [24]2 years ago

5 0

Answer: x = 18

Step-by-step explanation:

First, since both angles shown are equal to each other, you can immediately set your equation.

5x + 5 = 95

Next, move 5 to the opposite side of the equation to isolate your x value (5x). This step is simply subtracting the hanging 5.

5x + 5 (-5) = 95 (-5)

5x = 90

Finally, you need to divide both sides of the equation by 5 (the 5 next to the x) to *fully* isolate the x value.

5x / 5 = 90 / 5

x = 18

final answer: x = 18

Hopefully this helps with the remainder of your lesson! :)

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In Triangle XYZ, measure of angle X = 49° , XY = 18°, and

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There are two choices for angle Y: $Y \approx 54.987^{\circ}$ for $XZ \approx 15.193$ , $Y \approx 27.008^{\circ}$ for $XZ \approx 8.424$ .

Step-by-step explanation:

There are mistakes in the statement, correct form is now described:

<em>In triangle XYZ, measure of angle X = 49°, XY = 18 and YZ = 14. Find the measure of angle Y:</em>

The line segment XY is opposite to angle Z and the line segment YZ is opposite to angle X. We can determine the length of the line segment XZ by the Law of Cosine:

$YZ^{2} = XZ^{2} + XY^{2} -2\cdot XY\cdot XZ \cdot \cos X$ (1)

If we know that $X = 49^{\circ}$ , $XY = 18$ and $YZ = 14$ , then we have the following second order polynomial:

$14^{2} = XZ^{2} + 18^{2} - 2\cdot (18)\cdot XZ\cdot \cos 49^{\circ}$

$XZ^{2}-23.618\cdot XZ +128 = 0$ (2)

By the Quadratic Formula we have the following result:

$XZ \approx 15.193\,\lor\,XZ \approx 8.424$

There are two possible triangles, we can determine the value of angle Y for each by the Law of Cosine again:

$XZ^{2} = XY^{2} + YZ^{2} - 2\cdot XY \cdot YZ \cdot \cos Y$

$\cos Y = \frac{XY^{2}+YZ^{2}-XZ^{2}}{2\cdot XY\cdot YZ}$

$Y = \cos ^{-1}\left(\frac{XY^{2}+YZ^{2}-XZ^{2}}{2\cdot XY\cdot YZ} \right)$

1) $XZ \approx 15.193$

$Y = \cos^{-1}\left[\frac{18^{2}+14^{2}-15.193^{2}}{2\cdot (18)\cdot (14)} \right]$

$Y \approx 54.987^{\circ}$

2) $XZ \approx 8.424$

$Y = \cos^{-1}\left[\frac{18^{2}+14^{2}-8.424^{2}}{2\cdot (18)\cdot (14)} \right]$

$Y \approx 27.008^{\circ}$

There are two choices for angle Y: $Y \approx 54.987^{\circ}$ for $XZ \approx 15.193$ , $Y \approx 27.008^{\circ}$ for $XZ \approx 8.424$ .

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