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3 years ago

Need help with the questions?

Mathematics

2 answers:

Alekssandra [29.7K]3 years ago

4 0

Hello!

A whole circle is 360°. Since there are 60 minutes in an hour, each minute is 6°. Since we need to get to 32, we multiply by 6.

32(6)=192
The answer to a is 192°

For B we are given the number 210. If we divide by 6, therefore the answer to b is 35 minutes, or about 0.583 hours.

For C both hands will be pointing at the 6, so it will create no angle; the answer to c is 0°.

I hope this helps!

gulaghasi [49]3 years ago

3 0

Alright think of each number as 30 multiplied its number so 2 is 60 because 30 times two is 60 or 12 is 360. A is 72 degrees

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

marissa [1.9K]

Answer:

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:

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

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) $XZ \approx 8.424$

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

$Y \approx 27.008^{\circ}$

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