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SpyIntel [72]
3 years ago
11

What is the angle supplementary to the angle measuring 165°12′?

Mathematics
2 answers:
kicyunya [14]3 years ago
3 0
<span>Two angles are supplementary if their sum is 180 degrees. One degree is made up of 60 minutes. So the angle supplementary to an angle measuring 165d12m is 180d - 165d12m, which gives us 14d48m. So the answer is C. </span>
masya89 [10]3 years ago
3 0

Answer:

Option C is correct.

14^{\circ}48' is the angle supplementary to the angle measuring 165^{\circ}12'

Step-by-step explanation:

To find the angle supplementary to the angle measuring 165^{\circ}12'

Let A be the angle supplementary to the angle measuring 165^{\circ}12'.

Supplementary Angles states that the two Angles are Supplementary when they add up to 180 degrees.

Use the conversion:

1 degree = 60 minute.

Then, we have the given angle 165^{\circ}12' = 165\frac{12}{60} =165\frac{1}{5} =165.2^{\circ}

Now, by definition of supplementary angle;

\angle A + 165.2^{\circ}= 180^{\circ}

Subtract 165.2 on both sides we get;

\angle A= 180^{\circ} - 165.2^{\circ} =14.8^{\circ} =14^{\circ}48'

Therefore, the angle supplementary to the angle measuring 165^{\circ}12' is, 14^{\circ}48'




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Given the following trigonometric ratio, enumerate the meaning ratio ​
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\sin \theta = \frac{AC}{\sqrt{AC^{2} + BC^{2}}}

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Step-by-step explanation:

From Trigonometry we know the following definitions for each trigonometric ratio:

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\sin \theta = \frac{y}{h} (1)

Cosine

\cos \theta = \frac{x}{h} (2)

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\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{y}{x} (3)

Cotangent

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Where:

x - Adjacent leg.

y - Opposite leg.

h - Hypotenuse.

The length of the hypotenuse is determined by the Pythagorean Theorem:

h = \sqrt{x^{2}+y^{2}}

If y = AC and x = BC, then the trigonometric ratios are presented below:

\sin \theta = \frac{AC}{\sqrt{AC^{2} + BC^{2}}}

\cos \theta = \frac{BC}{\sqrt{AC^{2} + BC^{2}}}

\cot \theta = \frac{BC}{AC}

\sec \theta = \frac{\sqrt{AC^{2}+BC^{2}}}{BC}

\csc \theta = \frac{\sqrt{AC^{2}+BC^{2}}}{AC}

6 0
3 years ago
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