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zysi [14]
3 years ago
14

The support for a window air-conditioning unit forms a triangle and an exterior angle. What is the measure of the exterior angle

?

Mathematics
2 answers:
Shalnov [3]3 years ago
8 0

Let the exterior angle be y.

We know that the exterior angle is the sum of opposite two interior angles.

Therefore, y = 3x + 90            --- (1)

Also, y and 5x - 6 are linear pair.

Therefore, y + (5x - 6) = 180

Substituting y = 3x + 90 from (1), we get,

                 (3x + 90) + (5x - 6) = 180

                 8x + 84 = 180

                 8x = 180 - 84 = 96

                  x = 12 degrees

Hence, exterior angle y = 3(12) + 90 = 126 degrees.

Brums [2.3K]3 years ago
6 0

In the diagram is shown right triangle with angles of measures 3x^{\circ} and (5x-6)^{\circ}. These two angles are complementary, then

3x^{\circ}+(5x-6)^{\circ}=90^{\circ}.

Solve this equation:

3x+5x-6=90,\\ \\8x=90+6,\\ \\8x=96,\\ \\x=\dfrac{96}{8}=12.

Therefore, 3x^{\circ}=36^{\circ} and (5x-6)^{\circ}=54^{\circ}.

Now find measures of exterior angles:

  • the measure of exterior angle that is adjacent to the angle 36^{\circ} is 180^{\circ}-36^{\circ}=144^{\circ};
  • the measure of exterior angle that is adjacent to the angle 54^{\circ} is 180^{\circ}-54^{\circ}=126^{\circ}.
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Answer:

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Step-by-step explanation:

First, let's determine which quadrant our angle θ lies in.

Remember ASTC, where:

Everything is positive in QI,

Only sine (and cosecant) is positive in QII,

Only tangent (and cotangent) is positive in QIII,

And only cosine (and secant) is positive in QIV.

Since our tangent is negative, and our cosine is positive, this means that our θ <em>must</em> be in QIV.

In QIV, sine is negative, tangent is negative, and cosine is positive.

With that, let's figure out the remaining trig ratios.

We know that:

\tan(\theta)=-1/2

Remember that tangent is the ratio of the opposite side to the adjacent side.

Let's figure out our hypotenuse using the Pythagorean Theorem:

a^2+b^2=c^2

Substitute 1 for a and 2 for b (we can ignore the negative since we're squaring anyways). This yields:

(1)^2+(2)^2=c^2

Square:

1+4=c^2

Add:

c^2=5

Take the square root:

c=\sqrt{5}

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Sine and Cosecant:

Remember that:

\sin(\theta)=opp/hyp

Substitute 1 for the opposite and √5 for the hypotenuse. This yields:

\sin(\theta)=1/\sqrt5

Rationalize:

\sin(\theta)=\sqrt5/5

And since our angle is in QIV, we add a negative:

\sin(\theta)=-\sqrt5/5

Cosecant is simply the reciprocal of sine. So:

\csc(\theta)=-\sqrt5

Cosine and Secant:

Remember that:

\cos(\theta)=adj/hyp

Substitute 2 for the adjacent and √5 for the hypotenuse. This yields:

\cos(\theta)=2/\sqrt5

Rationalize:

\cos(\theta)=2\sqrt5/5

Since our angle is in QIV, cosine stays positive.

Secant is the reciprocal of cosine. So:

\sec(\theta)=\sqrt5/2

Tangent and Cotangent:

We were given that:

\tan(\theta)=-1/2

To find cotangent, flip:

\cot(\theta)=-2

And we're done!

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