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xz_007 [3.2K]
3 years ago
8

Tell whether the angels are complementary or supplementary then find x

Mathematics
1 answer:
NNADVOKAT [17]3 years ago
8 0

Answer:

Supplementary angles

x = 31

Step-by-step explanation:

Since, (3x + 25)° and 2x° are straight line angles. Therefore they are supplementary angles

(3x + 25)° + 2x° =180°

(5x + 25)° = 180°

5x + 25 = 180

5x = 180 - 25

5x = 155

x = 155/5

x = 31

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Answer:

\displaystyle \sin\Big(\frac{x}{2}\Big) = \frac{7\sqrt{58} }{ 58 }

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Step-by-step explanation:

We are given that:

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First, recall that sine is the ratio of the opposite side to the hypotenuse. Therefore, the adjacent side is:

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So, with respect to x, the opposite side is 21, the adjacent side is 20, and the hypotenuse is 29.

Since x is in QIII, sine is negative, cosine is negative, and tangent is also negative.

And if x is in QIII, this means that:

180

So:

\displaystyle 90 < \frac{x}{2} < 135

Thus, x/2 will be in QII, where sine is positive, cosine is negative, and tangent is negative.

1)

Recall that:

\displaystyle \sin\Big(\frac{x}{2}\Big)=\pm\sqrt{\frac{1 - \cos(x)}{2}}

Since x/2 is in QII, this will be positive.

Using the above information, cos(x) is -20/29. Therefore:

\displaystyle \sin\Big(\frac{x}{2}\Big)=\sqrt{\frac{1 +  20/29}{2}

Simplify:

\displaystyle \sin\Big(\frac{x}{2}\Big)=\sqrt{\frac{49/29}{2}}=\sqrt{\frac{49}{58}}=\frac{7}{\sqrt{58}}=\frac{7\sqrt{58}}{58}

2)

Likewise:

\displaystyle  \cos \Big( \frac{x}{2} \Big) =\pm \sqrt{ \frac{1+\cos(x)}{2} }

Since x/2 is in QII, this will be negative.

Using the above information, cos(x) is -20/29. Therefore:

\displaystyle  \cos \Big( \frac{x}{2} \Big) =-\sqrt{ \frac{1- 20/29}{2} }

Simplify:

\displaystyle \cos\Big(\frac{x}{2}\Big)=-\sqrt{\frac{9/29}{2}}=-\sqrt{\frac{9}{58}}=-\frac{3}{\sqrt{58}}=-\frac{3\sqrt{58}}{58}

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\displaystyle \tan\Big(\frac{x}{2}\Big) = \frac{\sin(x/2)}{\cos(x/2)}

Therefore:

\displaystyle \tan\Big(\frac{x}{2}\Big)=\frac{7\sqrt{58}/58}{-3\sqrt{58}/58}=-\frac{7}{3}

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