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Makovka662 [10]
3 years ago
13

Find the tangent ratio of angle Θ. Hint: Use the slash symbol ( / ) to represent the fraction bar, and enter the fraction with n

o spaces. (4 points)

Mathematics
2 answers:
Setler79 [48]3 years ago
6 0

Answer:

tan theta = 12/5

Step-by-step explanation:

Assuming this is a right triangle

tan theta = opp/ adjacent

tan theta = 12/5

Nadya [2.5K]3 years ago
3 0

Answer:

tanθ=12/5

Step-by-step explanation:

The tangent of an angle in a right triangle is the ratio of the opposite side to the adjacent side.

tanθ= opposite/adajcent

The side opposite of the angle θ is 12.

tanθ= 12/adjacent

The side next to, or adjacent, to angleθ is 5.

tanθ=12/5

The tangent ratio of angle θ is 12/5.

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In the diagram above, angle 1 = 2x+20 and<br>angle 2 = 3x+10. Find the measure of angle 1. ​
anzhelika [568]

Answer:

angle 1 is equal to 40°

Step-by-step explanation:

angles 1 and 2 are congruent due to alternate interior angles of parallel lines. This means that 2x+20=3x+10, solving this leaves us with x=10. Since angle 1 = 2x +20 and x=10, solve to get 2(10)+20 =20+20 = 40

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3 years ago
Rationalize <br><img src="https://tex.z-dn.net/?f=1%20%5Cdiv%20%20%5Csqrt%7B%7D3%20-%20%20%5Csqrt%7B2%7D%20" id="TexFormula1" ti
Marat540 [252]

Answer:

1 \div \sqrt{}3 - \sqrt{2}

\frac{1 -  \sqrt{2} \sqrt{3}  }{ \sqrt{3} }

\frac{(1 -  \sqrt{6}) \sqrt{3}  }{ \sqrt{3} \sqrt{3}  }

\frac{ \sqrt{3}  -  \sqrt{6 \times 3} }{3}

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8 0
3 years ago
What is the exact value of sin(theta + beta)?
NISA [10]

Answer:  \frac{-3\sqrt{13}+4\sqrt{3}}{20}

This is the single fraction of -3*sqrt(13)+4*sqrt(3) up top all over 20.

sqrt = square root

=======================================================

Explanation:

Angle theta is between pi and 3pi/2. This places the angle in quadrant Q3 where both cosine and sine are negative

Use the pythagorean trig identity to get the following:

\sin^2 \theta + \cos^2 \theta = 1\\\\\sin^2 \theta + \left(-\frac{\sqrt{3}}{4}\right)^2 = 1\\\\\sin^2 \theta + \frac{3}{16} = 1\\\\\sin^2 \theta = 1 - \frac{3}{16}\\\\\sin^2 \theta = \frac{16}{16} - \frac{3}{16}\\\\\sin^2 \theta = \frac{16-3}{16}\\\\\sin^2 \theta = \frac{13}{16}\\\\\sin \theta = -\sqrt{\frac{13}{16}} \ \text{ ... sine is negative in Q3}\\\\\sin \theta = -\frac{\sqrt{13}}{\sqrt{16}}\\\\\sin \theta = -\frac{\sqrt{13}}{4}\\\\

Angle beta is in Q1 where sine and cosine are positive.

Draw a right triangle with legs 3 and 4. The hypotenuse is 5 through the pythagorean theorem. In other words, we have a 3-4-5 right triangle.

Since \tan \beta = \frac{3}{4}, this means \sin \beta = \frac{3}{5} \ \text{ and } \ \cos \beta = \frac{4}{5}

Use these ideas:

  • sin = opposite/hypotenuse
  • cos = adjacent/hypotenuse
  • tan = opposite/adjacent

In this case we have: opposite = 3, adjacent = 4, hypotenuse = 5.

-------------------------------------

To recap:

\cos \theta = -\frac{\sqrt{3}}{4}\\\\\sin \theta = -\frac{\sqrt{13}}{4}\\\\\cos \beta = \frac{3}{5}\\\\\sin \beta = \frac{4}{5}\\\\

They lead to this

\sin\left(\theta + \beta\right) = \sin \theta * \cos \beta - \cos \theta * \sin \beta\\\\\sin\left(\theta + \beta\right) = -\frac{\sqrt{13}}{4} * \frac{3}{5} - \left(-\frac{\sqrt{3}}{4}\right) * \frac{4}{5}\\\\\sin\left(\theta + \beta\right) = -\frac{3\sqrt{13}}{20}+\frac{4\sqrt{3}}{20}\\\\\sin\left(\theta + \beta\right) = \frac{-3\sqrt{13}+4\sqrt{3}}{20}\\\\

6 0
2 years ago
Look at pic to answer ! 20 points given
luda_lava [24]

Answer: do you mean 10 points smh

But the answer is A

Step-by-step explanation:

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