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bija089 [108]
3 years ago
10

Allison claims that the ABQ is similar to RPQ, given that AB and PR are parallel.

Mathematics
1 answer:
Marta_Voda [28]3 years ago
7 0
Let's see the important definitions you find in your options:- alternate interior angles are found when two parallel lines are crossed by a transversal, and they are inside the two lines on opposite sides of the transversal;- vertically opposite angles are found when two lines cross and they are the ones facing each other;- corresponding angles are found when two parallel lines are crossed by a transversal, and they are on matching corners.
Looking at the picture we can say that:a) 1 and 2 are vertically opposite angles;b) ABQ and QPR are alternate interior angles;c) BAQ and QRP are alternate interior angles.
Hence, Allison's correct claims are:1 = 2 because they are vertically opposite angles. BAQ = QRP because they are alternate interior angles. 
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Given:

Right triangle

To find:

The six trigonometric functions of θ

Solution:

Hypotenuse = 18

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Opposite side to θ = ?

Using Pythagoras theorem:

\text {Hypotenuse}^2 = \text{adjacent}^2+\text{opposite}^2

18^2 =10^2+\text{opposite}^2

324=100+\text{opposite}^2

Subtract 100 from both sides.

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Taking square root on both sides.

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Using trigonometric ratio formula:

$\sin\theta =\frac{\text{opposite }}{\text{hypotenuse}}

$\sin\theta =\frac{4\sqrt{14} }{18}

$\csc\theta =\frac{\text{hypotenuse}}{\text{opposite }}

$\csc\theta =\frac{18}{4\sqrt{14} }

$\cos \theta=\frac{\text { adjacent side }}{\text { hypotenuse }}

$\cos \theta=\frac{10}{18}

$\sec \theta=\frac{\text { hypotenuse }}{\text { adjacent side }}

$\sec \theta=\frac{18 }{10}

$\tan \theta=\frac{\text { opposite side }}{\text { adjacent side }}

$\tan \theta=\frac{4\sqrt{14} }{10}

$\cot \theta=\frac{\text { adjacent side }}{\text { opposite side }}

$\cot \theta=\frac{10}{4\sqrt{14} }

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