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3 years ago

Sv is an angle bisector of rst. If mrsv = (3x+5) and mrst = (8x-14), find x.

Mathematics

1 answer:

MA_775_DIABLO [31]3 years ago

5 0

m∠RSV = 3x + 5, m∠RST = 8x - 14 given

m∠RSV ≅ m∠VST definition of segment bisector

m∠RSV + m∠VST = m∠RST angle addition postulate

(3x + 5) + (3x + 5) = (8x - 14) substitution

6x + 10 = 8x - 14 simplify (added like terms)

10 = 2x - 14 subtraction property of equality

24 = 2x addition property of equality

12 = x division property of equality

Answer: x = 12

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Given tan theta =9, use trigonometric identities to find the exact value of each of the following:_______

Ludmilka [50]

Answer:

$(a)\ \sec^2(\theta) = 82$

$(b)\ \cot(\theta) = \frac{1}{9}$

$(c)\ \cot(\frac{\pi}{2} - \theta) = 9$

$(d)\ \csc^2(\theta) = \frac{82}{81}$

Step-by-step explanation:

Given

$\tan(\theta) = 9$

Required

Solve (a) to (d)

Using tan formula, we have:

$\tan(\theta) = \frac{Opposite}{Adjacent}$

This gives:

$\frac{Opposite}{Adjacent} = 9$

Rewrite as:

$\frac{Opposite}{Adjacent} = \frac{9}{1}$

Using a unit ratio;

$Opposite = 9; Adjacent = 1$

Using Pythagoras theorem, we have:

$Hypotenuse^2 = Opposite^2 + Adjacent^2$

$Hypotenuse^2 = 9^2 + 1^2$

$Hypotenuse^2 = 81 + 1$

$Hypotenuse^2 = 82$

Take square roots of both sides

$Hypotenuse =\sqrt{82}$

So, we have:

$Opposite = 9; Adjacent = 1$

$Hypotenuse =\sqrt{82}$

Solving (a):

$\sec^2(\theta)$

This is calculated as:

$\sec^2(\theta) = (\sec(\theta))^2$

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

Where:

$\cos(\theta) = \frac{Adjacent}{Hypotenuse}$

$\cos(\theta) = \frac{1}{\sqrt{82}}$

So:

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

$\sec^2(\theta) = (\frac{1}{\frac{1}{\sqrt{82}}})^2$

$\sec^2(\theta) = (\sqrt{82})^2$

$\sec^2(\theta) = 82$

Solving (b):

$\cot(\theta)$

This is calculated as:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

Where:

$\tan(\theta) = 9$ ---- given

So:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

$\cot(\theta) = \frac{1}{9}$

Solving (c):

$\cot(\frac{\pi}{2} - \theta)$

In trigonometry:

$\cot(\frac{\pi}{2} - \theta) = \tan(\theta)$

Hence:

$\cot(\frac{\pi}{2} - \theta) = 9$

Solving (d):

$\csc^2(\theta)$

This is calculated as:

$\csc^2(\theta) = (\csc(\theta))^2$

$\csc^2(\theta) = (\frac{1}{\sin(\theta)})^2$

Where:

$\sin(\theta) = \frac{Opposite}{Hypotenuse}$

$\sin(\theta) = \frac{9}{\sqrt{82}}$

So:

$\csc^2(\theta) = (\frac{1}{\frac{9}{\sqrt{82}}})^2$

$\csc^2(\theta) = (\frac{\sqrt{82}}{9})^2$

$\csc^2(\theta) = \frac{82}{81}$

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Answer:

The sample space is:

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(T, W): table height and white

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(B, W): bar height and white

(B, K): bar height and black

(X, N): XL height and brown

(X, W): XL height and white

(W, K): XL height and black

Explanation:

The sample space is the set of all the possible outputs, i.e. the possible different combinations that can be choosen.

Use the letters T, B, and X to represent, respectively, table height, bar height, and XL height,

Use letters N, W, and K to represent, respectively, the colors brown, white and black.

Each combination consists of a height (T, B or X) and a color (N, W, K); thus, your sample space shall have 3 × 3 different combinations. These are:

(T,N): table height and brown

(T, W): table height and white

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(B, N): bar height and brown

(B, W): bar height and white

(B, K): bar height and black

(X, N): XL height and brown

(X, W): XL height and white

(W, K): XL height and black

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Step-by-step explanation:

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