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likoan [24]
3 years ago
8

Can someone please explain how to do this!!!

Mathematics
1 answer:
DedPeter [7]3 years ago
8 0

Hi there!

So we are given that:-

  • tan theta = 7/24 and is on the third Quadrant.

In the third Quadrant or Quadrant III, sine and cosine both are negative, which makes tangent positive.

Since we want to find the value of cos theta. cos must be less than 0 or in negative.

To find cos theta, we can either use the trigonometric identity or Pythagorean Theorem. Here, I will demonstrate two ways to find cos.

<u>U</u><u>s</u><u>i</u><u>n</u><u>g</u><u> </u><u>t</u><u>h</u><u>e</u><u> </u><u>I</u><u>d</u><u>e</u><u>n</u><u>t</u><u>i</u><u>t</u><u>y</u>

\large \displaystyle{ {tan}^{2}  \theta + 1 =  {sec}^{2}  \theta}

Substitute tan theta = 7/24 in.

\large \displaystyle{ {(  \frac{7}{24}) }^{2}  + 1 =  {sec}^{2} \theta }

Evaluate.

\large \displaystyle{ \frac{49}{576}  + 1 =  {sec}^{2}  \theta} \\  \large \displaystyle{ \frac{625}{576}  =  {sec}^{2}  \theta}

Reminder -:

\large \displaystyle{ sec \theta =  \frac{1}{cos \theta} }

Hence,

\large \displaystyle{ \frac{576}{625}  =  {cos}^{2}  \theta} \\  \large \displaystyle{ \sqrt{ \frac{576}{625} }  = cos  \theta} \\  \large \displaystyle{ \frac{24}{25}  = cos \theta}

Because in QIII, cos<0. Hence,

\large \displaystyle \boxed{ \blue{cos \theta =  -  \frac{24}{25} }}

<u>U</u><u>s</u><u>i</u><u>n</u><u>g</u><u> </u><u>t</u><u>h</u><u>e</u><u> </u><u>P</u><u>y</u><u>t</u><u>h</u><u>a</u><u>g</u><u>o</u><u>r</u><u>e</u><u>a</u><u>n</u><u> </u><u>T</u><u>h</u><u>e</u><u>o</u><u>r</u><u>e</u><u>m</u>

\large \displaystyle{ {a}^{2}  +  {b}^{2}  =  {c}^{2} }

Define c as our hypotenuse while a or b can be adjacent or opposite.

Because tan theta = opposite/adjacent. Therefore:-

\large \displaystyle{ {7}^{2}  +  {24}^{2}  =  {c}^{2} } \\  \large \displaystyle{49 + 576 =  {c}^{2} } \\  \large \displaystyle{625 =  {c}^{2} } \\  \large \displaystyle{25 = c}

Thus, the hypotenuse side is 25. Using the cosine ratio:-

\large \displaystyle{cos \theta =  \frac{adjacent}{hypotenuse}}

Therefore:-

\large \displaystyle{cos \theta =  \frac{24}{25} }

Because cos<0 in Q3.

\large \displaystyle \boxed{ \red{cos \theta =  -  \frac{24}{25} }}

Hence, the value of cos theta is -24/25.

Let me know if you have any questions!

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\qquad\qquad\huge\underline{{\sf Answer}}

Here's the solution ~

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HELPPPPPPPPPPP!!!!!!!!! Oh
Luba_88 [7]

Answer:

We conclude that:

3\cdot \:2\begin{pmatrix}-2&3&0\\ -4&2&6\\ 6&-5&6\end{pmatrix}=\begin{pmatrix}-12&18&0\\ -24&12&36\\ 36&-30&36\end{pmatrix}

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Step-by-step explanation:

Given the expression

3\times 2\begin{pmatrix}-2&3&0\\ \:-4&2&6\\ \:6&-5&6\end{pmatrix}

solving

3\times 2\begin{pmatrix}-2&3&0\\ \:-4&2&6\\ \:6&-5&6\end{pmatrix}

Scalar Multiplication: Multiply each of the matrix elements by a scalar

=\begin{pmatrix}3\cdot \:2\left(-2\right)&3\cdot \:2\cdot \:3&3\cdot \:2\cdot \:0\\ 3\cdot \:2\left(-4\right)&3\cdot \:2\cdot \:2&3\cdot \:2\cdot \:6\\ 3\cdot \:2\cdot \:6&3\cdot \:2\left(-5\right)&3\cdot \:2\cdot \:6\end{pmatrix}

Simplify each element

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Therefore, we conclude that:

3\cdot \:2\begin{pmatrix}-2&3&0\\ -4&2&6\\ 6&-5&6\end{pmatrix}=\begin{pmatrix}-12&18&0\\ -24&12&36\\ 36&-30&36\end{pmatrix}

Hence, option B is correct.

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