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olga55 [171]
3 years ago
14

Solve for x . Leave your answer in the simplest form

Mathematics
1 answer:
Alisiya [41]3 years ago
3 0

Hello!

Here we are given a composite shape, meaning that we can divide this shape up into ways that we are given formulas to solve for legs and such.

I can see here that this can be divided into a triangle and a rectangle using a horizontal line.

The leg lengths of this triangle would be 9, because it is congruent to the bottom side of the rectangle, and 7, because it is the left side minus the right side.

This would also make a right triangle, meaning we could solve for the hypotenuse utilizing the Pythagorean Theorem.

a^2+b^2=c^2

6^2+7^2=c^2

36+49=c^2

c^2=85

c=\sqrt{85}

Which is in simplest radical form.

Hope this helps!

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DedPeter [7]

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!

8 0
3 years ago
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