Question

HELP ASAP

Answer 1

8sqr113
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   113

Answer 2

Answer:

Given : The terminal side of $\theta$ passes through the point (8, -7).

As we measure the value of $\cos \theta$ from the triangle that is formed in the quadrant in which the terminal side finish.

From the diagram as shown below, we can see that the terminal side is in the IV quadrant.

Use the triangle ABC:

here, AC = x = 8 units and BC = y = -7

Using Pythagoras theorem:

$AB^2 = AC^2+BC^2$

$AB^2 = 8^2+(-7)^2 = 64 + 49 = 113$

or

$AB = \sqrt{113}$ units.

To find the exact value of $\cos \theta$ in simplified form.

$\cos \theta = \frac{Adjacent side}{Hypotenuse side}$

$\cos \theta = \frac{AC}{AB}$

Substitute the value of AC = 8 units and AB = $\sqrt{113}$ units.

$\cos \theta =\frac{8}{\sqrt{113}}$

or

$\cos \theta = \frac{8}{\sqrt{113} }\times \frac{\sqrt{113} }{\sqrt{113} } =\frac{8\sqrt{113} }{(\sqrt{113})^2 }$

Simplify:

$\cos \theta =\frac{8\sqrt{113} }{113}$

Therefore, the exact value of $\cos \theta$ in simplified form is;

$\frac{8\sqrt{113} }{113}$