How to find the cosine of an angle - ACT Math

Card 0 of 54

Question

What is cos θ?

Answer

cos = adjacent/hypotenuse = $\frac{5}{\sqrt{29}}$

To get the radical out of the denominator, we multiply:

$\frac{\sqrt{29}}{\sqrt{29}}\ast\frac{5}{\sqrt{29}}$

$=\frac{5\sqrt{29}}{29}$

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Question

If $\tan A=\frac{b}{c}$ , where $b>0\ and\ c>0$ and $\frac{\pi }{2}<A<\pi$ ,

then what is $cos\ A$ ?

Answer

In the triangle below, the tangent of $\angle A\ is\ \frac{b}{c}$ , or the opposite side of the angle divided by the adjacent side of the angle. According to the Pythagorean

Theorem, the $hypotenuse^{2}=c^{2}+b^{2}$

Thus the hypotenuse equals $\sqrt{b^{2}+c^{2}}$ .

The cosine of an angle is the adjacent side of the angle divided by the hypotenuse of the triangle, giving us $\frac{c}{\sqrt{c^{2}+b^{2}}}$ .

However, since $tanA$ is $\frac{sinA}{cosA}$ , and when $A$ is between $\frac{\pi }{2}\ and\ \pi$ , $sinA$ is positive while $cosA$ is negative. Thus, $c$ is negative, giving us the final answer of $-\frac{c}{\sqrt{c^{2}+b^{2}}}$ .

Compare your answer with the correct one above

Question

What is the cosine of the angle formed between the -axis and the line passing through with a slope of ? Round to the nearest hundredth.

Answer

You do not even need to compute the line for this question. All that you need to do is note that you can make a little right triangle with a height of and a base of , which you get from the slope of the line. So, to compute the cosine, you will need to find the hypotenuse using the Pythagorean theorem:

$h=\sqrt{2.5^2+1^2}=\sqrt{7.25}$

So, our little triangle looks like:

The cosine will be the adjacent side, , divided by the hypotenuse, $\sqrt{7.25}$ :

$\frac{1}{\sqrt{7.25}}$ , or approximately .

Compare your answer with the correct one above

Question

What is the $\cos$ of an angle with $\sin \left ( \frac{4}{5} \right )$ ?

Answer

Since the sine of the angle is $\frac{4}{5}$ , that means the opposite side of the triangle is and the hypotenuse is . Automatically, you know this is a special 3-4-5 right triangle, and that the missing side is 3. If not, you could also find the 3rd side by doing Pythagorean Theorem. This gives you your answer of $\cos=\frac{3}{5}$

Compare your answer with the correct one above

Question

Right triangle $\Delta ABC$ has sides , and $AC = \sqrt{74}$ . What is the cosine of $\angle A$ ?

Answer

SOHCAHTOA tells us that $\textup{cos} = \frac{\textup{adjacent}}{\textup{hypotenuse}}$ , and we know the hypotenuse is the longest side of the triangle, $\sqrt(74)$ . Our adjacent side will be the other side that has as a vertex, .

Thus, $\textup{cos} (A) = \frac{AB}{AC} = \frac{5}{\sqrt{74}}$ .

Compare your answer with the correct one above

Question

Right triangle $\Delta ABC$ has sides , $BC = 5\sqrt{11}$ and . What is the cosine of $\angle A$ ?

Answer

SOHCAHTOA tells us that $\textup{cos} = \frac{\textup{adjacent}}{\textup{hypotenuse}}$ , and we know the hypotenuse is the longest side of the triangle, . Our adjacent side will be the other side that has as a vertex, .

Thus, $\textup{cos} (A) = \frac{AB}{AC} = \frac{7}{18}$ .

Compare your answer with the correct one above

Question

What is cos θ?

Answer

cos = adjacent/hypotenuse = $\frac{5}{\sqrt{29}}$

To get the radical out of the denominator, we multiply:

$\frac{\sqrt{29}}{\sqrt{29}}\ast\frac{5}{\sqrt{29}}$

$=\frac{5\sqrt{29}}{29}$

Compare your answer with the correct one above

Question

If $\tan A=\frac{b}{c}$ , where $b>0\ and\ c>0$ and $\frac{\pi }{2}<A<\pi$ ,

then what is $cos\ A$ ?

Answer

In the triangle below, the tangent of $\angle A\ is\ \frac{b}{c}$ , or the opposite side of the angle divided by the adjacent side of the angle. According to the Pythagorean

Theorem, the $hypotenuse^{2}=c^{2}+b^{2}$

Thus the hypotenuse equals $\sqrt{b^{2}+c^{2}}$ .

The cosine of an angle is the adjacent side of the angle divided by the hypotenuse of the triangle, giving us $\frac{c}{\sqrt{c^{2}+b^{2}}}$ .

However, since $tanA$ is $\frac{sinA}{cosA}$ , and when $A$ is between $\frac{\pi }{2}\ and\ \pi$ , $sinA$ is positive while $cosA$ is negative. Thus, $c$ is negative, giving us the final answer of $-\frac{c}{\sqrt{c^{2}+b^{2}}}$ .

Compare your answer with the correct one above

Question

What is the cosine of the angle formed between the -axis and the line passing through with a slope of ? Round to the nearest hundredth.

Answer

You do not even need to compute the line for this question. All that you need to do is note that you can make a little right triangle with a height of and a base of , which you get from the slope of the line. So, to compute the cosine, you will need to find the hypotenuse using the Pythagorean theorem:

$h=\sqrt{2.5^2+1^2}=\sqrt{7.25}$

So, our little triangle looks like:

The cosine will be the adjacent side, , divided by the hypotenuse, $\sqrt{7.25}$ :

$\frac{1}{\sqrt{7.25}}$ , or approximately .

Compare your answer with the correct one above

Question

What is the $\cos$ of an angle with $\sin \left ( \frac{4}{5} \right )$ ?

Answer

Since the sine of the angle is $\frac{4}{5}$ , that means the opposite side of the triangle is and the hypotenuse is . Automatically, you know this is a special 3-4-5 right triangle, and that the missing side is 3. If not, you could also find the 3rd side by doing Pythagorean Theorem. This gives you your answer of $\cos=\frac{3}{5}$

Compare your answer with the correct one above

Question

Right triangle $\Delta ABC$ has sides , and $AC = \sqrt{74}$ . What is the cosine of $\angle A$ ?

Answer

SOHCAHTOA tells us that $\textup{cos} = \frac{\textup{adjacent}}{\textup{hypotenuse}}$ , and we know the hypotenuse is the longest side of the triangle, $\sqrt(74)$ . Our adjacent side will be the other side that has as a vertex, .

Thus, $\textup{cos} (A) = \frac{AB}{AC} = \frac{5}{\sqrt{74}}$ .

Compare your answer with the correct one above

Question

Right triangle $\Delta ABC$ has sides , $BC = 5\sqrt{11}$ and . What is the cosine of $\angle A$ ?

Answer

SOHCAHTOA tells us that $\textup{cos} = \frac{\textup{adjacent}}{\textup{hypotenuse}}$ , and we know the hypotenuse is the longest side of the triangle, . Our adjacent side will be the other side that has as a vertex, .

Thus, $\textup{cos} (A) = \frac{AB}{AC} = \frac{7}{18}$ .

Compare your answer with the correct one above

Question

What is cos θ?

Answer

cos = adjacent/hypotenuse = $\frac{5}{\sqrt{29}}$

To get the radical out of the denominator, we multiply:

$\frac{\sqrt{29}}{\sqrt{29}}\ast\frac{5}{\sqrt{29}}$

$=\frac{5\sqrt{29}}{29}$

Compare your answer with the correct one above

Question

If $\tan A=\frac{b}{c}$ , where $b>0\ and\ c>0$ and $\frac{\pi }{2}<A<\pi$ ,

then what is $cos\ A$ ?

Answer

In the triangle below, the tangent of $\angle A\ is\ \frac{b}{c}$ , or the opposite side of the angle divided by the adjacent side of the angle. According to the Pythagorean

Theorem, the $hypotenuse^{2}=c^{2}+b^{2}$

Thus the hypotenuse equals $\sqrt{b^{2}+c^{2}}$ .

The cosine of an angle is the adjacent side of the angle divided by the hypotenuse of the triangle, giving us $\frac{c}{\sqrt{c^{2}+b^{2}}}$ .

However, since $tanA$ is $\frac{sinA}{cosA}$ , and when $A$ is between $\frac{\pi }{2}\ and\ \pi$ , $sinA$ is positive while $cosA$ is negative. Thus, $c$ is negative, giving us the final answer of $-\frac{c}{\sqrt{c^{2}+b^{2}}}$ .

Compare your answer with the correct one above

Question

What is the cosine of the angle formed between the -axis and the line passing through with a slope of ? Round to the nearest hundredth.

Answer

You do not even need to compute the line for this question. All that you need to do is note that you can make a little right triangle with a height of and a base of , which you get from the slope of the line. So, to compute the cosine, you will need to find the hypotenuse using the Pythagorean theorem:

$h=\sqrt{2.5^2+1^2}=\sqrt{7.25}$

So, our little triangle looks like:

The cosine will be the adjacent side, , divided by the hypotenuse, $\sqrt{7.25}$ :

$\frac{1}{\sqrt{7.25}}$ , or approximately .

Compare your answer with the correct one above

Question

What is the $\cos$ of an angle with $\sin \left ( \frac{4}{5} \right )$ ?

Answer

Since the sine of the angle is $\frac{4}{5}$ , that means the opposite side of the triangle is and the hypotenuse is . Automatically, you know this is a special 3-4-5 right triangle, and that the missing side is 3. If not, you could also find the 3rd side by doing Pythagorean Theorem. This gives you your answer of $\cos=\frac{3}{5}$

Compare your answer with the correct one above

Question

Right triangle $\Delta ABC$ has sides , and $AC = \sqrt{74}$ . What is the cosine of $\angle A$ ?

Answer

SOHCAHTOA tells us that $\textup{cos} = \frac{\textup{adjacent}}{\textup{hypotenuse}}$ , and we know the hypotenuse is the longest side of the triangle, $\sqrt(74)$ . Our adjacent side will be the other side that has as a vertex, .

Thus, $\textup{cos} (A) = \frac{AB}{AC} = \frac{5}{\sqrt{74}}$ .

Compare your answer with the correct one above

Question

Right triangle $\Delta ABC$ has sides , $BC = 5\sqrt{11}$ and . What is the cosine of $\angle A$ ?

Answer

SOHCAHTOA tells us that $\textup{cos} = \frac{\textup{adjacent}}{\textup{hypotenuse}}$ , and we know the hypotenuse is the longest side of the triangle, . Our adjacent side will be the other side that has as a vertex, .

Thus, $\textup{cos} (A) = \frac{AB}{AC} = \frac{7}{18}$ .

Compare your answer with the correct one above

Question

What is cos θ?

Answer

cos = adjacent/hypotenuse = $\frac{5}{\sqrt{29}}$

To get the radical out of the denominator, we multiply:

$\frac{\sqrt{29}}{\sqrt{29}}\ast\frac{5}{\sqrt{29}}$

$=\frac{5\sqrt{29}}{29}$

Compare your answer with the correct one above

Question

If $\tan A=\frac{b}{c}$ , where $b>0\ and\ c>0$ and $\frac{\pi }{2}<A<\pi$ ,

then what is $cos\ A$ ?

Answer

In the triangle below, the tangent of $\angle A\ is\ \frac{b}{c}$ , or the opposite side of the angle divided by the adjacent side of the angle. According to the Pythagorean

Theorem, the $hypotenuse^{2}=c^{2}+b^{2}$

Thus the hypotenuse equals $\sqrt{b^{2}+c^{2}}$ .

The cosine of an angle is the adjacent side of the angle divided by the hypotenuse of the triangle, giving us $\frac{c}{\sqrt{c^{2}+b^{2}}}$ .

However, since $tanA$ is $\frac{sinA}{cosA}$ , and when $A$ is between $\frac{\pi }{2}\ and\ \pi$ , $sinA$ is positive while $cosA$ is negative. Thus, $c$ is negative, giving us the final answer of $-\frac{c}{\sqrt{c^{2}+b^{2}}}$ .

Compare your answer with the correct one above

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