How to find an angle with cosine - ACT Math

Card 0 of 63

Question

In the above triangle, and . Find $\theta$ .

Answer

With right triangles, we can use SOH CAH TOA to solve for unknown side lengths and angles. For this problem, we are given the adjacent and hypotenuse sides of the triangle with relation to the angle. With this information, we can use the cosine function to find the angle.

$\cos = \frac{adjacent}{hypotenuse}$

$\cos\left ( \theta\right ) = \frac{15}{25} = 0.6$

$\arccos\left ( 0.6\right ) = \theta$

$53.1^{\circ}= \theta$

Compare your answer with the correct one above

Question

For the above triangle, and . Find $\theta$ .

Answer

With right triangles, we can use SOH CAH TOA to solve for unknown side lengths and angles. For this problem, we are given the adjacent and hypotenuse sides of the triangle with relation to the angle. With this information, we can use the cosine function to find the angle.

$\cos = \frac{adjacent}{hypotenuse}$

$\cos\left ( \theta\right ) = \frac{12}{30} = 0.4$

$\arccos\left ( 0.4\right ) = \theta$

$66.4^{\circ}= \theta$

Compare your answer with the correct one above

Question

For the above triangle, and . Find $\theta$ .

Answer

With right triangles, we can use SOH CAH TOA to solve for unknown side lengths and angles. For this problem, we are given the adjacent and hypotenuse sides of the triangle with relation to the angle. However, if we plug the given values into the formula for cosine, we get:

$\cos = \frac{adjacent}{hypotenuse}$

$\cos\left ( \theta\right ) = \frac{17}{13} = 1.3$

$\arccos\left ( 1.3\right ) = \theta = \textup{no solution}$

This problem does not have a solution. The sides of a right triangle must be shorter than the hypotenuse. A triangle with a side longer than the hypotenuse cannot exist. Similarly, the domain of the arccos function is $\left [ -1, 1\right ]$ . It is not defined at 1.3.

Compare your answer with the correct one above

Question

A $50\textup{-foot}$ rope is thrown down from a building to the ground and tied up at a distance of $27\textup{ feet}$ from the base of the building. What is the angle measure between the rope and the ground? Round to the nearest hundredth of a degree_._

Answer

You can draw your scenario using the following right triangle:

Recall that the cosine of an angle is equal to the ratio of the adjacent side to the hypotenuse of the triangle. You can solve for the angle by using an inverse cosine function:

$\small \small \Theta = cos^-^1(\frac{27}{50})=57.31636115374206$ or $\small 57.32$ degrees.

Compare your answer with the correct one above

Question

What is the value of $\small \Theta$ in the right triangle above? Round to the nearest hundredth of a degree.

Answer

Recall that the cosine of an angle is equal to the ratio of the adjacent side to the hypotenuse of the triangle. You can solve for the angle by using an inverse cosine function:

$\small \Theta = cos^-^1(\frac{47}{140})=70.38402086459453$ or $70.38$^{\circ}$$ .

Compare your answer with the correct one above

Question

A support beam (buttress) lies against a building under construction. If the beam is feet long and strikes the building at a point feet up the wall, what angle does the beam strike the building at? Round to the nearest degree.

Answer

Our answer lies in inverse functions. If the buttress is feet long and is feet up the ladder at the desired angle, then:

$\textup{cos }x^{\circ} = \frac{24}{25}$

Thus, using inverse functions we can say that $x = \textup{cos}^{-1}\frac{24}{25} \approx 16.26$

Thus, our buttress strikes the buliding at approximately a $16^{\circ}$ angle.

Compare your answer with the correct one above

Question

A stone monument stands as a tourist attraction. A tourist wants to catch the sun at just the right angle to "sit" on top of the pillar. The tourist lies down on the ground meters away from the monument, points the camera at the top of the monument, and the camera's display reads "DISTANCE -- METERS". To the nearest degree, what angle is the sun at relative to the horizon?

Answer

Our answer lies in inverse functions. If the monument is meters away and the camera is meters from the monument's top at the desired angle, then:

$\textup{cos} x^{\circ} = \frac{8}{12}$

Thus, using inverse functions we can say that $x = \textup{cos}^{-1}\frac{8}{12} \approx 48.189$

Thus, our buttress strikes the buliding at approximately a $48.19^{\circ}$ angle.

Compare your answer with the correct one above

Question

In the above triangle, and . Find $\theta$ .

Answer

With right triangles, we can use SOH CAH TOA to solve for unknown side lengths and angles. For this problem, we are given the adjacent and hypotenuse sides of the triangle with relation to the angle. With this information, we can use the cosine function to find the angle.

$\cos = \frac{adjacent}{hypotenuse}$

$\cos\left ( \theta\right ) = \frac{15}{25} = 0.6$

$\arccos\left ( 0.6\right ) = \theta$

$53.1^{\circ}= \theta$

Compare your answer with the correct one above

Question

For the above triangle, and . Find $\theta$ .

Answer

With right triangles, we can use SOH CAH TOA to solve for unknown side lengths and angles. For this problem, we are given the adjacent and hypotenuse sides of the triangle with relation to the angle. With this information, we can use the cosine function to find the angle.

$\cos = \frac{adjacent}{hypotenuse}$

$\cos\left ( \theta\right ) = \frac{12}{30} = 0.4$

$\arccos\left ( 0.4\right ) = \theta$

$66.4^{\circ}= \theta$

Compare your answer with the correct one above

Question

For the above triangle, and . Find $\theta$ .

Answer

With right triangles, we can use SOH CAH TOA to solve for unknown side lengths and angles. For this problem, we are given the adjacent and hypotenuse sides of the triangle with relation to the angle. However, if we plug the given values into the formula for cosine, we get:

$\cos = \frac{adjacent}{hypotenuse}$

$\cos\left ( \theta\right ) = \frac{17}{13} = 1.3$

$\arccos\left ( 1.3\right ) = \theta = \textup{no solution}$

This problem does not have a solution. The sides of a right triangle must be shorter than the hypotenuse. A triangle with a side longer than the hypotenuse cannot exist. Similarly, the domain of the arccos function is $\left [ -1, 1\right ]$ . It is not defined at 1.3.

Compare your answer with the correct one above

Question

A $50\textup{-foot}$ rope is thrown down from a building to the ground and tied up at a distance of $27\textup{ feet}$ from the base of the building. What is the angle measure between the rope and the ground? Round to the nearest hundredth of a degree_._

Answer

You can draw your scenario using the following right triangle:

Recall that the cosine of an angle is equal to the ratio of the adjacent side to the hypotenuse of the triangle. You can solve for the angle by using an inverse cosine function:

$\small \small \Theta = cos^-^1(\frac{27}{50})=57.31636115374206$ or $\small 57.32$ degrees.

Compare your answer with the correct one above

Question

What is the value of $\small \Theta$ in the right triangle above? Round to the nearest hundredth of a degree.

Answer

Recall that the cosine of an angle is equal to the ratio of the adjacent side to the hypotenuse of the triangle. You can solve for the angle by using an inverse cosine function:

$\small \Theta = cos^-^1(\frac{47}{140})=70.38402086459453$ or $70.38$^{\circ}$$ .

Compare your answer with the correct one above

Question

A support beam (buttress) lies against a building under construction. If the beam is feet long and strikes the building at a point feet up the wall, what angle does the beam strike the building at? Round to the nearest degree.

Answer

Our answer lies in inverse functions. If the buttress is feet long and is feet up the ladder at the desired angle, then:

$\textup{cos }x^{\circ} = \frac{24}{25}$

Thus, using inverse functions we can say that $x = \textup{cos}^{-1}\frac{24}{25} \approx 16.26$

Thus, our buttress strikes the buliding at approximately a $16^{\circ}$ angle.

Compare your answer with the correct one above

Question

A stone monument stands as a tourist attraction. A tourist wants to catch the sun at just the right angle to "sit" on top of the pillar. The tourist lies down on the ground meters away from the monument, points the camera at the top of the monument, and the camera's display reads "DISTANCE -- METERS". To the nearest degree, what angle is the sun at relative to the horizon?

Answer

Our answer lies in inverse functions. If the monument is meters away and the camera is meters from the monument's top at the desired angle, then:

$\textup{cos} x^{\circ} = \frac{8}{12}$

Thus, using inverse functions we can say that $x = \textup{cos}^{-1}\frac{8}{12} \approx 48.189$

Thus, our buttress strikes the buliding at approximately a $48.19^{\circ}$ angle.

Compare your answer with the correct one above

Question

In the above triangle, and . Find $\theta$ .

Answer

With right triangles, we can use SOH CAH TOA to solve for unknown side lengths and angles. For this problem, we are given the adjacent and hypotenuse sides of the triangle with relation to the angle. With this information, we can use the cosine function to find the angle.

$\cos = \frac{adjacent}{hypotenuse}$

$\cos\left ( \theta\right ) = \frac{15}{25} = 0.6$

$\arccos\left ( 0.6\right ) = \theta$

$53.1^{\circ}= \theta$

Compare your answer with the correct one above

Question

For the above triangle, and . Find $\theta$ .

Answer

With right triangles, we can use SOH CAH TOA to solve for unknown side lengths and angles. For this problem, we are given the adjacent and hypotenuse sides of the triangle with relation to the angle. With this information, we can use the cosine function to find the angle.

$\cos = \frac{adjacent}{hypotenuse}$

$\cos\left ( \theta\right ) = \frac{12}{30} = 0.4$

$\arccos\left ( 0.4\right ) = \theta$

$66.4^{\circ}= \theta$

Compare your answer with the correct one above

Question

For the above triangle, and . Find $\theta$ .

Answer

With right triangles, we can use SOH CAH TOA to solve for unknown side lengths and angles. For this problem, we are given the adjacent and hypotenuse sides of the triangle with relation to the angle. However, if we plug the given values into the formula for cosine, we get:

$\cos = \frac{adjacent}{hypotenuse}$

$\cos\left ( \theta\right ) = \frac{17}{13} = 1.3$

$\arccos\left ( 1.3\right ) = \theta = \textup{no solution}$

This problem does not have a solution. The sides of a right triangle must be shorter than the hypotenuse. A triangle with a side longer than the hypotenuse cannot exist. Similarly, the domain of the arccos function is $\left [ -1, 1\right ]$ . It is not defined at 1.3.

Compare your answer with the correct one above

Question

A $50\textup{-foot}$ rope is thrown down from a building to the ground and tied up at a distance of $27\textup{ feet}$ from the base of the building. What is the angle measure between the rope and the ground? Round to the nearest hundredth of a degree_._

Answer

You can draw your scenario using the following right triangle:

Recall that the cosine of an angle is equal to the ratio of the adjacent side to the hypotenuse of the triangle. You can solve for the angle by using an inverse cosine function:

$\small \small \Theta = cos^-^1(\frac{27}{50})=57.31636115374206$ or $\small 57.32$ degrees.

Compare your answer with the correct one above

Question

What is the value of $\small \Theta$ in the right triangle above? Round to the nearest hundredth of a degree.

Answer

Recall that the cosine of an angle is equal to the ratio of the adjacent side to the hypotenuse of the triangle. You can solve for the angle by using an inverse cosine function:

$\small \Theta = cos^-^1(\frac{47}{140})=70.38402086459453$ or $70.38$^{\circ}$$ .

Compare your answer with the correct one above

Question

A support beam (buttress) lies against a building under construction. If the beam is feet long and strikes the building at a point feet up the wall, what angle does the beam strike the building at? Round to the nearest degree.

Answer

Our answer lies in inverse functions. If the buttress is feet long and is feet up the ladder at the desired angle, then:

$\textup{cos }x^{\circ} = \frac{24}{25}$

Thus, using inverse functions we can say that $x = \textup{cos}^{-1}\frac{24}{25} \approx 16.26$

Thus, our buttress strikes the buliding at approximately a $16^{\circ}$ angle.

Compare your answer with the correct one above

Tap the card to reveal the answer