Understand that by similarity, side ratios in right triangles are properties of the angles in the triangle, leading to definitions of trigonometric ratios for acute angles - Geometry

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Question

Assume the following two triangles are similar. Using the fact that their side ratios are $\frac{AB}{DE}= \frac{AC}{DF}$ , what trigonomic function could this represent for angles and ?

Answer

We must begin by manipulating our equation of the side ratios to get fractions that include both of the sides of the same triangles:

$\frac{AB}{DE}= \frac{AC}{DF}$

$\begin{tabular}{lcc} \ $\implies AB = \frac{DE*AC}{DF}$ & $multiply ; both ; sides ; by; DE$ \ $\implies \frac{AB}{AC} = \frac{DE}{DF}$ & $divide ; both ; sides ; by ; AC$ \ \end{tabular}$

If we look at angle , we see that $\frac{AB}{AC}$ is the $\frac{opposite}{hypotenuse}$ . Looking at angle we see that $\frac{DE}{DF}$ is also the $\frac{opposite}{hypotenuse}$ . This is the definition of the sine function. So:

$sin(C) = \frac{AB}{AC} = \frac{DE}{DF} = sin(F)$

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Question

Using the information of the side lengths in the triangle below, use the side ratios to find the .

Answer

The definition of cosine of an acute angle is $cos(angle) = \frac{adjacent}{hypotenuse}$ . So here we must determine what sides are adjacent and which is the hypotenuse. The hypotenuse is the easiest to pick out. This is the side that is directly across from the right angle in a right triangle. Our hypotenuse is . Now we must choose the adjacent side. Adjacent means next to, and since $\bar{AC}$ is our hypotenuse then $\bar{AB}$ must be our adjacent side. So:

$cos(A) = \frac{adjacent}{hypotenuse}$

$\implies cos(A) = \frac{AB}{AC}$

$\implies cos(A) = \frac{3}{4}$

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Question

Assume the two triangles below are similar. Using the fact that their sides ratios are $\frac{BC}{EF} = \frac{AC}{DF}$ , what trigonometric function could this represent for angles and

Answer

We must begin by manipulating our equation of the side ratios to get fractions that include both of the sides of the same triangles:

$\frac{BC}{EF} = \frac{AC}{DF}$

$\implies BC = \frac{EF*AC}{DF}$ multiply both sides by

$\implies \frac{BC}{AC} = \frac{EF}{DF}$ divide both sides by

If we look at angle we see that $\frac{BC}{AC}$ is the $\frac{adjacent}{hypotenuse}$ . Looking at angle we see that is also $\frac{adjacent}{hypotenuse}$ . This is the definition of the cosine function of an angle. So:

$cos(C) = \frac{BC}{AC} = \frac{EF}{DF} = cos(F)$

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Question

Using the sides of a right triangle, what is the definition of $tan(\theta)$ ?

Answer

We can see this to be true by looking at the following triangle.

We are considering $\angle C$ . The opposite side to $\angle C$ is $\bar{AB}$ . $\angle C$ has two adjacent sides, but since $\bar{AC}$ is opposite to the right angle, this is the hypotenuse of the triangle. So the adjacent side to $\angle C$ must be $\bar{BC}$ . We can set up the following equation to check and make sure that $tan(\theta)=\frac{opposite}{adjacent}$ .

$tan(\theta)=\frac{opposite}{adjacent}$

$tan(36.87) = \frac{3}{4}$

This shows that $tan(\theta)=\frac{opposite}{adjacent}$ .

Compare your answer with the correct one above

Question

Assume the two triangles below are similar. Using the fact that their side ratios are $\frac{AB}{DE}=\frac{BC}{EF}$ , what trigonomic function could this represent for angles and .

Answer

We must begin by manipulating our equation of the side ratios to get fractions that include both of the sides of the same triangles:

$\frac{AB}{DE}=\frac{BC}{EF}$

$\implies AB = \frac{DE*BC}{EF}$ multiply both sides by

$\implies \frac{AB}{BC} = \frac{DE}{EF}$ divide both sides by

If we look at angle we see that $\frac{AB}{BC}$ is the $\frac{opposite}{adjacent}$ . Looking at angle we see that $\frac{DE}{EF}$ is also the $\frac{opposite}{adjacent}$ . This is the definition of the tangent of an angle. So:

$tan(C) = \frac{AB}{BC} = \frac{DE}{EF} = tan(F)$

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Question

True or False: By similarity, side ratios in right triangles are properties of the angles in the triangle.

Answer

Similar right triangles are two right triangles that differ in side lengths but have congruent corresponding angles. This means that if you have an angle, $\theta_1$ , in the first triangle and an angle, $\theta_2$ , in the second triangle. So $\theta_1 = \theta_2$ .If we are considering the cosine of these two angles.

$cos(\theta_1) = \cos(\theta_2)$

$\frac{adjacent_1}{hypotenuse_1} = \frac{adjacent_2}{hypotenuse_2}$

Side ratios would also follow from computing the sine and tangent of the angles using their sides as well. This shows that the side ratios are properties of the angles in the triangles.

Compare your answer with the correct one above

Question

Consider the following right triangle. Use trigonometric ratios to solve for sides and .

Answer

We are given angle and we need to find sides and . So first we need to think about what relation these sides are to angle . Side is the hypotenuse of the triangle since it is opposite of the right angle. So side must be the side adjacent to angle . Recall that the trigonometric ratio that corresponds to cosine is $\frac{adjacent}{hypotenuse}$ . We can solve for the missing sides by solving for the following equation.

$cos(C) = \frac{adjacent}{hypotenuse}$

$cos(30) = \frac{BC}{AC}$

$\frac{\sqrt{3}}{2} = \frac{BC}{AC}$

It follows that:

$\sqrt{3} = BC$

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Question

Consider the following right triangle. Use trigonometric ratios to solve for and .

Answer

We are given angle and we need to find sides and . So first we need to think about what relation these sides are to angle . Side is the hypotenuse of the triangle since it is the opposite of the right angle. Side is opposite angle . Recall that the trigonometric ratio that corresponds to sine is $\frac{opposite}{hypotenuse}$ . We can solve for the missing sides by solving for the following equation.

$sin(C) = \frac{opposite}{hypotenuse}$

$sin(30) = \frac{AB}{AC}$

$\frac{1}{2} = \frac{AB}{AC}$

It follows that:

Compare your answer with the correct one above

Question

Find angle using trigonometric ratios.

Answer

We are given the adjacent side to angle and the hypotenuse of the triangle. We can use this to set up the trigonometric ratio $\frac{adjacent}{hypotenuse}$ which we know to be the definition of cosine. We can solve for angle using the following equation.

$cos(A) = \frac{adjacent}{hypotenuse}$

$cos(A) = \frac{\sqrt{2}}{2}$

$A = cos^{-1}(\frac{\sqrt{2}}{2})$

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Question

Using the information from the triangle below, use the side ratios to find .

Answer

The definition of the sine of an angle is $\frac{opposite}{hypotenuse}$ . So here we must determine which side is opposite of angle and which side is the hypotenuse of this triangle. We know that side is the hypotenuse since it is opposite of the right angle. Side is directly opposite of angle . So:

$sin(A) = \frac{opposite}{hypotenuse}$

$sin(A) = \frac{BC}{AC}$

$sin(A) = \frac{6}{10}$

Compare your answer with the correct one above

Question

Assume the following two triangles are similar. Using the fact that their side ratios are $\frac{AB}{DE}= \frac{AC}{DF}$ , what trigonomic function could this represent for angles and ?

Answer

We must begin by manipulating our equation of the side ratios to get fractions that include both of the sides of the same triangles:

$\frac{AB}{DE}= \frac{AC}{DF}$

$\begin{tabular}{lcc} \ $\implies AB = \frac{DE*AC}{DF}$ & $multiply ; both ; sides ; by; DE$ \ $\implies \frac{AB}{AC} = \frac{DE}{DF}$ & $divide ; both ; sides ; by ; AC$ \ \end{tabular}$

If we look at angle , we see that $\frac{AB}{AC}$ is the $\frac{opposite}{hypotenuse}$ . Looking at angle we see that $\frac{DE}{DF}$ is also the $\frac{opposite}{hypotenuse}$ . This is the definition of the sine function. So:

$sin(C) = \frac{AB}{AC} = \frac{DE}{DF} = sin(F)$

Compare your answer with the correct one above

Question

Using the information of the side lengths in the triangle below, use the side ratios to find the .

Answer

The definition of cosine of an acute angle is $cos(angle) = \frac{adjacent}{hypotenuse}$ . So here we must determine what sides are adjacent and which is the hypotenuse. The hypotenuse is the easiest to pick out. This is the side that is directly across from the right angle in a right triangle. Our hypotenuse is . Now we must choose the adjacent side. Adjacent means next to, and since $\bar{AC}$ is our hypotenuse then $\bar{AB}$ must be our adjacent side. So:

$cos(A) = \frac{adjacent}{hypotenuse}$

$\implies cos(A) = \frac{AB}{AC}$

$\implies cos(A) = \frac{3}{4}$

Compare your answer with the correct one above

Question

Assume the two triangles below are similar. Using the fact that their sides ratios are $\frac{BC}{EF} = \frac{AC}{DF}$ , what trigonometric function could this represent for angles and

Answer

We must begin by manipulating our equation of the side ratios to get fractions that include both of the sides of the same triangles:

$\frac{BC}{EF} = \frac{AC}{DF}$

$\implies BC = \frac{EF*AC}{DF}$ multiply both sides by

$\implies \frac{BC}{AC} = \frac{EF}{DF}$ divide both sides by

If we look at angle we see that $\frac{BC}{AC}$ is the $\frac{adjacent}{hypotenuse}$ . Looking at angle we see that is also $\frac{adjacent}{hypotenuse}$ . This is the definition of the cosine function of an angle. So:

$cos(C) = \frac{BC}{AC} = \frac{EF}{DF} = cos(F)$

Compare your answer with the correct one above

Question

Using the sides of a right triangle, what is the definition of $tan(\theta)$ ?

Answer

We can see this to be true by looking at the following triangle.

We are considering $\angle C$ . The opposite side to $\angle C$ is $\bar{AB}$ . $\angle C$ has two adjacent sides, but since $\bar{AC}$ is opposite to the right angle, this is the hypotenuse of the triangle. So the adjacent side to $\angle C$ must be $\bar{BC}$ . We can set up the following equation to check and make sure that $tan(\theta)=\frac{opposite}{adjacent}$ .

$tan(\theta)=\frac{opposite}{adjacent}$

$tan(36.87) = \frac{3}{4}$

This shows that $tan(\theta)=\frac{opposite}{adjacent}$ .

Compare your answer with the correct one above

Question

Assume the two triangles below are similar. Using the fact that their side ratios are $\frac{AB}{DE}=\frac{BC}{EF}$ , what trigonomic function could this represent for angles and .

Answer

We must begin by manipulating our equation of the side ratios to get fractions that include both of the sides of the same triangles:

$\frac{AB}{DE}=\frac{BC}{EF}$

$\implies AB = \frac{DE*BC}{EF}$ multiply both sides by

$\implies \frac{AB}{BC} = \frac{DE}{EF}$ divide both sides by

If we look at angle we see that $\frac{AB}{BC}$ is the $\frac{opposite}{adjacent}$ . Looking at angle we see that $\frac{DE}{EF}$ is also the $\frac{opposite}{adjacent}$ . This is the definition of the tangent of an angle. So:

$tan(C) = \frac{AB}{BC} = \frac{DE}{EF} = tan(F)$

Compare your answer with the correct one above

Question

True or False: By similarity, side ratios in right triangles are properties of the angles in the triangle.

Answer

Similar right triangles are two right triangles that differ in side lengths but have congruent corresponding angles. This means that if you have an angle, $\theta_1$ , in the first triangle and an angle, $\theta_2$ , in the second triangle. So $\theta_1 = \theta_2$ .If we are considering the cosine of these two angles.

$cos(\theta_1) = \cos(\theta_2)$

$\frac{adjacent_1}{hypotenuse_1} = \frac{adjacent_2}{hypotenuse_2}$

Side ratios would also follow from computing the sine and tangent of the angles using their sides as well. This shows that the side ratios are properties of the angles in the triangles.

Compare your answer with the correct one above

Question

Consider the following right triangle. Use trigonometric ratios to solve for sides and .

Answer

We are given angle and we need to find sides and . So first we need to think about what relation these sides are to angle . Side is the hypotenuse of the triangle since it is opposite of the right angle. So side must be the side adjacent to angle . Recall that the trigonometric ratio that corresponds to cosine is $\frac{adjacent}{hypotenuse}$ . We can solve for the missing sides by solving for the following equation.

$cos(C) = \frac{adjacent}{hypotenuse}$

$cos(30) = \frac{BC}{AC}$

$\frac{\sqrt{3}}{2} = \frac{BC}{AC}$

It follows that:

$\sqrt{3} = BC$

Compare your answer with the correct one above

Question

Consider the following right triangle. Use trigonometric ratios to solve for and .

Answer

We are given angle and we need to find sides and . So first we need to think about what relation these sides are to angle . Side is the hypotenuse of the triangle since it is the opposite of the right angle. Side is opposite angle . Recall that the trigonometric ratio that corresponds to sine is $\frac{opposite}{hypotenuse}$ . We can solve for the missing sides by solving for the following equation.

$sin(C) = \frac{opposite}{hypotenuse}$

$sin(30) = \frac{AB}{AC}$

$\frac{1}{2} = \frac{AB}{AC}$

It follows that:

Compare your answer with the correct one above

Question

Find angle using trigonometric ratios.

Answer

We are given the adjacent side to angle and the hypotenuse of the triangle. We can use this to set up the trigonometric ratio $\frac{adjacent}{hypotenuse}$ which we know to be the definition of cosine. We can solve for angle using the following equation.

$cos(A) = \frac{adjacent}{hypotenuse}$

$cos(A) = \frac{\sqrt{2}}{2}$

$A = cos^{-1}(\frac{\sqrt{2}}{2})$

Compare your answer with the correct one above

Question

Using the information from the triangle below, use the side ratios to find .

Answer

The definition of the sine of an angle is $\frac{opposite}{hypotenuse}$ . So here we must determine which side is opposite of angle and which side is the hypotenuse of this triangle. We know that side is the hypotenuse since it is opposite of the right angle. Side is directly opposite of angle . So:

$sin(A) = \frac{opposite}{hypotenuse}$

$sin(A) = \frac{BC}{AC}$

$sin(A) = \frac{6}{10}$

Compare your answer with the correct one above

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