SAT Right Triangles & Trigonometry Question 7 | Free SAT Prep

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Question 7·Easy·Right Triangles and Trigonometry

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Points $A$ , $B$ , and $C$ are shown on the $xy$ -plane. Segment $\overline{AB}$ lies on the positive $x$ -axis, and $\angle ABC$ is a right angle. If $\theta$ is the angle at point $A$ between $\overline{AB}$ and $\overline{AC}$ , what is the value of $\tan\theta$ ?

If $\theta$ is the angle a segment makes with the positive $x$ -axis, compute $\tan\theta$ as slope = rise/run using the coordinates (or, in a right triangle, opposite/adjacent). Reduce the fraction and compare to the options.

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Connect tangent to slope

When an angle $\theta$ is measured from the positive $x$ -axis, $\tan\theta$ equals the slope (rise/run) of the line forming that angle.

Use the coordinates

Compute the rise as $y_C-y_A$ and the run as $x_C-x_A$ .

Simplify the fraction

After you form the rise/run fraction, reduce it to lowest terms.

Desmos Guide

Plot the points

In Desmos, enter the points A=(0,0) and C=(9,12).

Compute the slope

Compute (12-0)/(9-0) (or generally (y_C-y_A)/(x_C-x_A)).

Match to a choice

Simplify the result and select the answer choice that matches that fraction.

Step-by-step Explanation

Relate $\tan\theta$ to rise and run

Since $\theta$ is measured from the positive $x$ -axis at $A$ , $\tan\theta$ equals the slope of $\overline{AC}$ :

\tan\theta = \frac{\text{rise from }A\text{ to }C}{\text{run from }A\text{ to }C}.

Compute the ratio

From the coordinates, the rise from $A(0,0)$ to $C(9,12)$ is $12-0=12$ and the run is $9-0=9$ .

\tan\theta=\frac{12}{9}=\frac{4}{3}.

So, $\tan\theta$ is $\dfrac{4}{3}$ .

Strategy

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation