From a point on level ground, the angle of elevation to the top of a vertical tower is . After is moved feet directly toward the base of the tower, the angle of elevation becomes . Approximately how tall, in feet, is the tower?

Solution available with step-by-step explanation

SAT Right Triangles & Trigonometry Question 9 | Free SAT Prep | Aniko

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Question 9·Hard·Right Triangles and Trigonometry

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From a point $P$ on level ground, the angle of elevation to the top of a vertical tower is $35^\circ$ . After $P$ is moved $100$ feet directly toward the base of the tower, the angle of elevation becomes $50^\circ$ . Approximately how tall, in feet, is the tower?

For angle-of-elevation problems on the SAT, always start by sketching a right triangle and labeling the vertical height and horizontal distances clearly. Use the tangent ratio, since angle of elevation from level ground relates the opposite (height) and adjacent (ground distance) sides. When there are two different observation points, set up a tangent equation for each, express both in terms of the same height, and then set them equal to solve for the unknown horizontal distance. Use your calculator in degree mode to evaluate tangent accurately, avoid over-rounding intermediate values, and then plug back in to find the requested height.

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Visualize the geometry

Draw a vertical tower, a point $P$ on the ground some distance away, and a line of sight from $P$ to the top of the tower forming a right triangle. Then draw a second right triangle after moving 100 feet closer.

Pick the correct trig ratio

In each right triangle, you know an angle and want to relate the vertical height of the tower to the horizontal distance on the ground. Which trig function uses opposite and adjacent sides?

Set up and connect the equations

Write one equation using the 35° angle and another using the 50° angle, both involving the same height $h$ but different horizontal distances. How can you use the fact that these two expressions both equal $h$ to solve for the unknown distances?

Solve step by step

After you set the two expressions for $h$ equal, isolate $x$ with algebra, then plug that value back into one of your tangent equations to find the height.

Desmos Guide

Use degree mode

In Desmos, make sure the calculator is set to degree mode so that tan(35) and tan(50) are interpreted as 35 degrees and 50 degrees.

Graph both sides of the equation to find x

Enter one function as f(x) = tan(35)x and another as g(x) = tan(50)(x - 100). Use the intersection tool (or click where the graphs cross) to find the x-value where f(x) = g(x); this x-value is the original horizontal distance from point P to the tower.

Compute the tower's height from x

In a new expression line, type tan(35)*[that x-value from the intersection]. The numerical result that Desmos shows is the approximate height of the tower in feet.

Step-by-step Explanation

Draw and label the right triangles

Visualize a vertical tower of height $h$ and a point $P$ on level ground at a horizontal distance $x$ from the base of the tower.

From $P$ , the angle of elevation to the top of the tower is $35^\circ$ , forming a right triangle with height $h$ and base $x$ .
Then you move 100 feet directly toward the tower, so the new horizontal distance to the base is $x-100$ .
From this closer point, the angle of elevation is $50^\circ$ , forming a second right triangle with the same height $h$ and base $x-100$ .

Write tangent equations for each position

In a right triangle, $\tan(\text{angle}) = \dfrac{\text{opposite}}{\text{adjacent}}$ .

From the original point $P$ (farther away): $\tan 35^\circ = \dfrac{h}{x}$ , so $h = x \tan 35^\circ$ .
From the new point 100 feet closer: $\tan 50^\circ = \dfrac{h}{x-100}$ , so $h = (x-100) \tan 50^\circ$ .

Set the two expressions for the height equal

Both expressions equal the same height $h$ , so set them equal to each other:

x \tan 35^\circ = (x - 100) \tan 50^\circ

Now expand the right-hand side:

x \tan 35^\circ = x \tan 50^\circ - 100 \tan 50^\circ

Solve the equation for the horizontal distance x

Collect the $x$ terms on one side:

x \tan 35^\circ - x \tan 50^\circ = -100 \tan 50^\circ

Factor out $x$ :

x (\tan 35^\circ - \tan 50^\circ) = -100 \tan 50^\circ

Solve for $x$ :

x = \dfrac{-100 \tan 50^\circ}{\tan 35^\circ - \tan 50^\circ} = \dfrac{100 \tan 50^\circ}{\tan 50^\circ - \tan 35^\circ}

Use your calculator in degree mode:

$\tan 35^\circ \approx 0.700$
$\tan 50^\circ \approx 1.192$

x \approx \dfrac{100 \cdot 1.192}{1.192 - 0.700} \approx \dfrac{119.2}{0.492} \approx 242 \text{ feet (approximately)}.

Find the height of the tower and choose the closest answer

Use $h = x \tan 35^\circ$ with $x \approx 242$ :

h \approx 242 \cdot 0.700 \approx 169.4 \text{ feet}.

This is approximately $170$ feet, so the tower is about 170 feet tall. The closest answer choice is C) 170.

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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