Question 40·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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In the figure above, point $B$ is $6$ units above the ground directly above point $D$ , and $\overline{AD}=18$ . What is the height $h$ of the tower?

Two angles of elevation to the same tower give two right triangles, but their vertical legs start at different heights. Use tangent, which compares vertical rise with horizontal distance, to write an equation for each angle. Then graph both equations in Desmos and click their intersection. The key trap is treating the point above the ground as though it were on the ground.

Hints

Hint 1
Because $D$ lies between $C$ and $A$ , $CD$ is $18$ less than $CA$ . If you call $CA$ $x$ , what expression gives the ground distance $CD$ ?
Hint 2
An angle of elevation is measured upward from a horizontal line. Tangent compares vertical rise with horizontal distance. For the angle at $B$ , measure the rise from $B$ , not from the ground.
Hint 3
The two tangent equations describe pairs of possible distances and heights. Their intersection satisfies both angles. If $y$ represents the tower’s height, which coordinate of that point should you compare with the choices?

Step-by-step

Approach 1: Graph the two tangent equations
Step 1
Express the shorter ground distance
Let $x=CA$ , the ground distance from the tower to $A$ , and let $y=h$ , the tower’s height. Since $D$ is between $C$ and $A$ and $AD=18$ , the shorter ground distance is $CD=x-18$ .
Step 2
Measure the rise from B
$B$ is $6$ units above the ground, while $T$ is $y$ units above it. So the vertical rise from $B$ to $T$ is $y-6$ , not $y$ .
Step 3
Write the angle equation at A
From the $30^\circ$ angle at $A$ , the height $y$ is the opposite leg and $x$ is the adjacent leg. Tangent is opposite over adjacent, so $\tan(30^\circ)=\frac{y}{x}$ . Set Desmos to Degrees and type $\tan(30)=\frac{y}{x}$ . It graphs the pairs $(x,y)$ that fit this angle.
Step 4
Write the angle equation at B
The horizontal line through $B$ is parallel to the ground, so the $60^\circ$ angle uses horizontal distance $CD=x-18$ . Its vertical rise is $y-6$ because the angle starts at $B$ . Tangent gives $\tan(60^\circ)=\frac{y-6}{x-18}$ . Add $\tan(60)=\frac{y-6}{x-18}$ in Degrees mode. Desmos graphs a second line crossing the first.
Step 5
Match the height to an exact choice
Click the intersection. Desmos shows about $(21.804,12.588)$ ; the second coordinate is $y=h$ . Type each choice to compare: $9\sqrt3-3\approx12.588$ , while the others give about $15.588$ , $9.588$ , and $7.392$ . So the tower’s height is $9\sqrt3-3$ units. Choice A.

Approach 2: Find the exact height with special-angle ratios
Step 1
Turn the angle at A into a distance
A $30^\circ$ - $60^\circ$ - $90^\circ$ triangle has $\tan(30^\circ)=\frac{1}{\sqrt3}$ . Use the equation from $A$ : $\frac{1}{\sqrt3}=\frac{y}{x}$ . Multiply by $x$ : $\frac{x}{\sqrt3}=y$ . Multiply by $\sqrt3$ : $x=\sqrt3y$ . This makes the longer ground distance exact.
Step 2
Turn the angle at B into a distance
The same special-triangle ratio gives $\tan(60^\circ)=\sqrt3$ . Use the equation from $B$ : $\sqrt3=\frac{y-6}{x-18}$ . Multiply by $x-18$ : $\sqrt3(x-18)=y-6$ . Divide by $\sqrt3$ : $x-18=\frac{y-6}{\sqrt3}$ . This is the ground distance measured from $D$ , not $A$ .
Step 3
Use the 18-unit gap
The two ground distances differ by $AD=18$ , so $CA-CD=18$ . Substitute their expressions in terms of $y$ :
$\sqrt3y-\frac{y-6}{\sqrt3}=18.$
Step 4
Solve for the tower’s height
Multiply by $\sqrt3$ : $3y-(y-6)=18\sqrt3$ . Distribute the minus sign: $3y-y+6=18\sqrt3$ . Combine like terms: $2y+6=18\sqrt3$ . Subtract $6$ : $2y=18\sqrt3-6$ . Divide by $2$ : $y=9\sqrt3-3$ . Because $y=h$ , the tower’s height is $9\sqrt3-3$ units. Choice A.

Question 40·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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