Question 4·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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A vertical antenna $RT$ is mounted on the roof $R$ of a vertical building whose base is $B$ . Point $P$ is on level ground with $B$ . From $P$ , the angle of elevation to $R$ is $\theta$ , and the angle of elevation to $T$ is $\phi$ . It is known that $\tan \theta=\frac{5}{12}$ , $\tan \phi=\frac{3}{4}$ , and $PT=130$ meters. What is the length, in meters, of antenna $RT$ ?

Your answer

Two angles of elevation from one ground point create right triangles with a shared horizontal leg. Use tangent, the ratio of the opposite leg to the adjacent leg, and a familiar right-triangle side pattern to find the heights to the top and to the roof. Then subtract: the height to the top of the antenna includes the building.

Hints

Hint 1
The level ground and vertical building make a right triangle with each sightline. Tangent compares the vertical leg with the horizontal leg. For the angle to $T$ , which leg is opposite and which is adjacent?
Hint 2
The upper triangle’s legs have ratio $3:4$ . In a $3$ - $4$ - $5$ right triangle, the hypotenuse is $5$ parts, so the 130-meter sightline tells you the size of each part.
Hint 3
The two triangles have the same ground distance from $P$ to $B$ . Use it with the lower tangent ratio to find the roof height. The antenna is the part of the total height above that roof.

Step-by-step

Find both heights, then subtract
Step 1
Read the upper tangent ratio
Ground segment $PB$ is horizontal, and $BT$ is vertical, so they form a right triangle. Tangent means opposite leg divided by adjacent leg. For the angle to $T$ , $BT$ is opposite and $PB$ is adjacent, so $\frac{BT}{PB}=\tan\phi=\frac{3}{4}$ .
Step 2
Scale the upper right triangle
Legs in the ratio $3:4$ form a $3$ - $4$ - $5$ right triangle, where the hypotenuse is $5$ parts. Here $PT=130$ is the hypotenuse, so each part is $130/5=26$ . Scale the two legs: $BT=3(26)=78,\qquad PB=4(26)=104$ .
Step 3
Use the shared ground distance
The sightline to the roof forms a smaller right triangle with the same $PB=104$ . Its opposite leg is the building height $BR$ , not the antenna. The given lower tangent ratio becomes: $\frac{BR}{104}=\tan\theta=\frac{5}{12}$ .
Step 4
Isolate the building height
Multiply both sides by $104$ to isolate $BR$ : $BR=\frac{5}{12}\cdot104$ . This is the height to the roof, which you'll subtract from the height to $T$ .
Step 5
Evaluate the roof height
Type $\frac{5}{12}\cdot104$ in Desmos. It shows $43.333\ldots$ ; use the fraction button to see $BR=\frac{130}{3}$ meters.
Step 6
Subtract to get only the antenna
Add $78-\frac{5}{12}\cdot104$ in Desmos. It shows $34.666\ldots$ ; the fraction button gives $\frac{104}{3}$ . Subtract the roof height from the height to $T$ , because $BT=BR+RT$ . The antenna is $\frac{104}{3}$ meters long. Grid in 104/3.

Question 4·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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