Question 176·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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A rectangular poster has corners labeled $A$ (bottom left), $B$ (bottom right), $C$ (top right), and $D$ (top left). The diagonal $\overline{AC}$ is drawn, and point $P$ lies on $\overline{AC}$ such that $PC=20$ inches. If $AB=28$ inches and $BC=21$ inches, which choice is equal to $\tan \angle PBA$ ?

A point partway along a rectangle’s diagonal creates similar triangles: triangles with matching angles and proportional sides. Find the whole diagonal, then use the point’s position to find the vertical and horizontal legs of the smaller right triangle. For tangent, you need opposite over adjacent. The tempting mistake is to use the rectangle’s full width instead of the leg beside the requested angle.

Hints

Hint 1
The rectangle’s diagonal is the hypotenuse, the side across from the right angle. The sides $21$ and $28$ fit a $3$ - $4$ - $5$ right triangle scaled by $7$ . How long is the whole diagonal?
Hint 2
Draw a perpendicular from $P$ to $AB$ , meeting it at $E$ . Triangles $AEP$ and $ABC$ share angle $A$ and each have a right angle, so they are similar. Which part of the diagonal is $AP$ ?
Hint 3
From the angle at $B$ , tangent compares the vertical leg across from the angle with the horizontal leg beside it. That horizontal leg is $BE$ , not the whole base $AB$ .

Step-by-step

Scale the legs from the diagonal
Step 1
Find the whole diagonal
The rectangle makes $\triangle ABC$ a right triangle with legs $AB=28$ and $BC=21$ . These are $7$ times the legs $4$ and $3$ of a $3$ - $4$ - $5$ right triangle, so the hypotenuse, the side across from the right angle, is the whole diagonal: $AC=5(7)=35$ .
Step 2
Find the diagonal segment from A to P
Since $P$ lies between $A$ and $C$ , subtract the given piece $PC=20$ from the whole diagonal: $AP=35-20=15$ . The $20$ -inch piece starts at $C$ , not at $A$ .
Step 3
Match the smaller triangle to the rectangle
Let $E$ be directly below $P$ on $AB$ . Triangles $AEP$ and $ABC$ have right angles at $E$ and $B$ , and they share angle $A$ . Two matching angles make them similar, so matching sides have the same ratio. Height matches height: $\frac{PE}{21}=\frac{AP}{AC}=\frac{15}{35}$ .
Step 4
Calculate P’s height above the base
The proportion gives $PE=21(15/35)$ . Type $21(15/35)$ in Desmos; it shows $9$ . So the vertical leg is $PE=9$ inches.
Step 5
Calculate the horizontal leg from B
The segment $PC$ is $20/35$ of the diagonal, so its horizontal run is $20/35$ of the full $28$ -inch width. Type $28(20/35)$ on the next Desmos line; it shows $16$ . Since $C$ is directly above $B$ , that run equals $BE$ , not $AB$ . $BE=16$ inches.
Step 6
Take tangent from the angle at B
In right triangle $BEP$ , the opposite leg across from $\angle PBA$ is $PE=9$ , and the adjacent leg beside it is $BE=16$ . Use the opposite and adjacent legs for the angle at $B$ . Tangent is opposite over adjacent, so $\tan\angle PBA=\frac{PE}{BE}=\frac{9}{16}$ . The tangent of $\angle PBA$ is $\frac{9}{16}$ . Choice A.

Question 176·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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