Question 472·Hard·Math

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In right triangle $ABC$ , $\angle C = 90^{\circ}$ . The altitude from $C$ to the hypotenuse $\overline{AB}$ has length $12$ and divides $\overline{AB}$ into two segments whose lengths differ by $7$ units, with the segment closer to $B$ being longer.

What is $\tan A$ ?

An altitude to the hypotenuse creates two smaller right triangles that are similar. Name the two pieces of the hypotenuse by where they sit, then use the altitude’s square to relate their lengths. A one-variable Desmos regression finds the positive length you need. For tangent, use the small triangle containing the requested angle, and don’t swap the two pieces.

Hints

Hint 1
The altitude ends at a point on the hypotenuse; call it $D$ . In the small right triangle at $A$ , tangent means opposite leg over adjacent leg. Which of $CD$ and $AD$ has each role?
Hint 2
The two small triangles are similar: they have the same three angle sizes. Their matching sides give $CD^2=AD\cdot DB$ . This links the altitude to the two pieces of the hypotenuse.
Hint 3
Let $x$ be the shorter piece, next to $A$ . The piece next to $B$ is $x+7$ , not $x-7$ . Put both pieces into the altitude relationship, then find the positive length.

Step-by-step

Use the altitude and the small right triangle
Step 1
Identify the tangent ratio
Call the point where the altitude meets $AB$ point $D$ . Because $CD$ is perpendicular to $AB$ , triangle $ACD$ is right at $D$ . Use the small right triangle containing the angle you want. From $A$ , $CD$ is opposite and $AD$ is the adjacent leg, so tangent is opposite over adjacent: $\tan A=\frac{CD}{AD}=\frac{12}{AD}$ .
Step 2
Relate the altitude to both pieces
Triangles $ACD$ and $CBD$ are similar, meaning they have the same angle sizes: both are right at $D$ , and $\angle A$ matches $\angle BCD$ because both complement $\angle B$ . Matching sides gives $\frac{CD}{AD}=\frac{DB}{CD}$ . Cross-multiply to get the altitude relationship: $CD^2=AD\cdot DB$ .
Step 3
Turn the difference into an equation
Let $x=AD$ . The piece closer to $B$ is longer by $7$ , so $DB=x+7$ . Substitute those lengths and $CD=12$ into the altitude relationship: $12^2=x(x+7)$ . Using $x-7$ for $DB$ would put the longer piece next to the wrong vertex.
Step 4
Find the shorter piece in Desmos
Type $x_1(x_1+7)\sim12^2\{x_1>0\}$ in Desmos. The $\sim$ asks Desmos to find a value, and the restriction keeps $x_1$ positive, as a length must be. Under PARAMETERS, Desmos shows $x_1=9$ , so $AD=9$ .
Step 5
Finish the tangent ratio
Use $AD=9$ in the ratio from triangle $ACD$ : $\tan A=\frac{12}{9}=\frac{4}{3}$ . The altitude is opposite $A$ in that small triangle, so the tangent of $A$ is $\frac{4}{3}$ . Choice B.

Question 472·Hard·Math

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