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Question 18·Hard·Right Triangles and Trigonometry

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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 $\tan A$ ?

For right triangles with an altitude drawn to the hypotenuse, immediately label the foot of the altitude and assign variables to the two hypotenuse segments. Use the key relationship $\text{altitude}^2 = (\text{segment 1})(\text{segment 2})$ plus any given information (like a sum or difference) to solve for those segment lengths. Then, instead of finding all three sides of the large triangle, look for a smaller right triangle that actually contains the requested angle; use the definition of tangent (opposite over adjacent) directly in that smaller triangle to get the ratio quickly and avoid extra computation.

Hints

Click me!

Set up the picture and name the new point

Draw right triangle $ABC$ with $\angle C = 90^\circ$ . Drop an altitude from $C$ to $\overline{AB}$ and call the foot of the altitude $D$ . Label $CD = 12$ , and let $AD$ and $DB$ be the two pieces of the hypotenuse that differ by 7 units.

Use the altitude-to-hypotenuse relationship

Let $AD = x$ and $DB = x+7$ . In a right triangle, the altitude from the right angle to the hypotenuse is the geometric mean of the two hypotenuse segments. Translate that into an equation involving $CD$ , $AD$ , and $DB$ .

Solve for the shorter segment and think about tangent

After you use the altitude relationship, solve the resulting quadratic to find $AD$ and $DB$ . Then, to find $\tan A$ , consider which smaller right triangle contains angle $A$ and use "opposite over adjacent" in that smaller triangle.

Do you really need the big triangle’s legs?

Instead of computing the long legs $AC$ and $BC$ , notice that angle $A$ appears in triangle $ADC$ , where you already know $CD$ and $AD$ . Use that triangle directly to compute $\tan A$ as a ratio of those two sides.

Desmos Guide

Solve for the hypotenuse segment using a graph

In Desmos, enter the equation y = x^2 + 7x - 144. Then find the positive x-intercept by clicking where the graph crosses the x-axis. The positive x-value is the length of $AD$ .

Use the segment and altitude to compute the tangent

Take the positive x-value you found (for $AD$ ) and, in a new expression line, type 12 / (that x-value). The resulting decimal is the value of $\tan A$ . Compare that decimal to the decimal equivalents of the answer choices to decide which fraction matches.

Step-by-step Explanation

Introduce variables and use the altitude formula

Let $D$ be the foot of the altitude from $C$ to $\overline{AB}$ .

Let $AD = x$ .
Then $DB = x + 7$ (because the two segments differ by 7 units).

In any right triangle, if an altitude is drawn from the right angle to the hypotenuse, then

\text{(altitude)}^2 = (\text{one hypotenuse segment}) \times (\text{the other segment}).

Here, that means

CD^2 = AD \cdot DB.

Substitute the given length $CD = 12$ and our expressions for $AD$ and $DB$ :

12^2 = x(x + 7).

144 = x(x + 7).

Solve for the hypotenuse segments

Start from

144 = x(x + 7).

Rewrite as a quadratic equation:

x^2 + 7x - 144 = 0.

Factor this:

(x + 16)(x - 9) = 0.

So $x = -16$ or $x = 9$ . A length must be positive, so $x = 9$ .

Therefore:

$AD = 9$
$DB = x + 7 = 9 + 7 = 16$

Relate $\tan A$ to the smaller right triangle

Look at the smaller right triangle $\triangle ADC$ :

It is a right triangle with right angle at $D$ .
Angle $A$ in $\triangle ADC$ is the same as angle $A$ in the original triangle $ABC$ .

In $\triangle ADC$ :

The side opposite angle $A$ is $CD$ .
The side adjacent to angle $A$ is $AD$ .

By definition of tangent in a right triangle,

\tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac{CD}{AD}.

Substitute the values we know:

\tan A = \frac{12}{9}.

Simplify the tangent and match the answer choice

We have

\tan A = \frac{12}{9}.

Simplify this fraction by dividing numerator and denominator by 3:

\tan A = \frac{12 \div 3}{9 \div 3} = \frac{4}{3}.

So $\tan A = \dfrac{4}{3}$ , which corresponds to choice B) 4/3.

Question 18·Hard·Right Triangles and Trigonometry

Strategy

Hints

Desmos Guide

Step-by-step Explanation

Study Hints

Strategy

Hints

Desmos Guide

Step-by-step Explanation