In right triangle , . The altitude from to the hypotenuse has length and divides into two segments whose lengths differ by units, with the segment closer to being longer. What is ?

Solution available with step-by-step explanation

SAT Right Triangles & Trigonometry Question 18 | Free SAT Prep | Aniko

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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.

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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.

Strategy

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation