In right triangle , angle is the right angle. The altitude from to the hypotenuse meets at point , dividing the hypotenuse so that and . What is ?

Solution available with step-by-step explanation

SAT Right Triangles & Trigonometry Question 48 | Free SAT Prep | Aniko

00:00

Question 48·Hard·Right Triangles and Trigonometry

0 / 50

In right triangle $\triangle ABC$ , angle $C$ is the right angle. The altitude from $C$ to the hypotenuse $\overline{AB}$ meets $\overline{AB}$ at point $D$ , dividing the hypotenuse so that $AD = 4$ and $DB = 9$ . What is $\sin A$ ?

For right-triangle problems involving an altitude to the hypotenuse, quickly add the segments on the hypotenuse to get its full length, then use similarity-based relationships instead of re-deriving everything: each leg squared equals the product of the hypotenuse and the adjacent hypotenuse segment. Identify which segment goes with which leg, compute the needed leg, and then apply the basic trig definition (sine = opposite over hypotenuse, cosine = adjacent over hypotenuse). This avoids setting up extra variables or long systems and lets you move efficiently from geometry to the trig ratio the question asks for.

Hints

Click me!

Start with what you can find immediately

Use the given segments $AD = 4$ and $DB = 9$ to find the entire hypotenuse $AB$ . Then think about which side is opposite $\angle A$ .

Use the special property of an altitude to the hypotenuse

The altitude from the right angle to the hypotenuse creates two smaller right triangles similar to the original. In such a setup, each leg of the big triangle is related to the whole hypotenuse and one of the segments $AD$ or $DB$ . Focus on how leg $BC$ relates to segment $DB$ and hypotenuse $AB$ .

Connect the leg to a product

There is a formula: in this situation, the square of a leg equals the product of the full hypotenuse and the part of the hypotenuse next to that leg. Use it to find $BC$ , then plug into the definition $\sin A = \dfrac{\text{opposite}}{\text{hypotenuse}}$ .

Desmos Guide

Compute the hypotenuse

In a Desmos expression line, type AB = 4 + 9 and see that $AB$ is the sum of the given segments on the hypotenuse.

Compute the leg BC using the altitude property

In the next line, use the relationship $BC^2 = DB \cdot AB$ by typing BC = sqrt(9*AB). The numerical value that appears is the length of side $BC$ .

Compute sin(A) as a ratio of sides

Now type sinA = BC/AB. The resulting decimal is the value of $\sin A$ . To match it to an answer choice, type each option into Desmos (for example, sqrt(13)/3, 4/13, etc.) and see which expression gives the same decimal as sinA.

Step-by-step Explanation

Understand the setup and find the hypotenuse

Draw (or imagine) right triangle $\triangle ABC$ with $\angle C = 90^\circ$ and hypotenuse $\overline{AB}$ . The altitude from $C$ meets $AB$ at $D$ , with $AD = 4$ and $DB = 9$ .

First, find the full length of the hypotenuse:

$AB = AD + DB = 4 + 9 = 13$ .

We will eventually use $\sin A = \dfrac{\text{opposite}}{\text{hypotenuse}}$ , so we need the length of the side opposite $\angle A$ , which is $BC$ , and we already know the hypotenuse $AB = 13$ .

Use the altitude-to-hypotenuse property to find leg BC

When you drop an altitude from the right angle to the hypotenuse in a right triangle, it creates two smaller right triangles that are similar to the original triangle. From this similarity, an important relationship follows:

Each leg squared equals the product of the hypotenuse and the adjacent segment of the hypotenuse.

For leg $BC$ , the adjacent segment on the hypotenuse is $DB$ , so:

BC^2 = DB \cdot AB.

Substitute the known values $DB = 9$ and $AB = 13$ :

BC^2 = 9 \cdot 13 = 117.

BC = \sqrt{117} = \sqrt{9 \cdot 13} = 3\sqrt{13}.

Now you know both $BC$ and $AB$ .

Apply the definition of sine for angle A

In right triangle trigonometry,

$\sin A = \dfrac{\text{opposite side to } A}{\text{hypotenuse}}$ .

For $\angle A$ :

The side opposite $\angle A$ is $BC$ .
The hypotenuse is $AB$ .

\sin A = \frac{BC}{AB} = \frac{3\sqrt{13}}{13}.

Therefore, the correct answer is $\boxed{\dfrac{3\sqrt{13}}{13}}$ .

Strategy

Hints

Desmos Guide

Step-by-step Explanation

Study Hints

Strategy

Hints

Desmos Guide

Step-by-step Explanation