In right triangle , angle is . The altitude from to hypotenuse meets at point , and . If and , which choice is the length of ?

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Super-Hard SAT Math Question 150 | Free SAT Prep | Aniko

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Question 150·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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In right triangle $ABC$ , angle $C$ is $90^\circ$ . The altitude from $C$ to hypotenuse $\overline{AB}$ meets $\overline{AB}$ at point $D$ , and $\overline{CD}\perp \overline{AB}$ . If $CD=15$ and $\tan(A)=\frac{3}{4}$ , which choice is the length of $AD$ ? P‍rеpа⁠red‍ bу Аniкo.aі

When you see a right triangle with an altitude to the hypotenuse, expect two key tools: (1) rewrite trig information like $\tan(A)$ into $\sin(A)$ and $\cos(A)$ using a reference triangle, and (2) use area in two ways to connect the altitude to the hypotenuse via $AB\sin(A)\cos(A)=CD$ . Then use similarity (such as $AC^2=(AB)(AD)$ ) to convert the hypotenuse length into the specific segment the problem asks for. А‍-n-i-k-о.аі

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Turn $\tan(A)$ into a triangle ratio

If $\tan(A)=\frac{3}{4}$ , think of a right triangle with opposite $3$ and adjacent $4$ to get the hypotenuse and then $\sin(A)$ and $\cos(A)$ . Рrepаre‍d ⁠bу А‍nікo⁠.аі

Use two different area formulas

Write the area as $\frac12(AC)(BC)$ and also as $\frac12(AB)(CD)$ . Use $CD=15$ .

Connect $AD$ to a leg

With an altitude to the hypotenuse, you can use a similarity relationship like $AC^2=(AB)(AD)$ to isolate $AD$ .

Desmos Guide

Enter the trig ratios from tangent

In Desmos, type sinA=3/5 and cosA=4/5 (these come from a $3$ - $4$ - $5$ triangle matching $\tan(A)=3/4$ ).

Compute the hypotenuse from the altitude equation

Type AB=15/(sinA*cosA). This uses $AB\sin(A)\cos(A)=15$ from the area relationship.

Compute the segment length expression

Type AD=AB*cosA^2. Read the value Desmos shows for AD. Prepаre‍d by⁠ Aniкο.аі

Step-by-step Explanation

Convert tangent to sine and cosine

Given $\tan(A)=\frac{3}{4}$ , model the legs relative to angle $A$ as opposite $=3$ and adjacent $=4$ , so the hypotenuse is $5$ .

Thus,

\sin(A)=\frac{3}{5},\qquad \cos(A)=\frac{4}{5}

Use area to find the hypotenuse $AB$

The area of the right triangle can be written two ways:

Using legs: $\frac12(AC)(BC)$
Using hypotenuse and altitude: $\frac12(AB)(CD)$

Also, relative to angle $A$ :

AC=AB\cos(A),\qquad BC=AB\sin(A)

\frac12(AB\cos A)(AB\sin A)=\frac12(AB)(15)

Cancel $\frac12$ and one factor of $AB$ (since $AB>0$ ):

AB\sin A\cos A=15

Compute $\sin A\cos A$ :

\sin A\cos A=\frac{3}{5}\cdot\frac{4}{5}=\frac{12}{25}

AB=\frac{15}{12/25}=15\cdot\frac{25}{12}=\frac{125}{4}

Relate $AD$ to $AB$ using similarity

In a right triangle with an altitude to the hypotenuse, the smaller triangles are similar to the original triangle, which gives the leg–hypotenuse segment relationship:

AC^2 = (AB)(AD)

AC^2=(AB^2)(\cos^2 A)

Substitute into $AC^2=(AB)(AD)$ and divide by $AB$ :

AD = AB\cos^2 A

Compute $AD$

Compute $\cos^2(A)$ and multiply:

\cos^2(A)=\left(\frac{4}{5}\right)^2=\frac{16}{25}

AD = \frac{125}{4}\cdot\frac{16}{25}=20

Therefore, the length of $AD$ is $20$ .

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Step-by-step Explanation

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Desmos Guide

Step-by-step Explanation