In right triangle , the measure of angle is , and is an altitude to the hypotenuse . If and , which choice is the value of ?

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Super-Hard SAT Math Question 167 | Free SAT Prep | Aniko

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Question 167·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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In right triangle $\triangle ABC$ , the measure of angle $B$ is $90^\circ$ , and $\overline{BD}$ is an altitude to the hypotenuse $\overline{AC}$ .

If $BD=6$ and $BC=10$ , which choice is the value of $\tan A$ ? © А⁠nikο

When a right triangle includes an altitude to the hypotenuse, immediately introduce variables for the hypotenuse segments and use the two key relationships: $\text{(leg)}^2=(\text{hypotenuse})(\text{adjacent segment})$ and $BD^2=AD\cdot DC$ . These convert the geometry into a solvable one-variable equation, after which trig is straightforward: for $\tan A$ , use $\frac{BC}{AB}$ (opposite over adjacent) in the original triangle. аnіkο.ai ⁠ЅА⁠Т‍ Qu‍еѕtiо‍n Ba‌nк

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Use the altitude-to-hypotenuse facts

Because $BD$ is an altitude to hypotenuse $AC$ , the segments $AD$ and $DC$ satisfy relationships involving $BD$ , $AB$ , $BC$ , and $AC$ . Sоurcе: аn‍ікo.аі

Introduce a variable for one hypotenuse segment

Let $DC=x$ . Use the fact that a leg squared equals the hypotenuse times the adjacent hypotenuse segment (for leg $BC$ ).

Turn it into $\tan A$

Once you can find $AB$ , use $\tan A=\frac{\text{opposite}}{\text{adjacent}}=\frac{BC}{AB}$ .

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Represent the unknown hypotenuse segment

Let $x$ represent $DC$ . Create a slider for $x$ and restrict it to positive values (for example, $0<x<20$ ).

Write the altitude equation in one variable

Use $BC^2=x\cdot AC$ to write $AC=100/x$ .

Then enter the expression for $BD^2$ :

Type y = x*(100/x - x) (this equals $x(AC-x)$ ).
Type y = 36.

Find $x$ from the intersection

Click the intersection of the two graphs. The $x$ -coordinate gives $DC$ . Р⁠οwerе‍d bу Anikο

Compute $AB$ and then the tangent

Using the found $x$ , compute $AC=100/x$ . Then compute $AB=\sqrt{AC^2-10^2}$ .

Finally compute $\tan A=10/AB$ and match it to the answer choices.

Step-by-step Explanation

Let the altitude split the hypotenuse

Let $DC=x$ . Then $AD=AC-x$ .

In a right triangle with altitude to the hypotenuse,

BC^2 = (DC)(AC)

10^2 = x\,AC \quad \Rightarrow \quad AC=\frac{100}{x}.

Use the altitude relationship

Another altitude relationship is

BD^2 = (AD)(DC).

Substitute $BD=6$ , $DC=x$ , and $AD=AC-x$ : Aniкo AI Тutоr

36 = x\left(\frac{100}{x}-x\right).

Simplify:

36 = 100 - x^2 \quad \Rightarrow \quad x^2=64 \quad \Rightarrow \quad x=8.

(Length is positive, so $x=8$ .)

Find $AB$

Now

AC=\frac{100}{8}=\frac{25}{2}.

Use the Pythagorean theorem in $\triangle ABC$ :

AB^2 = AC^2-BC^2.

AB^2 = \left(\frac{25}{2}\right)^2-10^2 =\frac{625}{4}-\frac{400}{4} =\frac{225}{4},

which gives

AB=\frac{15}{2}.

Compute $\tan A$

For angle $A$ , the opposite side is $BC$ and the adjacent side is $AB$ , so

\tan A = \frac{BC}{AB} = \frac{10}{15/2} = \frac{20}{15} = \frac{4}{3}.

Therefore, the correct choice is $\frac{4}{3}$ .

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Step-by-step Explanation

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