A right-triangular metal bracket has hypotenuse length centimeters and area square centimeters. Let be the acute angle opposite the longer leg of the bracket. Which choice is the value of ?

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SAT Right Triangles & Trigonometry Question 15 | Free SAT Prep | Aniko

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Question 15·Hard·Right Triangles and Trigonometry

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A right-triangular metal bracket has hypotenuse length $26$ centimeters and area $120$ square centimeters. Let $A$ be the acute angle opposite the longer leg of the bracket.

Which choice is the value of $\tan A$ ?

When a right triangle gives you the hypotenuse and the area, immediately write two equations for the legs: $x^2+y^2=c^2$ and $xy=2(\text{area})$ . To get $\tan$ of an acute angle, set $t=\frac{y}{x}$ (opposite over adjacent), rewrite $y=tx$ , and substitute to form a quadratic in $t$ . If both $t$ and $\frac{1}{t}$ appear, use a clue like “opposite the longer leg” to choose the root greater than 1.

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Use area and the Pythagorean Theorem

If the legs are $x$ and $y$ , then $\frac{1}{2}xy=120$ and $x^2+y^2=26^2$ .

Introduce $t=\tan A$

Because $\tan A=\frac{y}{x}$ , you can write $y=tx$ and substitute into both equations.

Two possible tangents

You’ll get a quadratic in $t$ with two positive solutions that are reciprocals. Use “angle $A$ is opposite the longer leg” to choose the correct one.

Desmos Guide

Solve for $t$ with a quadratic

In Desmos, enter the equation

60t^2-169t+60=0.

Find the two solutions

Use the graph’s $x$ -intercepts (or the built-in solver) to get the two positive solutions for $t$ .

Select the solution consistent with the description

Because angle $A$ is opposite the longer leg, $\tan A>1$ . Choose the solution greater than $1$ , which is the correct answer.

Step-by-step Explanation

Write equations for the legs

Let the perpendicular legs have lengths $x$ and $y$ , where $y$ is opposite angle $A$ .

Area:

\frac{1}{2}xy=120 \quad\Rightarrow\quad xy=240.

Hypotenuse $26$ :

x^2+y^2=26^2=676.

Express everything using $t=\tan A$

Since $\tan A=\frac{\text{opposite}}{\text{adjacent}}$ , let

t=\tan A=\frac{y}{x} \quad\Rightarrow\quad y=tx.

Use $xy=240$ :

x(tx)=240 \Rightarrow t x^2=240 \Rightarrow x^2=\frac{240}{t}.

Substitute into the Pythagorean equation

From $x^2+y^2=676$ and $y=tx$ :

x^2+(tx)^2=676 \Rightarrow x^2(1+t^2)=676.

Substitute $x^2=\frac{240}{t}$ :

\frac{240}{t}(1+t^2)=676.

Multiply by $t$ :

240(1+t^2)=676t \Rightarrow 240t^2-676t+240=0.

Divide by $4$ :

60t^2-169t+60=0.

Factor:

(12t-5)(5t-12)=0,

so $t=\frac{12}{5}$ or $t=\frac{5}{12}$ .

Use the “longer leg” condition

Angle $A$ is opposite the longer leg, so $A$ is the larger acute angle, which means $\tan A>1$ .

Therefore, $\tan A=\frac{12}{5}$ .

Strategy

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Step-by-step Explanation

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Strategy

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Step-by-step Explanation