In right triangle , angle is a right angle, the measure of angle is , and the length of the altitude from to hypotenuse is units. If the length of can be written in the form , where is a constant, what is the value of ?

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SAT Right Triangles & Trigonometry Question 30 | Free SAT Prep | Aniko

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Question 30·Hard·Right Triangles and Trigonometry

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In right triangle $ABC$ , angle $C$ is a right angle, the measure of angle $A$ is $35^\circ$ , and the length of the altitude from $C$ to hypotenuse $\overline{AB}$ is $6$ units. If the length of $\overline{AB}$ can be written in the form $k\csc 70^\circ$ , where $k$ is a constant, what is the value of $k$ ?

Your answer

(Express the answer as an integer)

When a right triangle problem gives an altitude to the hypotenuse and an acute angle, a fast method is to use the area in two ways: (1) as $\tfrac{1}{2}$ times the product of the legs, expressed via the hypotenuse and trig (using sine and cosine of the given angle), and (2) as $\tfrac{1}{2}$ times the hypotenuse times the altitude. Equating the two area expressions typically produces an equation like $c\sin\theta\cos\theta = h$ , which you can simplify quickly with the double-angle identity $\sin(2\theta)=2\sin\theta\cos\theta$ and then solve for the hypotenuse; finally, match the result to any required form (such as $k\csc 70^\circ$ ) to extract the constant.

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Use area formulas

Think about the area of triangle $ABC$ . You know the altitude to the hypotenuse is 6, and you can also write the area using the two legs of the right triangle.

Relate the legs to the hypotenuse with trigonometry

Let the hypotenuse $AB$ be $x$ . With angle $A = 35^\circ$ , express $AC$ and $BC$ in terms of $x$ using $\sin 35^\circ$ and $\cos 35^\circ$ .

Set the two area expressions equal

Write the area once as $\tfrac{1}{2}AC\cdot BC$ and once as $\tfrac{1}{2}AB\cdot 6$ , then set them equal and solve for $AB$ .

Use a trig identity with 70°

When you get an expression involving $\sin 35^\circ \cos 35^\circ$ , remember the double-angle identity $\sin(2\theta) = 2\sin\theta\cos\theta$ and note that $2\cdot 35^\circ = 70^\circ$ .

Desmos Guide

Compute the hypotenuse from the altitude and angle

In Desmos, enter the expression 2*6/sin(70°) (make sure Desmos is in degrees). The numeric output is the length of $AB$ .

Find the constant k in the form k·csc(70°)

In a new line, divide that hypotenuse value by csc(70°) (for example, type (2*6/sin(70°))/csc(70°)). The resulting number is the value of $k$ you are solving for.

Step-by-step Explanation

Express the legs in terms of the hypotenuse

Let the hypotenuse $\overline{AB}$ have length $x$ .

Relative to angle $A=35^\circ$ :

$\overline{AC}$ is adjacent to angle $A$ , so $AC = x\cos 35^\circ$ .
$\overline{BC}$ is opposite angle $A$ , so $BC = x\sin 35^\circ$ .

Write the area in two different ways

First, use the two legs $AC$ and $BC$ :

Area using legs:

\text{Area} = \frac{1}{2} \cdot AC \cdot BC = \frac{1}{2} \cdot (x\cos 35^\circ) \cdot (x\sin 35^\circ) = \frac{1}{2} x^2 \sin 35^\circ \cos 35^\circ.

Second, use the hypotenuse $AB$ as the base and the altitude from $C$ (which is 6) as the height:

Area using hypotenuse and altitude:

\text{Area} = \frac{1}{2} \cdot AB \cdot 6 = \frac{1}{2} x \cdot 6.

Set the areas equal and simplify

Since both expressions represent the same area, set them equal:

\frac{1}{2} x^2 \sin 35^\circ \cos 35^\circ = \frac{1}{2} x \cdot 6.

Cancel $\tfrac{1}{2}$ from both sides and (assuming $x\neq 0$ ) divide both sides by $x$ :

x \sin 35^\circ \cos 35^\circ = 6.

Now use the double-angle identity:

\sin 70^\circ = \sin(2\cdot 35^\circ) = 2\sin 35^\circ \cos 35^\circ,

\sin 35^\circ \cos 35^\circ = \frac{1}{2}\sin 70^\circ.

Substitute this into the equation:

x \cdot \frac{1}{2} \sin 70^\circ = 6.

Solve for the hypotenuse and match the given form

From

\frac{1}{2} x \sin 70^\circ = 6,

solve for $x$ :

\frac{x \sin 70^\circ}{2} = 6 \\[0.7em] x \sin 70^\circ = 12 \\[0.7em] x = \frac{12}{\sin 70^\circ} = 12\csc 70^\circ.

We are told $AB$ (which is $x$ ) can be written as $k\csc 70^\circ$ , so comparing $AB = 12\csc 70^\circ$ with $AB = k\csc 70^\circ$ shows that $k = 12$ . This is the required value.

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