3.3: Right Triangle Relationships | Geometry Honors

How it works

Pythagoras applies to right triangles. The hypotenuse is opposite the right angle. Trigonometric ratios connect an acute angle with side ratios; special triangles give exact values.

a^2+b^2=c^2,\quad\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}

Worked examples

Example 1

A right triangle has legs $\displaystyle 6$ and $\displaystyle 8$ . Find the hypotenuse.

$\displaystyle c=\sqrt{36+64}=10$ .

Example 2

A right triangle has hypotenuse $\displaystyle 12$ and an acute angle $\displaystyle 30^\circ$ . Find the opposite leg.

The opposite leg is $\displaystyle 12\sin30^\circ=6$ .

Problems

Problem 3.3.1

In a $\displaystyle 3$ - $\displaystyle 4$ - $\displaystyle 5$ right triangle, find sine, cosine, and tangent of the angle opposite the side of length $\displaystyle 3$ .

Triangle with vertices A at 0,3, B at 0,0, C at 4,0, and side lengths 3,4,5. — The right angle is at B; the hypotenuse AC has length 5.

Show worked solution

$\displaystyle \sin\theta=3/5$ , $\displaystyle \cos\theta=4/5$ , and $\displaystyle \tan\theta=3/4$ .

Triangle with vertices A at 0,3, B at 0,0, C at 4,0, and side lengths 3,4,5. The solution adds labeled relationships. — For angle C, opposite = 3, adjacent = 4, and hypotenuse = 5.

Problem 3.3.2

A ladder $\displaystyle 13$ ft long rests against a wall with its base $\displaystyle 5$ ft from the wall. How high does it reach?

Show worked solution

The height is $\displaystyle \sqrt{13^2-5^2}=\sqrt{144}=12$ ft.

Problem 3.3.3

An isosceles right triangle has legs $\displaystyle 7$ . Find its hypotenuse exactly.

Show worked solution

$\displaystyle c=\sqrt{7^2+7^2}=7\sqrt{2}$ .

Lesson Problems PDF Solutions PDF Unit MCQ

How it works

Worked examples

Example 1

Example 2

Problems

Problem 3.3.1

Problem 3.3.2

Problem 3.3.3

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