MAT 142: Right Triangle Trigonometry

Dr. Gilbert

June 12, 2026

Reminders

Last class we studied the graphs of trigonometric functions, inverse trigonometric functions, and solved trigonometric equations. Try the following warm-up problems.

Problem 1: Evaluate \(\displaystyle{\arccos\!\left(\frac{-1}{2}\right)}\) and \(\displaystyle{\arcsin\!\left(\frac{-1}{2}\right)}\).

Problem 2: Find all solutions to \(\displaystyle{\sqrt{3}\tan\!\left(5x - 3\right) + 1 = 0}\).

Objectives

Today we close the loop. We return to the projectile problem from Day 23 — and now we have everything we need to finish it.

After today’s class meeting, you should be able to:

Apply the Pythagorean Theorem to find missing side lengths in right triangles.
Use \(\sin\left(\right)\), \(\cos\left(\right)\), and \(\tan\left(\right)\) to relate angles and side lengths in right triangles.
Use inverse trigonometric functions to find missing angles.
Solve applied problems involving angles of elevation, depression, and right-triangle geometry.

Right Triangle Labeling Convention

Convention: \(A\) and \(B\) denote the two non-right angles. Side \(a\) is opposite angle \(A\); side \(b\) is opposite angle \(B\); side \(c\) is the hypotenuse.

The Pythagorean Theorem

Theorem: For any right triangle with legs \(a\) and \(b\) and hypotenuse \(c\):

\[a^2 + b^2 = c^2\]

Example 1: Find the missing side length in each case.

\(\left(1\right)\) Legs \(a = 5\) and \(b = 7\). Find the hypotenuse \(c\).

\(\left(2\right)\) Leg \(a = 5\) and hypotenuse \(c = 10\). Find the missing leg \(b\).

(1)

\[\begin{align} a^2 + b^2 &= c^2 \end{align}\]

The Pythagorean Theorem

Theorem: For any right triangle with legs \(a\) and \(b\) and hypotenuse \(c\):

\[a^2 + b^2 = c^2\]

Example 1: Find the missing side length in each case.

\(\left(1\right)\) Legs \(a = 5\) and \(b = 7\). Find the hypotenuse \(c\).

\(\left(2\right)\) Leg \(a = 5\) and hypotenuse \(c = 10\). Find the missing leg \(b\).

(1)

\[\begin{align} a^2 + b^2 &= c^2\\ \implies 5^2 + 7^2 &= c^2 \end{align}\]

The Pythagorean Theorem

Theorem: For any right triangle with legs \(a\) and \(b\) and hypotenuse \(c\):

\[a^2 + b^2 = c^2\]

Example 1: Find the missing side length in each case.

\(\left(1\right)\) Legs \(a = 5\) and \(b = 7\). Find the hypotenuse \(c\).

\(\left(2\right)\) Leg \(a = 5\) and hypotenuse \(c = 10\). Find the missing leg \(b\).

(1)

\[\begin{align} a^2 + b^2 &= c^2\\ \implies 5^2 + 7^2 &= c^2\\ \implies 25 + 49 &= c^2 \end{align}\]

The Pythagorean Theorem

Theorem: For any right triangle with legs \(a\) and \(b\) and hypotenuse \(c\):

\[a^2 + b^2 = c^2\]

Example 1: Find the missing side length in each case.

\(\left(1\right)\) Legs \(a = 5\) and \(b = 7\). Find the hypotenuse \(c\).

\(\left(2\right)\) Leg \(a = 5\) and hypotenuse \(c = 10\). Find the missing leg \(b\).

(1)

\[\begin{align} a^2 + b^2 &= c^2\\ \implies 5^2 + 7^2 &= c^2\\ \implies 25 + 49 &= c^2\\ \implies c^2 &= 74 \end{align}\]

The Pythagorean Theorem

Theorem: For any right triangle with legs \(a\) and \(b\) and hypotenuse \(c\):

\[a^2 + b^2 = c^2\]

Example 1: Find the missing side length in each case.

\(\left(1\right)\) Legs \(a = 5\) and \(b = 7\). Find the hypotenuse \(c\).

\(\left(2\right)\) Leg \(a = 5\) and hypotenuse \(c = 10\). Find the missing leg \(b\).

(1)

\[\begin{align} a^2 + b^2 &= c^2\\ \implies 5^2 + 7^2 &= c^2\\ \implies 25 + 49 &= c^2\\ \implies c^2 &= 74\\ \implies c &= \boxed{~\sqrt{74}~} \end{align}\]

Note that we omit the “\(\pm\)” here since we know that \(c\) is a length.

(2)

\[\begin{align} a^2 + b^2 &= c^2 \end{align}\]

The Pythagorean Theorem

Theorem: For any right triangle with legs \(a\) and \(b\) and hypotenuse \(c\):

\[a^2 + b^2 = c^2\]

Example 1: Find the missing side length in each case.

\(\left(1\right)\) Legs \(a = 5\) and \(b = 7\). Find the hypotenuse \(c\).

\(\left(2\right)\) Leg \(a = 5\) and hypotenuse \(c = 10\). Find the missing leg \(b\).

(1)

\[\begin{align} a^2 + b^2 &= c^2\\ \implies 5^2 + 7^2 &= c^2\\ \implies 25 + 49 &= c^2\\ \implies c^2 &= 74\\ \implies c &= \boxed{~\sqrt{74}~} \end{align}\]

Note that we omit the “\(\pm\)” here since we know that \(c\) is a length.

(2)

\[\begin{align} a^2 + b^2 &= c^2\\ \implies 5^2 + b^2 &= 10^2 \end{align}\]

The Pythagorean Theorem

Theorem: For any right triangle with legs \(a\) and \(b\) and hypotenuse \(c\):

\[a^2 + b^2 = c^2\]

Example 1: Find the missing side length in each case.

\(\left(1\right)\) Legs \(a = 5\) and \(b = 7\). Find the hypotenuse \(c\).

\(\left(2\right)\) Leg \(a = 5\) and hypotenuse \(c = 10\). Find the missing leg \(b\).

(1)

\[\begin{align} a^2 + b^2 &= c^2\\ \implies 5^2 + 7^2 &= c^2\\ \implies 25 + 49 &= c^2\\ \implies c^2 &= 74\\ \implies c &= \boxed{~\sqrt{74}~} \end{align}\]

Note that we omit the “\(\pm\)” here since we know that \(c\) is a length.

(2)

\[\begin{align} a^2 + b^2 &= c^2\\ \implies 5^2 + b^2 &= 10^2\\ \implies 25 + b^2 &= 100 \end{align}\]

The Pythagorean Theorem

Theorem: For any right triangle with legs \(a\) and \(b\) and hypotenuse \(c\):

\[a^2 + b^2 = c^2\]

Example 1: Find the missing side length in each case.

\(\left(1\right)\) Legs \(a = 5\) and \(b = 7\). Find the hypotenuse \(c\).

\(\left(2\right)\) Leg \(a = 5\) and hypotenuse \(c = 10\). Find the missing leg \(b\).

(1)

\[\begin{align} a^2 + b^2 &= c^2\\ \implies 5^2 + 7^2 &= c^2\\ \implies 25 + 49 &= c^2\\ \implies c^2 &= 74\\ \implies c &= \boxed{~\sqrt{74}~} \end{align}\]

Note that we omit the “\(\pm\)” here since we know that \(c\) is a length.

(2)

\[\begin{align} a^2 + b^2 &= c^2\\ \implies 5^2 + b^2 &= 10^2\\ \implies 25 + b^2 &= 100\\ \implies b^2 &= 75 \end{align}\]

The Pythagorean Theorem

Theorem: For any right triangle with legs \(a\) and \(b\) and hypotenuse \(c\):

\[a^2 + b^2 = c^2\]

Example 1: Find the missing side length in each case.

\(\left(1\right)\) Legs \(a = 5\) and \(b = 7\). Find the hypotenuse \(c\).

\(\left(2\right)\) Leg \(a = 5\) and hypotenuse \(c = 10\). Find the missing leg \(b\).

(1)

\[\begin{align} a^2 + b^2 &= c^2\\ \implies 5^2 + 7^2 &= c^2\\ \implies 25 + 49 &= c^2\\ \implies c^2 &= 74\\ \implies c &= \boxed{~\sqrt{74}~} \end{align}\]

Note that we omit the “\(\pm\)” here since we know that \(c\) is a length.

(2)

\[\begin{align} a^2 + b^2 &= c^2\\ \implies 5^2 + b^2 &= 10^2\\ \implies 25 + b^2 &= 100\\ \implies b^2 &= 75\\ \implies b &= \sqrt{75} \end{align}\]

The Pythagorean Theorem

Theorem: For any right triangle with legs \(a\) and \(b\) and hypotenuse \(c\):

\[a^2 + b^2 = c^2\]

Example 1: Find the missing side length in each case.

\(\left(1\right)\) Legs \(a = 5\) and \(b = 7\). Find the hypotenuse \(c\).

\(\left(2\right)\) Leg \(a = 5\) and hypotenuse \(c = 10\). Find the missing leg \(b\).

(1)

\[\begin{align} a^2 + b^2 &= c^2\\ \implies 5^2 + 7^2 &= c^2\\ \implies 25 + 49 &= c^2\\ \implies c^2 &= 74\\ \implies c &= \boxed{~\sqrt{74}~} \end{align}\]

Note that we omit the “\(\pm\)” here since we know that \(c\) is a length.

(2)

\[\begin{align} a^2 + b^2 &= c^2\\ \implies 5^2 + b^2 &= 10^2\\ \implies 25 + b^2 &= 100\\ \implies b^2 &= 75\\ \implies b &= \sqrt{75}\\ \implies b &= ~\boxed{5\sqrt{3}~} \end{align}\]

Trig Ratios in Right Triangles

The unit circle definitions of \(\sin\), \(\cos\), and \(\tan\) connect directly to side ratios in any right triangle. For angle \(A\):

\[\sin\!\left(A\right) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{a}{c} \qquad \cos\!\left(A\right) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{b}{c} \qquad \tan\!\left(A\right) = \frac{\text{opposite}}{\text{adjacent}} = \frac{a}{b}\]

A useful mnemonic: SOH-CAH-TOA.

We can also find angles using the inverse functions:

\[A = \arcsin\!\left(\frac{a}{c}\right) \qquad A = \arccos\!\left(\frac{b}{c}\right) \qquad A = \arctan\!\left(\frac{a}{b}\right)\]

Returning to Non-Vertical Motion from Day 23

On Day 23, we asked: if a ball is launched at \(60°\) with speed \(64\) ft/sec, what is the maximum height? Now we can answer. The vertical component of velocity is \(64\sin\!\left(60°\right) = 64 \cdot \frac{\sqrt{3}}{2} = 32\sqrt{3}\) ft/sec. Substituting into \(h\left(t\right) = -16t^2 + 32\sqrt{3}t\) gives a maximum height of \(48\) feet, reached at \(t = \frac{\sqrt{3}}{1} \approx 1.73\) seconds.

Solving Right Triangles I

Example 2: The right triangle below has \(a = 8\) and \(b = 15\). Find all missing sides and angles.

Step 1: Find \(c\) using the Pythagorean Theorem.

\[\begin{align} 8^2 + 15^2 &= c^2 \end{align}\]

Solving Right Triangles I

Example 2: The right triangle below has \(a = 8\) and \(b = 15\). Find all missing sides and angles.

Step 1: Find \(c\) using the Pythagorean Theorem.

\[\begin{align} 8^2 + 15^2 &= c^2\\ \implies 64 + 225 &= c^2 \end{align}\]

Solving Right Triangles I

Example 2: The right triangle below has \(a = 8\) and \(b = 15\). Find all missing sides and angles.

Step 1: Find \(c\) using the Pythagorean Theorem.

\[\begin{align} 8^2 + 15^2 &= c^2\\ \implies 64 + 225 &= c^2\\ \implies 289 &= c^2\\ \implies c &= 17 \end{align}\]

Solving Right Triangles I

Example 2: The right triangle below has \(a = 8\) and \(b = 15\). Find all missing sides and angles.

Step 2: Find angle \(A\) using \(\sin\!\left(A\right) = \dfrac{a}{c}\).

\[\begin{align} \sin\!\left(A\right) &= \frac{8}{17} \end{align}\]

Solving Right Triangles I

Example 2: The right triangle below has \(a = 8\) and \(b = 15\). Find all missing sides and angles.

Step 2: Find angle \(A\) using \(\sin\!\left(A\right) = \dfrac{a}{c}\).

\[\begin{align} \sin\!\left(A\right) &= \frac{8}{17}\\ \implies A &= \arcsin\!\left(\frac{8}{17}\right)\\ \implies A &\approx 28.07° \end{align}\]

Solving Right Triangles I

Example 2: The right triangle below has \(a = 8\) and \(b = 15\). Find all missing sides and angles.

Step 3: Find angle \(B\) using the angle sum.

\[\begin{align} A + B + 90° &= 180° \end{align}\]

Solving Right Triangles I

Example 2: The right triangle below has \(a = 8\) and \(b = 15\). Find all missing sides and angles.

Step 3: Find angle \(B\) using the angle sum.

\[\begin{align} A + B + 90° &= 180°\\ \implies 28.07° + B + 90° &\approx 180°\\ \implies B &\approx 61.93° \end{align}\]

The triangle is fully solved: \(c = 17\), \(A \approx 28.07°\), \(B \approx 61.93°\).

Triangle Practice

Try It! 1: Find all missing sides and angles for a right triangle with \(A = 30°\) and \(a = 5\).

Try It! 2: Find all missing sides and angles for a right triangle with \(B = 50°\) and \(c = 7\).

Try It! 3: Find all missing sides and angles for a right triangle with \(A = 45°\) and \(c = 4\).

Try It! 4: Find all missing sides and angles for a right triangle with \(a = 5\) and \(b = 15\).

Applied Example 1: Tree Height

Problem: An arborist who is \(5'8''\) tall stands at the base of a tall tree and walks \(40\) feet away. Using an inclinometer, they measure an angle of \(57°\) from eye level to the treetop. Estimate the height of the tree.

The height from eye level to the treetop uses \(\tan\):

\[\begin{align} \tan\!\left(57°\right) &= \frac{h_{\text{eye}}}{40} \end{align}\]

Applied Example 1: Tree Height

The height from eye level to the treetop uses \(\tan\):

\[\begin{align} \tan\!\left(57°\right) &= \frac{h_{\text{eye}}}{40}\\ \implies h_{\text{eye}} &= 40\tan\!\left(57°\right)\\ &\approx 61.59~\text{ft} \end{align}\]

Add the arborist’s eye height (\(5'8'' \approx 5.67\) ft):

\[\begin{align} h_{\text{tree}} &\approx 61.59 + 5.67 \approx \boxed{~67~\text{ft}~} \end{align}\]

Applied Example 2: Antenna Height

Problem: From a point \(300\) ft from the base of a building, the angle of elevation to the top of the building is \(40°\) and to the top of a rooftop antenna is \(43°\). Find the height of the antenna.

Height of building top:

\[\begin{align} h_{\text{bldg}} &= 300\tan\!\left(40°\right) \approx 251.73~\text{ft} \end{align}\]

Height of antenna top:

\[\begin{align} h_{\text{top}} &= 300\tan\!\left(43°\right) \approx 279.75~\text{ft} \end{align}\]

Antenna height:

\[\begin{align} h_{\text{antenna}} &= 279.75 - 251.73 \approx \boxed{~28~\text{ft}~} \end{align}\]

Applied Example 3: Radio Tower

Problem: A radio tower is \(325\) ft from a building. From a window, the angle of elevation to the top of the tower is \(43°\) and the angle of depression to the base is \(31°\). How tall is the tower?

Height above the window:

\[\begin{align} h_{\text{above}} &= 325\tan\!\left(43°\right) \approx 303.07~\text{ft} \end{align}\]

Depth below the window:

\[\begin{align} h_{\text{below}} &= 325\tan\!\left(31°\right) \approx 195.28~\text{ft} \end{align}\]

Total tower height:

\[\begin{align} h_{\text{tower}} &= 303.07 + 195.28 \approx \boxed{~498~\text{ft}~} \end{align}\]

Exit Ticket Task

Navigate to our MAT142 Exit Ticket Form, answer the questions, and complete the task below.

Note. Today’s discussion is listed as 21. Right Triangle Trigonometry

Task: A surveyor standing at ground level measures the angle of elevation to the top of a cliff to be \(38°\). She then walks \(50\) feet closer to the cliff and measures the angle of elevation to be \(52°\). How tall is the cliff?

Summary and Congratulations

Ideas From Today

In a right triangle, \(\sin = \frac{\text{opp}}{\text{hyp}}\), \(\cos = \frac{\text{adj}}{\text{hyp}}\), \(\tan = \frac{\text{opp}}{\text{adj}}\) — SOH-CAH-TOA.
Use \(\arcsin\left(\right)\), \(\arccos\left(\right)\), or \(\arctan\left(\right)\) to recover angles from side ratios.
The angle sum of a triangle is \(180°\); once two angles are known, the third is determined.
Angles of elevation and depression create right triangles whose sides correspond to horizontal distances and heights.

Looking Ahead

This is the end of our content for MAT 142.
The thread from Day 1 to today: we started with rates of change and limits, built up every class of algebraic function, then added a few completely new function classes (exponential, logarithmic, and trigonometric). Right triangle trig brings us back to where we started — ratios and geometry — but now with the full machinery of functions behind us.

Thank you for a great semester.

Homework:
Complete Homework 12 on MyOpenMath
Prepare for Exams