How to Find Missing Sides and Angles of a Right Triangle

How to Find Missing Sides and Angles of a Right Triangle

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In a right triangle, when one side and an angle are given we can find a missing side.

Trigonometric Ratios

The following trigonometric ratios can be used to determine the length of the missing side in a right triangle.

\(Sin \ θ \ = \ \frac{Opposite \ Side}{Hypotenuse}\)

\(Cos \ θ \ = \ \frac{Adjacent \ Side}{Hypotenuse}\)

\(tan \ θ \ = \ \frac{Opposite \ Side}{Adjacent \ Side}\)

\(Cot \ θ \ = \ \frac{Adjacent \ Side }{Opposite \ Side}\)

\(Cosec \ θ \ = \ \frac{Hypotenuse}{Opposite \ Side}\)

\(Sec \ θ \ = \ \frac{Hypotenuse}{Adjacent \ Side \ to}\)

Example

Find the value of \(x\).

Missing Sides and Angles of a Right Triangle

Solution

  • Hypotenuse Side \(= \ AC\)
  • Opposite Side \(= \ AB \ = \ x\)
  • Adjacent Side \(= \ BC \ = \ 7\)
  • \(θ \ = \ 30\)

The trigonometric ratio \(tan\) involves the adjacent and opposing sides.

\(tan(θ) \ = \ \frac{Opposite \ Side}{Adjacent \ Side} \ = \ \frac{AB}{BC}\) \(⇒ \ \frac{\sqrt{3}}{3} \ = \ \frac{x}{7} \ ⇒ \ x \ = \ \frac{7\sqrt{3}}{3} \ ≈ \ 4.04\)

So, \(4.04\) is the measurement of the missing side.

Missing Sides and Angles of a Right Triangle

Think of this lesson as more than a rule to memorize. Missing Sides and Angles of a Right Triangle is about angles, triangles, radians, and circular motion. A strong student does not rush to the first formula on the page; they pause, identify the structure of the problem, and then choose the tool that matches that structure. That pause is what prevents most mistakes.

Trig ratios connect angles to side lengths. In a right triangle, \(\sin\theta=\frac{opposite}{hypotenuse}\), \(\cos\theta=\frac{adjacent}{hypotenuse}\), and \(\tan\theta=\frac{opposite}{adjacent}\).

Here is the teacher way to approach the topic. First, read the problem slowly and underline the information that is actually given. Next, name the target: are you finding a value, simplifying an expression, comparing two quantities, solving for a variable, or interpreting a graph? Once the target is clear, the calculation becomes much less mysterious because every step has a job.

  • Identify the input value or expression.
  • Substitute carefully using parentheses.
  • Simplify one operation at a time.
  • Check domain restrictions such as zero denominators or even roots.

A helpful habit is to say what each number represents before using it. For example, if a number is a denominator, a radius, a slope, a common difference, or a coefficient, it should not be treated like an ordinary loose number. Its role tells you where it belongs in the formula. This is especially important on ACT-style questions because many wrong answer choices come from using the right numbers in the wrong places.

Another good habit is to keep the work organized vertically. Write one clean line for substitution, one line for simplifying, and one line for the final answer. If the problem has units, keep the units attached. If the problem has signs, exponents, or parentheses, copy them carefully from one line to the next. Most errors in this topic are not caused by a hard idea; they are caused by dropping a negative sign, combining unlike terms, using the wrong denominator, or skipping a check.

When you finish, ask a quick reasonableness question. Should the answer be positive or negative? Should it be larger or smaller than the original number? Does it fit the graph, table, shape, or equation? Can you plug it back into the original problem? This final check turns the lesson from a procedure into understanding.

On a test, the goal is not to write the longest solution. The goal is to write enough clear work that you can see the structure, avoid traps, and recover quickly if one line goes wrong. Practice the examples below with that mindset: identify the type of problem, choose the matching rule, show the substitution, simplify carefully, and check the answer before moving on.

Free printable Worksheets

Exercises for Finding Missing Sides and Angles of a Right Triangle

1) A right triangle has legs \(3\) and \(4\). Find the hypotenuse.

2) A right triangle has hypotenuse \(13\) and one leg \(5\). Find the other leg.

3) In a right triangle, angle \(A=30°\) and the hypotenuse is \(10\). Find the side opposite \(A\).

4) In a right triangle, angle \(A=60°\) and the hypotenuse is \(12\). Find the side adjacent to \(A\).

5) A \(45°-45°-90°\) triangle has one leg \(7\). Find the hypotenuse.

6) In a right triangle, angle \(A=30°\) and the side opposite \(A\) is \(9\). Find the hypotenuse.

7) In a right triangle, angle \(A=60°\) and the side adjacent to \(A\) is \(5\). Find the opposite side.

8) A \(45°-45°-90°\) triangle has hypotenuse \(10\sqrt{2}\). Find each leg.

9) A \(20\)-foot ladder makes a \(60°\) angle with the ground. How high up the wall does it reach?

10) A flagpole casts a shadow. The angle of elevation is \(30°\), and the pole is \(12\) ft tall. Find the shadow length.

11) A right triangle has hypotenuse \(17\) and one leg \(8\). Find the other leg.

12) In a right triangle, angle \(A=35°\) and the hypotenuse is \(20\). Find the side opposite \(A\) to the nearest tenth.

13) In a right triangle, angle \(A=52°\) and the adjacent side is \(14\). Find the hypotenuse to the nearest tenth.

14) From a point \(80\) ft from a building, the angle of elevation to the top is \(40°\). Find the building height to the nearest tenth.

15) A right triangle has legs \(9\) and \(12\). If angle \(A\) is opposite the \(9\)-unit leg, find \(sin \ A\).

16) A right triangle has legs \(7\) and \(24\). If angle \(A\) is adjacent to the \(24\)-unit leg, find \(cos \ A\).

17) A right triangle has hypotenuse \(26\). For angle \(B\), the opposite side is \(10\). Find \(tan \ B\).

18) In a right triangle, angle \(A=28°\) and the opposite side is \(15\). Find the adjacent side to the nearest tenth.

19) A ramp rises \(4\) ft over a horizontal run of \(18\) ft. Find the angle the ramp makes with the ground to the nearest tenth.

20) A right triangle has hypotenuse \(41\) and one leg \(40\). Find the smaller acute angle to the nearest tenth.

 

1)A right triangle has legs \(3\) and \(4\). Find the hypotenuse.

Use the Pythagorean Theorem: \(c^2=3^2+4^2\).

\(c^2=9+16=25\), so \(c=5\).

\(\color{red}{5}\)

2)A right triangle has hypotenuse \(13\) and one leg \(5\). Find the other leg.

Use \(a^2+5^2=13^2\).

\(a^2=169-25=144\), so \(a=12\).

\(\color{red}{12}\)

3)In a right triangle, angle \(A=30°\) and the hypotenuse is \(10\). Find the side opposite \(A\).

Use \(sin \ A=\frac{opposite}{hypotenuse}\).

\(sin \ 30°=\frac{x}{10}\), so \(x=10\cdot\frac{1}{2}=5\).

\(\color{red}{5}\)

4)In a right triangle, angle \(A=60°\) and the hypotenuse is \(12\). Find the side adjacent to \(A\).

Use \(cos \ A=\frac{adjacent}{hypotenuse}\).

\(cos \ 60°=\frac{x}{12}\), so \(x=12\cdot\frac{1}{2}=6\).

\(\color{red}{6}\)

5)A \(45°-45°-90°\) triangle has one leg \(7\). Find the hypotenuse.

In a \(45°-45°-90°\) triangle, \(hypotenuse=leg\sqrt{2}\).

Substitute the leg length: \(7\sqrt{2}\).

\(\color{red}{7\sqrt{2}}\)

6)In a right triangle, angle \(A=30°\) and the side opposite \(A\) is \(9\). Find the hypotenuse.

Use \(sin \ 30°=\frac{9}{h}\).

\(\frac{1}{2}=\frac{9}{h}\), so \(h=18\).

\(\color{red}{18}\)

7)In a right triangle, angle \(A=60°\) and the side adjacent to \(A\) is \(5\). Find the opposite side.

Use \(tan \ A=\frac{opposite}{adjacent}\).

\(tan \ 60°=\frac{x}{5}\), so \(x=5\sqrt{3}\).

\(\color{red}{5\sqrt{3}}\)

8)A \(45°-45°-90°\) triangle has hypotenuse \(10\sqrt{2}\). Find each leg.

In a \(45°-45°-90°\) triangle, \(hypotenuse=leg\sqrt{2}\).

\(10\sqrt{2}=x\sqrt{2}\), so \(x=10\).

\(\color{red}{10}\)

9)A \(20\)-foot ladder makes a \(60°\) angle with the ground. How high up the wall does it reach?

The ladder is the hypotenuse, and the height is opposite the \(60°\) angle.

\(sin \ 60°=\frac{h}{20}\), so \(h=20\cdot\frac{\sqrt{3}}{2}=10\sqrt{3}\).

\(\color{red}{10\sqrt{3}\text{ ft}}\)

10)A flagpole casts a shadow. The angle of elevation is \(30°\), and the pole is \(12\) ft tall. Find the shadow length.

The pole height is opposite the angle, and the shadow is adjacent.

\(tan \ 30°=\frac{12}{x}\), so \(x=\frac{12}{\sqrt{3}/3}=12\sqrt{3}\).

\(\color{red}{12\sqrt{3}\text{ ft}}\)

11)A right triangle has hypotenuse \(17\) and one leg \(8\). Find the other leg.

Use \(a^2+8^2=17^2\).

\(a^2=289-64=225\), so \(a=15\).

\(\color{red}{15}\)

12)In a right triangle, angle \(A=35°\) and the hypotenuse is \(20\). Find the side opposite \(A\) to the nearest tenth.

Use \(sin \ A=\frac{opposite}{hypotenuse}\).

\(x=20\sin 35°\approx20(0.5736)=11.5\).

\(\color{red}{11.5}\)

13)In a right triangle, angle \(A=52°\) and the adjacent side is \(14\). Find the hypotenuse to the nearest tenth.

Use \(cos \ A=\frac{adjacent}{hypotenuse}\).

\(cos \ 52°=\frac{14}{h}\), so \(h=\frac{14}{cos \ 52°}\approx22.7\).

\(\color{red}{22.7}\)

14)From a point \(80\) ft from a building, the angle of elevation to the top is \(40°\). Find the building height to the nearest tenth.

The height is opposite the angle, and \(80\) ft is adjacent.

\(tan \ 40°=\frac{h}{80}\), so \(h=80tan \ 40°\approx67.1\).

\(\color{red}{67.1\text{ ft}}\)

15)A right triangle has legs \(9\) and \(12\). If angle \(A\) is opposite the \(9\)-unit leg, find \(sin \ A\).

First find the hypotenuse: \(c^2=9^2+12^2=225\), so \(c=15\).

\(sin \ A=\frac{opposite}{hypotenuse}=\frac{9}{15}=\frac{3}{5}\).

\(\color{red}{\frac{3}{5}}\)

16)A right triangle has legs \(7\) and \(24\). If angle \(A\) is adjacent to the \(24\)-unit leg, find \(cos \ A\).

Find the hypotenuse: \(c^2=7^2+24^2=625\), so \(c=25\).

\(cos \ A=\frac{adjacent}{hypotenuse}=\frac{24}{25}\).

\(\color{red}{\frac{24}{25}}\)

17)A right triangle has hypotenuse \(26\). For angle \(B\), the opposite side is \(10\). Find \(tan \ B\).

Find the adjacent side: \(a^2+10^2=26^2\), so \(a^2=576\) and \(a=24\).

\(tan \ B=\frac{opposite}{adjacent}=\frac{10}{24}=\frac{5}{12}\).

\(\color{red}{\frac{5}{12}}\)

18)In a right triangle, angle \(A=28°\) and the opposite side is \(15\). Find the adjacent side to the nearest tenth.

Use \(tan \ A=\frac{opposite}{adjacent}\).

\(tan \ 28°=\frac{15}{x}\), so \(x=\frac{15}{tan \ 28°}\approx28.2\).

\(\color{red}{28.2}\)

19)A ramp rises \(4\) ft over a horizontal run of \(18\) ft. Find the angle the ramp makes with the ground to the nearest tenth.

Use \(tan \theta=\frac{rise}{run}=\frac{4}{18}\).

\(\theta=tan^{-1}(\frac{4}{18})\approx12.5°\).

\(\color{red}{12.5°}\)

20)A right triangle has hypotenuse \(41\) and one leg \(40\). Find the smaller acute angle to the nearest tenth.

Find the other leg: \(a^2+40^2=41^2\), so \(a=9\). The smaller angle is opposite \(9\): \(\theta=\sin^{-1}(\frac{9}{41})\approx12.7°\).

\(\color{red}{12.7^{\circ}}\)

Missing Sides and Angles of a Right Triangle Practice Quiz