Solving right triangles
A triangle has three sides and three angles. To solve a triangle means to find all six. In a right triangle, one angle is already known: it is ${90^\circ}$. If you also know two more parts, at least one of them a side, you can find the rest. This page shows how, and then uses it to measure heights and distances.
Names for the parts
We name the angles with capital letters ${A}$, ${B}$, and ${C}$, with ${C}$ the right angle. Each side gets the small letter of the angle across from it.
So ${c}$ is the hypotenuse, ${a}$ is opposite ${A}$, and ${b}$ is opposite ${B}$. Three facts link the parts:
- The acute angles add up to ${90^\circ}$: ${A+B=90^\circ}$.
- The Pythagorean theorem: ${a^2+b^2=c^2}$.
- The trigonometric functions: for example, ${\sin A=\dfrac{a}{c}}$ and ${\tan A=\dfrac{a}{b}}$.
Finding sides
Example 1: Solve the right triangle with ${A=34^\circ}$ and ${c=10}$. Round sides to two decimal places.
Solution:
Angle ${B}$: The acute angles add up to ${90^\circ}$:
${B=90^\circ-34^\circ=56^\circ}$
Side ${a}$: It is opposite ${A}$, and we know the hypotenuse. So use sine:
$\begin{align*}\sin 34^\circ&=\dfrac{a}{10}\\a&=10\sin 34^\circ\approx 5.59\end{align*}$
Side ${b}$: It is adjacent to ${A}$. So use cosine:
$\begin{align*}\cos 34^\circ&=\dfrac{b}{10}\\b&=10\cos 34^\circ\approx 8.29\end{align*}$
Tip: When you can, use the numbers you were given, not ones you rounded along the way. Rounding errors add up. In Example 1, we found ${b}$ from ${c=10}$, not from the rounded value of ${a}$.
Finding angles
Sometimes you know the sides and want an angle. Suppose ${\sin A=0.6}$. Which angle has a sine of ${0.6}$? The calculator key ${\sin^{-1}}$ (often written "asin" or "arcsin") answers this. In degree mode, ${\sin^{-1}(0.6)\approx 36.87^\circ}$. So ${A\approx 36.87^\circ}$.
The keys ${\cos^{-1}}$ and ${\tan^{-1}}$ work the same way. They are the inverse trigonometric functions. They are studied fully in Chapter 4. For now, use them only for acute angles in right triangles.
Watch out: The ${-1}$ in ${\sin^{-1}x}$ is not an exponent. It does not mean ${\dfrac{1}{\sin x}}$. It means “the angle whose sine is ${x}$.”
Example 2: Solve the right triangle with ${a=7}$ and ${b=12}$. Round to two decimal places.
Solution:
Side ${c}$: Use the Pythagorean theorem:
$\begin{align*}c&=\sqrt{7^2+12^2}\\&=\sqrt{193}\approx 13.89\end{align*}$
Angle ${A}$: We know the sides opposite and adjacent to ${A}$. So use tangent:
${\tan A=\dfrac{7}{12}}$
So
$A=\tan^{-1}\dfrac{7}{12}\approx 30.26^\circ$
Angle ${B}$:
$B\approx 90^\circ-30.26^\circ=59.74^\circ$
Angles of elevation and depression
Imagine looking at an object. The straight line from your eye to the object is the line of sight.
- If the object is above you, the angle between the horizontal and your line of sight is the angle of elevation.
- If the object is below you, the angle between the horizontal and your line of sight is the angle of depression.
When two people look at each other, as in the figure, the angle of elevation from the lower one equals the angle of depression from the upper one. This is because the two horizontal lines are parallel.
Example 3: A tree casts a shadow ${25}$ m long. At that moment, the angle of elevation of the sun is ${52^\circ}$. How tall is the tree?
Solution:
Let ${h}$ be the height of the tree. From the ${52^\circ}$ angle, ${h}$ is the opposite side, and the shadow is the adjacent side. So use tangent:
$\begin{align*}\tan 52^\circ&=\dfrac{h}{25}\\h&=25\tan 52^\circ\\&\approx 32.0\end{align*}$
The tree is about ${32.0}$ m tall.
Example 4: From the top of a lighthouse ${40}$ m above the sea, the angle of depression of a boat is ${9^\circ}$. How far is the boat from the base of the lighthouse?
Solution:
The angle of elevation from the boat to the top equals the angle of depression, so the angle at the boat is ${9^\circ}$. Let ${d}$ be the distance. From the ${9^\circ}$ angle, ${40}$ is the opposite side and ${d}$ is the adjacent side:
${\tan 9^\circ=\dfrac{40}{d}}$
Multiply both sides by ${d}$, then divide by ${\tan 9^\circ}$:
$d=\dfrac{40}{\tan 9^\circ}\approx 252.6\text{ m}$
Example 5: From a point ${D}$ on the ground, the angle of elevation of the top of a building is ${30^\circ}$. From a point ${C}$, ${100}$ ft closer to the building, it is ${45^\circ}$. How tall is the building?
Solution:
Let ${h}$ be the height, and let ${x}$ be the distance from ${C}$ to the building. There are two right triangles, so we get two equations.
From ${C}$: ${\tan 45^\circ=\dfrac{h}{x}}$. Since ${\tan 45^\circ=1}$, this gives ${h=x}$.
From ${D}$: The distance is ${x+100}$, so
${\tan 30^\circ=\dfrac{h}{x+100}}$
Use ${\tan 30^\circ=\dfrac{1}{\sqrt{3}}}$ and ${x=h}$:
${\dfrac{1}{\sqrt{3}}=\dfrac{h}{h+100}}$
Cross-multiply, and solve for ${h}$:
$\begin{align*}h+100&=\sqrt{3}\,h\\100&=\sqrt{3}\,h-h\\100&=(\sqrt{3}-1)h\\h&=\dfrac{100}{\sqrt{3}-1}\\&\approx 136.6\end{align*}$
The building is about ${136.6}$ ft tall.
Bearings
In navigation and surveying, a direction is often given as a bearing. A bearing starts from north or south and turns toward east or west. For example:
- N ${40^\circ}$ E means: face north, then turn ${40^\circ}$ toward the east.
- S ${25^\circ}$ W means: face south, then turn ${25^\circ}$ toward the west.
The angle in a bearing is always between ${0^\circ}$ and ${90^\circ}$.
Example 6: A ship leaves port ${O}$ and sails ${20}$ miles on a bearing of S ${50^\circ}$ E to a point ${P}$. It then turns and sails ${15}$ miles on a bearing of N ${40^\circ}$ E to a point ${Q}$. How far is the ship from port? What is its bearing from port?
Solution:
Step 1: Find the right angle. The first course is ${50^\circ}$ from south toward east. The second is ${40^\circ}$ from north toward east. Together they turn ${50^\circ+40^\circ=90^\circ}$, so the angle at ${P}$ is a right angle.
Step 2: Find the distance. The legs are ${20}$ and ${15}$, so
${d=\sqrt{20^2+15^2}=\sqrt{625}=25}$ miles
Step 3: Find the angle at ${O}$. From ${O}$, the side ${15}$ is opposite and the side ${20}$ is adjacent:
$\angle POQ=\tan^{-1}\dfrac{15}{20}\approx 36.9^\circ$
Step 4: Find the bearing. The line ${OP}$ is ${50^\circ}$ from south. The line ${OQ}$ is another ${36.9^\circ}$ farther from south, toward the east. So it is ${50^\circ+36.9^\circ=86.9^\circ}$ from south.
The ship is ${25}$ miles from port, on a bearing of S ${86.9^\circ}$ E.
Summary
- To solve a right triangle, use ${A+B=90^\circ}$, the Pythagorean theorem, and the trigonometric ratios.
- To find a side, choose the function that links the side you want with a side you know.
- To find an angle from two sides, use ${\sin^{-1}}$, ${\cos^{-1}}$, or ${\tan^{-1}}$ on a calculator in degree mode.
- Angles of elevation and depression are measured from the horizontal, up or down.
- A bearing such as N ${40^\circ}$ E starts from north or south and turns toward east or west.