Right Triangles Let You Measure The "Unmeasurable"
- Right angles appear everywhere: walls meeting floors, books resting on tables, streets meeting at corners.
- Because right triangles are so common, a large part of geometry and trigonometry is built around them.
- The key idea is that if you know enough information about a right triangle, you can calculate the parts you cannot measure directly (for example, the height of a building or the width of a river).
Right triangle
A triangle with one angle equal to $90^\circ$.
In right triangle geometry, there are two main tools:
- the Pythagorean Theorem, which relates the three side lengths
- the trigonometric ratios (sine, cosine, tangent), which connect angles to side ratios
The Pythagorean Theorem Relates The Three Sides
Consider a right triangle with legs (the two shorter sides) of lengths $a$ and $b$, and hypotenuse (the longest side, opposite the right angle) of length $c$.
Hypotenuse
The side opposite the right angle in a right triangle, always the longest side.
Pythagorean theorem
In any right triangle, the squares of the two shorter sides add to the square of the hypotenuse: $a^2+b^2=c^2$.
This theorem is powerful because it lets you find a missing side length if you know the other two.
Finding A Missing Side Using Pythagoras
- If you know $a$ and $b$, then $$c=\sqrt{a^2+b^2}$$
- If you know $c$ and one leg (say $a$), then $$b=\sqrt{c^2-a^2}$$
A right triangle has legs $6\text{ cm}$ and $8\text{ cm}$. Find the hypotenuse.
Solution
$$\;c=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10\text{ cm}$$
- Always check that you are squaring the two legs and that $c$ is the hypotenuse.
- A common mistake is to treat a shorter side as $c$, which makes $c^2-a^2$ negative.
The Converse Helps You Test If A Triangle Is Right-Angled
Sometimes you are given three side lengths and need to decide whether the triangle is right-angled.
Converse
A statement formed by switching the “if” part and the “then” part of a conditional statement.
The converse of the Pythagorean Theorem is true:
If the side lengths of a triangle satisfy $a^2+b^2=c^2$ (where $c$ is the longest side), then the triangle is a right triangle.
This gives a practical test:
- identify the longest side and call it $c$
- compute $a^2+b^2$ and compare with $c^2$
Acute Or Obtuse? Comparing $a^2+b^2$ With $c^2$
For any triangle with longest side $c$:
- if $a^2+b^2=c^2$, the triangle is right-angled
- if $a^2+b^2>c^2$, the triangle is acute (all angles less than $90^\circ$)
- if $a^2+b^2<c^2$, the triangle is obtuse (one angle greater than $90^\circ$)
- This works because enlarging the angle opposite $c$ makes $c$ grow compared with $a$ and $b$.
- The right triangle case is the "boundary" between acute and obtuse.
Trigonometric Ratios Connect Angles To Side Ratios
- In a right triangle, pick one of the acute angles, call it $A$.
- Relative to angle $A$, each side has a name:
- opposite: the side directly across from angle $A$
- adjacent: the side next to angle $A$ (but not the hypotenuse)
- hypotenuse: always opposite the $90^\circ$ angle
Sine
For an acute angle $A$ in a right triangle, $\sin A=\dfrac{\text{opposite}}{\text{hypotenuse}}$.
Cosine
For an acute angle $A$ in a right triangle, $\cos A=\dfrac{\text{adjacent}}{\text{hypotenuse}}$.
Tangent
For an acute angle $A$ in a right triangle, $\tan A=\dfrac{\text{opposite}}{\text{adjacent}}$.
- These are called ratios because they compare side lengths.
- If you scale the triangle up or down, the ratios stay the same, so the trig ratios depend only on the angle.
- Think of a right triangle as a "shape recipe."
- Changing its size changes the ingredients (side lengths), but the recipe proportions (trig ratios) stay fixed for a given angle.
Choosing The Correct Ratio
To solve a right-triangle problem, you typically know one angle and one side, and want another side.
A useful checklist is:
- label the sides as opposite/adjacent/hypotenuse relative to your chosen angle
- choose the trig ratio that involves the side you know and the side you want
- rearrange the equation and solve
A right triangle has angle $A=35^\circ$ and hypotenuse $c=12\text{ m}$. Find the side opposite $A$.
Solution
Use sine: $$\sin 35^\circ=\dfrac{\text{opposite}}{12}$$
$$\text{opposite}=12\sin 35^\circ\approx 6.88\text{ m}$$
In exam questions, write the ratio statement before using your calculator, for example $$\sin 35^\circ=\dfrac{x}{12}$$
This shows clear method marks even if your final decimal is slightly off.
Special Right Triangles Give Exact Trig Values
Some angles occur so often that it is worth knowing their trig ratios exactly (as radicals, also called surds).
The $45^\circ$-$45^\circ$-$90^\circ$ Triangle
An isosceles right triangle has equal legs and two equal acute angles.
If each leg is $1$, then by Pythagoras the hypotenuse is
$$c=\sqrt{1^2+1^2}=\sqrt{2}$$
Thus: $$\sin 45^\circ=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$$ $$ \cos 45^\circ=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$$ $$ \tan 45^\circ=1$$
The $30^\circ$-$60^\circ$-$90^\circ$ Triangle
- Split an equilateral triangle of side length $2$ down the middle.
- This creates a right triangle with angles $30^\circ$, $60^\circ$, $90^\circ$, hypotenuse $2$, short leg $1$, and long leg $\sqrt{3}$.
- Thus: $$\sin 30^\circ=\frac12$$ $$ \cos 30^\circ=\frac{\sqrt{3}}{2} $$ $$ \tan 30^\circ=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$$ $$\sin 60^\circ=\frac{\sqrt{3}}{2}$$ $$ \cos 60^\circ=\frac12$$ $$ \tan 60^\circ=\sqrt{3}$$
Memorize the side ratios:
- $45$-$45$-$90$: $1:1:\sqrt2$
- $30$-$60$-$90$: $1:\sqrt3:2$ (short : long : hypotenuse)
From these, you can reconstruct all the trig values quickly.
- Be careful about which angle is $30^\circ$ and which is $60^\circ$.
- The $30^\circ$ angle is always opposite the shortest side.
Using Right Triangles In Coordinate Geometry: The Distance Formula
- Right triangle geometry connects directly to the coordinate plane.
- Suppose you have two points $P_1(x_1,y_1)$ and $P_2(x_2,y_2)$.
- Draw a point $Q$ that is horizontally level with $P_1$ and vertically aligned with $P_2$.
- Then $P_1QP_2$ is a right triangle because $P_1Q$ is horizontal and $QP_2$ is vertical, and horizontal and vertical lines meet at $90^\circ$.
- The horizontal distance is $|x_2-x_1|$ and the vertical distance is $|y_2-y_1|$. Using Pythagoras: $$d^2=(x_2-x_1)^2+(y_2-y_1)^2$$
- So the distance formula is: $$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$
Find the distance between $P_1(2,3)$ and $P_2(8,-1)$.
Solution
$$d=\sqrt{(8-2)^2+(-1-3)^2}=\sqrt{6^2+(-4)^2}$$
$$=\sqrt{36+16}=\sqrt{52}=2\sqrt{13}\approx 7.21$$
Angle Size And Side Size Move Together
- In any triangle (including right triangles), larger angles are opposite longer sides.
- For the special triangles above:
- the greatest angle is $90^\circ$, opposite the longest side (the hypotenuse)
- the smallest acute angle is opposite the shortest leg
- This is a useful sense-check: if your calculations suggest that the side opposite a small angle is longer than the side opposite a large angle, something has gone wrong.
- State the Pythagorean Theorem and its converse.
- In a right triangle, what side is "adjacent" to angle $A$?
- Without a calculator, write $\sin 60^\circ$ and $\tan 30^\circ$ in surd form.
- Two sides of a triangle are 7 and 24, and the longest side is 25. Is it a right triangle? Explain.