A skateboard ramp, a wheelchair access slope, the beam of a lighthouse sweeping the sea: hidden inside each of them is a right triangle. In this chapter you will learn to calculate missing lengths and angles with the Pythagorean theorem, special triangles, the three trigonometric ratios, and then to go beyond right angles with the Law of Sines and the Law of Cosines.
1. The Pythagorean Theorem and Its Converse
In a right triangle with legs \(a\) and \(b\) and hypotenuse \(c\) (the side opposite the right angle): \[ a^2 + b^2 = c^2 \]
If the side lengths \(a\), \(b\), \(c\) of a triangle, with \(c\) the longest, satisfy \(a^2 + b^2 = c^2\), then the triangle is a right triangle. More precisely, if \(a^2 + b^2 > c^2\) the triangle is acute, and if \(a^2 + b^2 < c^2\) it is obtuse.
A floor measures \(12\) ft by \(9\) ft. How long is a cable stretched along a diagonal?
The diagonal is the hypotenuse of a right triangle with legs \(12\) and \(9\): \(d^2 = 12^2 + 9^2 = 144 + 81 = 225\), so \(d = 15\) ft.
A triangle has sides \(9\), \(40\) and \(41\). Is it a right triangle?
The longest side is \(41\). We compare \(9^2 + 40^2 = 81 + 1600 = 1681\) with \(41^2 = 1681\). They are equal, so by the converse the triangle is a right triangle.
Classify the triangles with sides \(6, 8, 9\) and \(5, 7, 9\).
For \(6, 8, 9\): \(6^2 + 8^2 = 100 > 81 = 9^2\), so the triangle is acute. For \(5, 7, 9\): \(5^2 + 7^2 = 74 < 81 = 9^2\), so the triangle is obtuse.
Three whole numbers that satisfy the theorem form a Pythagorean triple. Any multiple of a triple is another triple, for example \(6, 8, 10\) comes from \(3, 4, 5\).
| Triple | Check |
|---|---|
| \(3,\ 4,\ 5\) | \(9 + 16 = 25\) |
| \(5,\ 12,\ 13\) | \(25 + 144 = 169\) |
| \(8,\ 15,\ 17\) | \(64 + 225 = 289\) |
| \(7,\ 24,\ 25\) | \(49 + 576 = 625\) |
| \(20,\ 21,\ 29\) | \(400 + 441 = 841\) |
Always put the longest side alone on one side of the equation. Writing \(a^2 + c^2 = b^2\) when \(c\) is the hypotenuse gives wrong answers.
2. Special Right Triangles
Two right triangles are so common that you should know their side ratios by heart. Both come from simple shapes: the 45°-45°-90° triangle is half of a square cut along a diagonal, and the 30°-60°-90° triangle is half of an equilateral triangle cut along an altitude.
- 45°-45°-90°: the legs are equal (length \(s\)) and the hypotenuse is \(s\sqrt{2}\).
- 30°-60°-90°: the short leg (opposite \(30^\circ\)) is \(s\), the long leg (opposite \(60^\circ\)) is \(s\sqrt{3}\), and the hypotenuse is \(2s\).
The diagonal of a square picture frame is \(18\) in. Find its side length.
The diagonal is the hypotenuse of a 45°-45°-90° triangle: \(s\sqrt{2} = 18\), so \(s = \dfrac{18}{\sqrt{2}} = 9\sqrt{2} \approx 12.73\) in.
A 30°-60°-90° triangle has hypotenuse \(14\) cm. Find both legs.
The hypotenuse is \(2s = 14\), so the short leg is \(s = 7\) cm. The long leg is \(7\sqrt{3} \approx 12.12\) cm.
3. The Geometric Mean in Right Triangles
The geometric mean of two positive numbers \(a\) and \(b\) is \(\sqrt{ab}\). For example, the geometric mean of \(4\) and \(9\) is \(\sqrt{36} = 6\).
When you draw the altitude from the right angle to the hypotenuse, you create two smaller right triangles that are similar to the original one. This similarity produces two neat relationships.
In right triangle \(ABC\) with the right angle at \(C\) and altitude \(CD\) to the hypotenuse \(AB\):
- The altitude is the geometric mean of the two segments of the hypotenuse: \(CD^2 = AD \cdot DB\).
- Each leg is the geometric mean of the hypotenuse and the segment next to it: \(AC^2 = AD \cdot AB\) and \(BC^2 = DB \cdot AB\).
With \(AD = 4\) and \(DB = 9\): \(CD^2 = 4 \cdot 9 = 36\), so \(CD = 6\). The hypotenuse is \(AB = 4 + 9 = 13\). Then \(AC^2 = 4 \cdot 13 = 52\), so \(AC = 2\sqrt{13} \approx 7.21\), and \(BC^2 = 9 \cdot 13 = 117\), so \(BC = 3\sqrt{13} \approx 10.82\). Check: \(52 + 117 = 169 = 13^2\).
4. Sine, Cosine, and Tangent
When you know one acute angle of a right triangle, the ratios of its sides depend only on that angle, not on the size of the triangle. These ratios have names.
For an acute angle \(\theta\) in a right triangle: \[ \sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}}, \quad \cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}, \quad \tan\theta = \dfrac{\text{opposite}}{\text{adjacent}} \]
Remember SOH-CAH-TOA: Sine is Opposite over Hypotenuse, Cosine is Adjacent over Hypotenuse, Tangent is Opposite over Adjacent. On my home planet we sing it!
A right triangle has legs \(7\) and \(24\) and hypotenuse \(25\). For the angle \(A\) opposite the leg \(7\): \(\sin A = \dfrac{7}{25}\), \(\cos A = \dfrac{24}{25}\), \(\tan A = \dfrac{7}{24}\).
A \(6\) m ladder leans against a wall and makes a \(72^\circ\) angle with the ground. How high does it reach, and how far is its foot from the wall?
Height (opposite side): \(\sin 72^\circ = \dfrac{h}{6}\), so \(h = 6\sin 72^\circ \approx 5.71\) m. Distance (adjacent side): \(d = 6\cos 72^\circ \approx 1.85\) m.
Two more facts help you check answers. The sine of an angle equals the cosine of its complement: \(\sin\theta = \cos(90^\circ - \theta)\). And \(\sin^2\theta + \cos^2\theta = 1\), \(\tan\theta = \dfrac{\sin\theta}{\cos\theta}\). Here are the exact values to memorize.
| Angle | \(\sin\) | \(\cos\) | \(\tan\) |
|---|---|---|---|
| \(30^\circ\) | \(\dfrac{1}{2}\) | \(\dfrac{\sqrt{3}}{2}\) | \(\dfrac{\sqrt{3}}{3}\) |
| \(45^\circ\) | \(\dfrac{\sqrt{2}}{2}\) | \(\dfrac{\sqrt{2}}{2}\) | \(1\) |
| \(60^\circ\) | \(\dfrac{\sqrt{3}}{2}\) | \(\dfrac{1}{2}\) | \(\sqrt{3}\) |
5. Inverse Trigonometric Functions
To find an angle instead of a side, use an inverse function: if \(\tan A = 0.5\), then \(A = \tan^{-1}(0.5)\).
- Mark the angle you use, then label the sides opposite, adjacent, and hypotenuse.
- Choose the ratio that contains the unknown and one known side (SOH, CAH, or TOA).
- Write the equation and solve it. Use \(\sin^{-1}\), \(\cos^{-1}\), or \(\tan^{-1}\) for an angle.
- The other acute angle is \(90^\circ\) minus the first one.
In the triangle of Example 7, find angle \(A\). Since \(\tan A = \dfrac{7}{24}\), \(A = \tan^{-1}\!\left(\dfrac{7}{24}\right) \approx 16.3^\circ\). The other acute angle is \(90^\circ - 16.3^\circ = 73.7^\circ\).
\(\sin^{-1}(x)\) is not \(\dfrac{1}{\sin x}\); it is the angle whose sine is \(x\). Also check that your calculator is in degree mode, not radian mode.
6. Angles of Elevation and Depression
The angle of elevation is measured from the horizontal up to a line of sight. The angle of depression is measured from the horizontal down to a line of sight.
Because the horizontal line at the top is parallel to the ground, the angle of depression from the top of a cliff to a boat equals the angle of elevation from the boat to the top of the cliff (alternate interior angles).
A surveyor stands \(85\) ft from the base of a water tower. Her eyes are \(5.5\) ft above the ground and the angle of elevation to the top is \(34^\circ\). How tall is the tower?
We know the adjacent side (\(85\) ft) and want the opposite side, so use the tangent: \(\tan 34^\circ = \dfrac{x}{85}\), giving \(x = 85\tan 34^\circ \approx 57.3\) ft. Add eye height: \(57.3 + 5.5 \approx 62.8\) ft.
7. The Law of Sines
Not every triangle has a right angle. In any triangle \(ABC\), use \(a\), \(b\), \(c\) for the sides opposite angles \(A\), \(B\), \(C\).
\[ \dfrac{\sin A}{a} = \dfrac{\sin B}{b} = \dfrac{\sin C}{c} \]
Use it when you know two angles and a side (ASA or AAS), or two sides and an angle opposite one of them (SSA).
In triangle \(ABC\), \(A = 40^\circ\), \(B = 75^\circ\) and \(a = 20\). Find \(b\) and \(c\).
First, \(C = 180^\circ - 40^\circ - 75^\circ = 65^\circ\). Then \(b = \dfrac{20\sin 75^\circ}{\sin 40^\circ} \approx 30.05\) and \(c = \dfrac{20\sin 65^\circ}{\sin 40^\circ} \approx 28.20\).
With two sides and a non-included angle (SSA), there can be zero, one, or two triangles. If \(\sin B\) is greater than 1 there is no triangle. If \(\sin B\) gives an acute angle, check whether its supplement \(180^\circ - B\) also fits in the triangle.
8. The Law of Cosines
\[ c^2 = a^2 + b^2 - 2ab\cos C \]
The same pattern works for \(a^2\) and \(b^2\). Use it when you know two sides and the angle between them (SAS) or all three sides (SSS). When \(C = 90^\circ\), \(\cos C = 0\) and you recover the Pythagorean theorem.
Two sides of a triangle measure \(13\) and \(17\), and the angle between them is \(57^\circ\). Find the third side.
\(c^2 = 13^2 + 17^2 - 2(13)(17)\cos 57^\circ = 458 - 442\cos 57^\circ \approx 217.27\), so \(c \approx 14.74\).
A triangle has sides \(7\), \(9\), and \(12\). Find its largest angle.
The largest angle \(C\) is opposite \(12\): \(\cos C = \dfrac{7^2 + 9^2 - 12^2}{2(7)(9)} = \dfrac{-14}{126} \approx -0.111\), so \(C \approx 96.4^\circ\). A negative cosine means an obtuse angle.
Which law should you pick? Use this quick guide.
| You know | Use |
|---|---|
| Right angle plus two sides | Pythagorean theorem or SOH-CAH-TOA |
| AAS or ASA | Law of Sines |
| SSA (angle opposite a known side) | Law of Sines (check the ambiguous case) |
| SAS or SSS | Law of Cosines |
Key takeaways
- Pythagorean theorem: \(a^2 + b^2 = c^2\). Its converse tests for a right angle; comparing \(c^2\) with \(a^2 + b^2\) tells you acute, right, or obtuse.
- Special triangles: 45°-45°-90° has sides \(s, s, s\sqrt{2}\); 30°-60°-90° has sides \(s, s\sqrt{3}, 2s\).
- Geometric mean: \(h^2 = xy\) for the altitude, and leg\(^2\) = (adjacent segment)(hypotenuse).
- SOH-CAH-TOA gives sides from angles; \(\sin^{-1}\), \(\cos^{-1}\), \(\tan^{-1}\) give angles from sides.
- The angle of depression from one point equals the angle of elevation from the other.
- Law of Sines for AAS, ASA, SSA; Law of Cosines for SAS and SSS.
Test yourself: quick challenge for Grade 10
Speed drill for Grade 10: how many in 60 seconds?
🚀 Keep exploring with Zyro
✏️ Math practiceRight Triangles and Trigonometry: math practice, Grade 10
📝 Math testsRight Triangles and Trigonometry: math test, Grade 10
🎯 Math quizzesRight Triangles and Trigonometry: math quiz, Grade 10
✏️ Math practiceCircles: math practice, Grade 10
✏️ Math practiceArea, Surface Area, and Volume: math practice, Grade 10
📝 Math testsCircles: math test, Grade 10