Pythagoras and Trigonometry

Pythagoras' theorem

$a^2 + b^2 = c^2$ — where $c$ is the hypotenuse

The hypotenuse is the longest side, always opposite the right angle. It never touches the right angle.

Identify the hypotenuse (opposite the right angle).
Decide whether you are finding the hypotenuse or one of the shorter sides.
Finding the hypotenuse: square, add, square root.
Finding a shorter side: square, subtract, square root.
Check: the hypotenuse must be the biggest answer.

Worked Example 1 — Finding the hypotenuse

A right-angled triangle has legs $7$ cm and $24$ cm. Find the hypotenuse.

①$c^2 = 7^2 + 24^2 = 49 + 576 = 625$

②$c = \sqrt{625} = 25$ cm

Worked Example 2 — Finding a shorter side

A right-angled triangle has hypotenuse $17$ cm and one leg $8$ cm. Find the other leg.

①Here we subtract, because we want a shorter side.

②$b^2 = 17^2 - 8^2 = 289 - 64 = 225$

③$b = \sqrt{225} = 15$ cm

Adding here would give $\sqrt{353} = 18.8$ — longer than the hypotenuse, which is impossible. Always sense-check.

Pythagorean triples worth recognising: $3,4,5$ · $5,12,13$ · $8,15,17$ · $7,24,25$ · $9,40,41$. Multiples also work: $6,8,10$ and $9,12,15$.

Worked Example 3 — Testing for a right angle

A triangle has sides $9$ cm, $12$ cm and $16$ cm. Is it right-angled?

①The longest side would be the hypotenuse: $16^2 = 256$.

②The other two: $9^2 + 12^2 = 81 + 144 = 225$.

③$225 \neq 256$, so it is not right-angled.

Since $225 \lt 256$, the largest angle is actually obtuse.

Pythagoras links three sides. Trigonometry brings angles into the picture.

SOH CAH TOA

$\sin\theta = \dfrac{O}{H}$ · $\cos\theta = \dfrac{A}{H}$ · $\tan\theta = \dfrac{O}{A}$

Labelling the sides — always relative to the angle you are using:
H — the hypotenuse, opposite the right angle (this never changes)
O — the side opposite the angle $\theta$
A — the side adjacent to (next to) $\theta$, which is not the hypotenuse

O and A swap if you use the other angle. The hypotenuse never moves, but "opposite" and "adjacent" always depend on which angle you are working from. Relabel the triangle every time.

Mark the angle you are given, then label the three sides H, O and A.
Circle the two sides involved: the one you know and the one you want.
Choose the ratio that uses exactly those two — SOH, CAH or TOA.
Write the equation and rearrange.
Calculate, checking your calculator is in degrees mode.

Worked Example 4 — Opposite side

A right-angled triangle has hypotenuse $20$ cm and an angle of $37^\circ$. Find the side opposite that angle, to 1 d.p.

①We know H and want O → use SOH.

②$\sin 37^\circ = \dfrac{O}{20}$

③$O = 20 \times \sin 37^\circ = 20 \times 0.60182 = 12.0$ cm

Worked Example 5 — When the unknown is on the bottom

A right-angled triangle has an angle of $28^\circ$ with an opposite side of $9$ cm. Find the hypotenuse, to 1 d.p.

①We know O and want H → use SOH.

②$\sin 28^\circ = \dfrac{9}{H}$

③Multiply both sides by $H$: $\;H \sin 28^\circ = 9$

④$H = \dfrac{9}{\sin 28^\circ} = \dfrac{9}{0.46947} = 19.2$ cm

If the unknown is underneath, swap it with the answer: $H = \dfrac{O}{\sin\theta}$.

Worked Example 6 — Using tan

A ladder leans against a wall at $65^\circ$ to the ground. Its foot is $1.8$ m from the wall. How far up the wall does it reach, to 2 d.p.?

①The $1.8$ m is adjacent to the angle; the height up the wall is opposite → use TOA.

②$\tan 65^\circ = \dfrac{h}{1.8}$

③$h = 1.8 \times \tan 65^\circ = 1.8 \times 2.14451 = 3.86$ m

Inverse functions

$\theta = \sin^{-1}\left(\dfrac{O}{H}\right)$ · $\theta = \cos^{-1}\left(\dfrac{A}{H}\right)$ · $\theta = \tan^{-1}\left(\dfrac{O}{A}\right)$

On your calculator, $\sin^{-1}$, $\cos^{-1}$ and $\tan^{-1}$ are usually the SHIFT (or 2nd function) of the ordinary keys. They are sometimes labelled "arcsin", "arccos", "arctan".

Worked Example 7 — Finding an angle

A right-angled triangle has an opposite side of $7$ cm and a hypotenuse of $11$ cm. Find the angle, to 1 d.p.

①O and H → use SOH.

②$\sin\theta = \dfrac{7}{11} = 0.63636$

③$\theta = \sin^{-1}(0.63636) = 39.5^\circ$

Worked Example 8 — Angle from two legs

A ramp rises $1.2$ m over a horizontal distance of $8$ m. Find the angle of elevation, to 1 d.p.

①Opposite $= 1.2$, adjacent $= 8$ → use TOA.

②$\tan\theta = \dfrac{1.2}{8} = 0.15$

③$\theta = \tan^{-1}(0.15) = 8.5^\circ$

Angle of elevation — measured upwards from the horizontal.
Angle of depression — measured downwards from the horizontal.
Both are measured from a horizontal line, never from the vertical.

They are equal. The angle of elevation from the ground to the top of a tower equals the angle of depression from the top of the tower back down — they are alternate angles between two horizontal (hence parallel) lines.

Worked Example 9 — Angle of depression

From the top of a $45$ m cliff, the angle of depression to a boat is $22^\circ$. How far is the boat from the base of the cliff, to the nearest metre?

①The angle of depression from the top equals the angle of elevation from the boat, so use $22^\circ$ at the boat.

②Opposite $= 45$ m (the cliff), adjacent $= d$ (the distance) → use TOA.

③$\tan 22^\circ = \dfrac{45}{d}$

④$d = \dfrac{45}{\tan 22^\circ} = \dfrac{45}{0.40403} = 111.4$

About $111$ m.

3D problems are always solved by finding a suitable right-angled triangle inside the solid and then using the 2D methods.

Space diagonal of a cuboid

$d = \sqrt{l^2 + w^2 + h^2}$

Sketch the solid and mark the line you want.
Find a right-angled triangle that contains it.
If a side of that triangle is unknown, find it first with a second right-angled triangle (usually on the base).
Apply Pythagoras or trigonometry.
Keep full accuracy between the two stages.

Worked Example 10 — 3D diagonal and angle

A cuboid measures $6$ cm by $8$ cm by $5$ cm. Find (a) the length of the space diagonal, (b) the angle the diagonal makes with the base.

①(a) First the base diagonal: $\sqrt{6^2 + 8^2} = \sqrt{100} = 10$ cm.

②Now the space diagonal, using that base diagonal and the height:

③$d = \sqrt{10^2 + 5^2} = \sqrt{125} = 11.18$ cm (2 d.p.)

④(b) In that vertical triangle, opposite $= 5$ (height), adjacent $= 10$ (base diagonal).

⑤$\tan\theta = \dfrac{5}{10} = 0.5$, so $\theta = \tan^{-1}(0.5) = 26.6^\circ$ (1 d.p.)

Check with the one-step formula: $\sqrt{6^2+8^2+5^2} = \sqrt{36+64+25} = \sqrt{125}$ ✓

Pythagoras

$a^2 + b^2 = c^2$, with $c$ the hypotenuse.

Hypotenuse

Square, add, square root.

Shorter side

Square, subtract, square root.

SOH CAH TOA

$\sin = \tfrac{O}{H}$, $\cos = \tfrac{A}{H}$, $\tan = \tfrac{O}{A}$.

Finding an angle

Use $\sin^{-1}$, $\cos^{-1}$ or $\tan^{-1}$.

Unknown on the bottom

Swap it with the answer: $H = \tfrac{O}{\sin\theta}$.

Elevation / depression

Measured from the horizontal; the two are equal.

3D

Find a right-angled triangle inside the solid; often two stages.

Calculator

Degrees mode. Round only at the very end.

Question 1

A right-angled triangle has legs $9$ cm and $40$ cm. Find the hypotenuse.

▶ Show solution

$c^2 = 9^2 + 40^2 = 81 + 1600 = 1681$

$c = \sqrt{1681} = 41$ cm

Question 2

A right-angled triangle has hypotenuse $26$ cm and one leg $10$ cm. Find the other leg.

▶ Show solution

$b^2 = 26^2 - 10^2 = 676 - 100 = 576$

$b = \sqrt{576} = 24$ cm

Question 3

Is a triangle with sides $8$ cm, $15$ cm and $17$ cm right-angled? Show your working.

▶ Show solution

$8^2 + 15^2 = 64 + 225 = 289$

$17^2 = 289$

They are equal, so yes, the triangle is right-angled (with the right angle between the $8$ and $15$ sides).

Question 4

A right-angled triangle has hypotenuse $15$ cm and an angle of $42^\circ$. Find the side adjacent to that angle, to 1 d.p.

▶ Show solution

A and H → use CAH.

$\cos 42^\circ = \dfrac{A}{15}$

$A = 15 \times \cos 42^\circ = 15 \times 0.74314 = 11.1$ cm

Question 5

In a right-angled triangle, the side opposite an angle is $12$ cm and the adjacent side is $5$ cm. Find the angle, to 1 d.p.

▶ Show solution

O and A → use TOA.

$\tan\theta = \dfrac{12}{5} = 2.4$

$\theta = \tan^{-1}(2.4) = 67.4^\circ$

Question 6

A ladder $6$ m long leans against a wall, reaching $5.2$ m up. Find the angle it makes with the ground, to 1 d.p.

▶ Show solution

The ladder is the hypotenuse ($6$ m); the height is opposite ($5.2$ m).

$\sin\theta = \dfrac{5.2}{6} = 0.86667$

$\theta = \sin^{-1}(0.86667) = 60.1^\circ$

Question 7

From a point $80$ m from the base of a tower, the angle of elevation to the top is $34^\circ$. Find the height of the tower, to 1 d.p.

▶ Show solution

Adjacent $= 80$, opposite $= h$ → use TOA.

$\tan 34^\circ = \dfrac{h}{80}$

$h = 80 \times \tan 34^\circ = 80 \times 0.67451 = 54.0$ m

Question 8

Find the distance between the points $A(2,\ 3)$ and $B(10,\ 9)$.

▶ Show solution

$\Delta x = 8$, $\Delta y = 6$ — these are the two legs of a right-angled triangle.

$AB = \sqrt{8^2 + 6^2} = \sqrt{100} = 10$

Question 9

An isosceles triangle has two equal sides of $13$ cm and a base of $10$ cm.

(a) Find its perpendicular height. (b) Find its area. (c) Find its apex angle, to 1 d.p.

▶ Show solution

(a) The perpendicular from the apex bisects the base, giving a right-angled triangle with hypotenuse $13$ and one leg $5$.

$h^2 = 13^2 - 5^2 = 169 - 25 = 144$, so $h = 12$ cm.

(b) Area $= \tfrac{1}{2} \times 10 \times 12 = 60\text{ cm}^2$.

(c) In the half-triangle, the angle at the apex has opposite $5$ and hypotenuse $13$.

$\sin\alpha = \dfrac{5}{13}$, so $\alpha = 22.62^\circ$.

The full apex angle is $2\alpha = \mathbf{45.2^\circ}$ (1 d.p.).

Question 10

A square-based pyramid has base edges $10$ cm and vertical height $12$ cm. $M$ is the centre of the base and $V$ is the apex.

(a) Find the distance from $M$ to a base corner. (b) Find the length of a sloping edge $VA$. (c) Find the angle a sloping edge makes with the base, to 1 d.p.

▶ Show solution

(a) The base diagonal $= \sqrt{10^2 + 10^2} = \sqrt{200} = 14.142$ cm.

$M$ is the centre, so $MA = \dfrac{14.142}{2} = 7.071$ cm.

(b) Triangle $VMA$ is right-angled at $M$, with $VM = 12$ and $MA = 7.071$.

$VA = \sqrt{12^2 + 7.071^2} = \sqrt{144 + 50} = \sqrt{194} = \mathbf{13.93}$ cm (2 d.p.).

(c) The angle at $A$ in triangle $VMA$: opposite $= 12$, adjacent $= 7.071$.

$\tan\theta = \dfrac{12}{7.071} = 1.6971$

$\theta = \tan^{-1}(1.6971) = \mathbf{59.5^\circ}$ (1 d.p.).