Named Triangles & Pythagoras

Section 1 of 3

Naming Sides in Triangle ABC

Read the letters

In a triangle with a right angle, use the given angle to label opp / adj / hyp, then choose the ratio as before. $|AB|$ means the length of the side from $A$ to $B$.

Worked example — two sides from the hypotenuse

In $\triangle ABC$, $|\angle ABC| = 90^\circ$, $|\angle ACB| = 33^\circ$ and $|AC| = 12\,\text{m}$. Find $|AB|$ and $|BC|$.

$|AC| = 12$ is the hypotenuse; the angle is at $C$

$|AB|$ is opposite: $12\sin 33 = 6.53 = 6.5\,\text{m}$

$|BC|$ is adjacent: $12\cos 33 = 10.06 = 10.1\,\text{m}$

$|AB| = 6.5\,\text{m},\ |BC| = 10.1\,\text{m}$

Worked example — finding a named angle

In $\triangle PQR$, $|\angle PQR| = 90^\circ$, $|PQ| = 5\,\text{cm}$ and $|QR| = 7\,\text{cm}$. Find $|\angle QPR|$.

At $P$: opposite $= |QR| = 7$, adjacent $= |PQ| = 5$ $\Rightarrow$ $\tan$

$\tan P = \dfrac{7}{5} \Rightarrow P = 54.46 = 54.5^\circ$

$|\angle QPR| = 54.5^\circ$

Section 2 of 3

Pythagoras with Trig

Missing the third side

If you have two sides but need the third, use Pythagoras: $H^2 = O^2 + A^2$ (hypotenuse squared = sum of the other two squared).

Worked example — Pythagoras then an angle

In $\triangle PQR$, $|\angle PQR| = 90^\circ$, $|PR| = 10$ (hypotenuse) and $|PQ| = 8$. Find $|QR|$ and $|\angle PRQ|$.

Pythagoras: $|QR|^2 + 8^2 = 10^2 \Rightarrow |QR|^2 = 36 \Rightarrow |QR| = 6$

At $R$: $\sin R = \dfrac{8}{10} \Rightarrow R = 53.1^\circ$

$|QR| = 6,\ |\angle PRQ| = 53.1^\circ$

Worked example — hypotenuse and angle

In $\triangle ABC$, $|\angle ABC| = 90^\circ$, $|AB| = 5$ and $|BC| = 6$. Find $|AC|$ and $|\angle ACB|$.

$|AC|^2 = 5^2 + 6^2 = 61 \Rightarrow |AC| = \sqrt{61} = 7.8$

At $C$: $\tan C = \dfrac{5}{6} \Rightarrow C = 39.8^\circ$

$|AC| = 7.8,\ |\angle ACB| = 39.8^\circ$

Section 3 of 3

Two Triangles Joined

Some questions have two right-angled triangles sharing a side. Find the shared side first, then move to the second triangle.

Worked example — a shared height

$|\angle ABC| = 90^\circ$, $|AB| = 7$ and $|\angle ACB| = 50^\circ$. Find $|AC|$ and $|BC|$.

$|AB| = 7$ is opposite $50^\circ$; $|AC|$ is the hypotenuse: $|AC| = \dfrac{7}{\sin 50} = 9.1$

$|BC|$ is adjacent: $|BC| = \dfrac{7}{\tan 50} = 5.9$

$|AC| = 9.1,\ |BC| = 5.9$

You try

$|\angle ABC| = 90^\circ$ with the side opposite $A$ equal to $12$ and the side adjacent to $A$ equal to $8$. Find $|\angle BAC|$.

$\tan A = \tfrac{12}{8}$.

$A = \tan^{-1}\!\left(\tfrac{12}{8}\right) = 56.3^\circ$

$56.3^\circ$

You try

$\triangle ABC$ has hypotenuse $|AB| = 10$, angle $|\angle ABC| = 20^\circ$, and an altitude drops from $A$ to $D$ making $|\angle ADC| = 56^\circ$. Find (i) $|AC|$; (ii) $|AD|$.

First triangle: $|AC| = 10\sin 20$. Then $|AD| = \dfrac{|AC|}{\sin 56}$.

(i) $|AC| = 10\sin 20 = 3.4$

(ii) $|AD| = \dfrac{3.4}{\sin 56} = 4.1$

$|AC| = 3.4,\ |AD| = 4.1$

You try

$|AC| = 8$, $|\angle ACB| = 50^\circ$, and the altitude $|AD|$ meets $BD$. If $|AB| = 7$, find $|BD|$.

$|AD| = 8\sin 50 = 6.1$, then Pythagoras in $\triangle ADB$: $6.1^2 + |BD|^2 = 7^2$.

$|AD| = 8\sin 50 = 6.1$

$6.1^2 + |BD|^2 = 7^2 \Rightarrow |BD|^2 = 11.79 \Rightarrow |BD| = 3.4$

$|BD| = 3.4$

That’s Class 5.

Named triangles, Pythagoras with trig, and two joined triangles. Class 6: the Sine Rule for triangles with no right angle.