Trigonometry — Mathslive.ie

Section 1 of 6

SOH CAH TOA

In a right-angled triangle, name the sides from the angle $A$: the hypotenuse is opposite the right angle, the opposite is across from $A$, the adjacent is beside $A$.

The three ratios

$\sin A = \dfrac{\text{opp}}{\text{hyp}}$ $\cos A = \dfrac{\text{adj}}{\text{hyp}}$ $\tan A = \dfrac{\text{opp}}{\text{adj}}$

Remember it

Silly Old Harry · Caught A Herring · Trawling Off America.

Section 2 of 6

Finding a Side

Worked example — opposite over hypotenuse

Find $x$ if $\sin 30^{\circ} = \dfrac{x}{10}$ (hypotenuse $10$, opposite $x$).

$x = 10\sin 30$

$x = 5$

Worked example — adjacent over hypotenuse

A right triangle has hypotenuse $12$, angle $47^{\circ}$, adjacent $x$. Find $x$ (1 d.p.).

$\cos 47 = \dfrac{x}{12} \Rightarrow x = 12\cos 47$

$x = 8.2$

Worked example — opposite over adjacent

A right triangle has adjacent $15$, angle $49^{\circ}$, opposite $x$. Find $x$ (1 d.p.).

$\tan 49 = \dfrac{x}{15} \Rightarrow x = 15\tan 49$

$x = 17.3$

Worked example — adjacent again

A right triangle has hypotenuse $16$, angle $37^{\circ}$, adjacent $x$. Find $x$ (1 d.p.).

$\cos 37 = \dfrac{x}{16} \Rightarrow x = 16\cos 37$

$x = 12.8$

Worked example — the unknown on the bottom

Find $x$ if $\sin 53^{\circ} = \dfrac{7}{x}$.

$x\sin 53 = 7 \Rightarrow x = \dfrac{7}{\sin 53}$

$x \approx 8.8$

Worked example — tan with the unknown below

Find $x$ if $\tan 73^{\circ} = \dfrac{8}{x}$.

$x = \dfrac{8}{\tan 73}$

$x = 2.4$

Section 3 of 6

Pythagoras

When you have two sides and no angle, use Pythagoras: $\text{hyp}^2 = \text{opp}^2 + \text{adj}^2$.

Worked example — find the hypotenuse

A right triangle has the two shorter sides $9$ and $12$. Find the hypotenuse.

$\text{hyp}^2 = 9^2 + 12^2 = 81 + 144 = 225$

$\text{hyp} = 15$

Worked example — an isosceles right triangle

$abc$ is isosceles with $|ac|=|bc|$, $|ab|=\sqrt{50}$ and $|\angle acb|=90^{\circ}$. Find $|bc|$.

$x^2 + x^2 = (\sqrt{50})^2 \Rightarrow 2x^2 = 50$

$x^2 = 25$

$|bc| = 5$

Section 4 of 6

Finding an Angle

When you know two sides, the ratio gives the angle — use the inverse ($\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$) on the calculator (in degrees).

Inverse on the calculator

$\sin A = 0.5 \Rightarrow A = 30^{\circ}$ $\cos A = 0.8 \Rightarrow A = 37^{\circ}$ $\tan A = \sqrt{3} \Rightarrow A = 60^{\circ}$

Worked example — tan of an angle

A right triangle has opposite $5$ and adjacent $7$. Find $A$ (1 d.p.).

$\tan A = \dfrac{5}{7} \Rightarrow A = \tan^{-1}(0.714)$

$A = 35.5^{\circ}$

Worked example — sin of an angle

A right triangle has opposite $7$ and hypotenuse $12$. Find $A$.

$\sin A = \dfrac{7}{12} \Rightarrow A = \sin^{-1}(0.583)$

$A = 35.7^{\circ}$

Section 5 of 6

Elevation & Depression

The angle of elevation is measured up from the horizontal; the angle of depression down from it. Draw the right triangle and use SOH CAH TOA.

Worked example — elevation, finding a height

The angle of elevation to the top of a house is $72^{\circ}$ from a point $5$ m away. Find the height of the house (1 d.p.).

$\tan 72 = \dfrac{x}{5} \Rightarrow x = 5\tan 72$

$x = 15.4$ m

Worked example — elevation, finding an angle (ladder)

A ladder $8$ m long rests against a vertical wall with its foot $3$ m from the wall. Find the angle of elevation $A$.

$\cos A = \dfrac{3}{8} \Rightarrow A = \cos^{-1}(0.375)$

$A = 68^{\circ}$

Worked example — a shorter ladder

A ladder $6$ m long leans against a $5$ m wall (reaching the top). Find the angle of elevation $A$.

$\sin A = \dfrac{5}{6} \Rightarrow A = \sin^{-1}(0.833)$

$A = 56^{\circ}$

Now you try

You try

From the top of a cliff a buoy is seen at an angle of depression of $51^{\circ}$. The buoy is $6.3$ m from the foot. How high is the cliff?

Pen and paper out — try it before you reveal.

Depression $=$ angle at the buoy (alternate). $\tan 51 = \dfrac{x}{6.3}$

$x = 6.3\tan 51$

$x = 7.8$ m

You try

From the top of a $12$ m cliff the angle of depression to a boat changes from $39^{\circ}$ to $43^{\circ}$. Find how far the boat moved, $|AB|$.

Pen and paper out — try it before you reveal.

Far: $\tan 51 = \dfrac{x}{12} \Rightarrow x = 12\tan 51 = 14.8$

Near: $\tan 47 = \dfrac{y}{12} \Rightarrow y = 12\tan 47 = 12.9$

$|AB| = 14.8 - 12.9$

$|AB| \approx 1.9$ m

Section 6 of 6

Longer Problems

You try

$abc$ has $|bc|=6$; $d$ on $[ab]$ with $cd \perp ab$, $|cd|=4$, $|ad|=9$. Find $|\angle cbd|$ and $|\angle cad|$ (nearest degree).

Pen and paper out — try it before you reveal.

$\sin(\angle cbd) = \dfrac{4}{6} \Rightarrow \angle cbd = 42^{\circ}$

$\tan(\angle cad) = \dfrac{4}{9} \Rightarrow \angle cad = 24^{\circ}$

$|\angle cbd| = 42^{\circ},\ |\angle cad| = 24^{\circ}$

You try

$abcd$ is a triathlon course: swim $9$ km $a\!\to\!b$, run $12$ km $b\!\to\!c$, cycle $c\!\to\!d\!\to\!a$. $cd \perp$ the altitude, $|\angle adc| = 36.87^{\circ}$, and the altitude from $a$ is $15$. Find $|ac|$, $|cd|$ and the total course length.

Pen and paper out — try it before you reveal.

$|ac|$: right triangle legs $9,12$, so $|ac|^2 = 9^2+12^2 = 225 \Rightarrow |ac| = 15$

$|cd|$: $\tan 36.87 = \dfrac{15}{|cd|} \Rightarrow |cd| = \dfrac{15}{\tan 36.87} = 20$

$|ad|$: $|ad|^2 = 15^2 + 20^2 = 625 \Rightarrow |ad| = 25$

Total $= 9 + 12 + 20 + 25$

$|ac|=15,\ |cd|=20,$ total $= 66$ km

That’s Trigonometry.

SOH CAH TOA for sides and angles, Pythagoras, elevation & depression, and full exam problems — the whole of Junior Cert trigonometry.