Year 10 Core — Trigonometric Ratios (6A)

Today's lesson

We're starting Chapter 6 — Trigonometry. By the end of today you should be able to:

Learning intentions

Identify the hypotenuse, opposite and adjacent sides relative to a given angle in a right-angled triangle
Know the three trig ratios — sine, cosine and tangent — and the SOHCAHTOA mnemonic
Pick the right ratio for a given triangle based on which sides and angles you know
Find an unknown side length in a right-angled triangle using a trig ratio

Part 1 — The three ratios (warm up, ~6 min)

In any right-angled triangle, label the three sides relative to a chosen angle $\theta$:

the hypotenuse is the long side opposite the right angle (it never changes)
the opposite side is across from $\theta$
the adjacent side is the remaining side, next to $\theta$

📺 Walkthrough: labelling the sides of a right triangle, then the three SOHCAHTOA ratios.

The three ratios — SOHCAHTOA

$\sin\theta=\dfrac{\text{opposite}}{\text{hypotenuse}}=\dfrac{O}{H}$ $\cos\theta=\dfrac{\text{adjacent}}{\text{hypotenuse}}=\dfrac{A}{H}$ $\tan\theta=\dfrac{\text{opposite}}{\text{adjacent}}=\dfrac{O}{A}$

SOH-CAH-TOA — Sine = Opp/Hyp, Cos = Adj/Hyp, Tan = Opp/Adj.

Building understanding — use a calculator (degree mode), 3 d.p.

$\cos 37^\circ$
$\sin 72^\circ$
$\tan 50^\circ$
$\cos 21.4^\circ$

a) $0.799$
b) $0.951$
c) $1.192$
d) $0.931$

Building understanding — choose the ratio for each triangle (don't solve yet):

a) angle 29°, hyp 7, find x (adj)

b) angle 53°, adj 2, find x (opp)

c) angle 75°, hyp 10.3, find x (opp)

a) hyp + adj → use $\cos$ b) adj + opp → use $\tan$ c) hyp + opp → use $\sin$

Part 2 — Solving for the numerator (Example 1, ~12 min)

When the unknown is on top of the fraction (i.e. $\cos 27^\circ=\dfrac{x}{11}$), you can solve in one step by multiplying both sides by the bottom number.

EXAMPLE 1a — Watch the walkthrough first

Find $x$, correct to 2 d.p.

📺 Pick the right ratio (cos), set up the equation, multiply through.

$x$ is adjacent to $27^\circ$, and $11$ is the hypotenuse → use $\cos$.
$\cos 27^\circ=\dfrac{x}{11}$
$x=11\times\cos 27^\circ=\boxed{9.80\text{ cm}}$ (2 d.p.)

Now you try: Hyp 3 cm, angle 24°, find adjacent $x$. Answer: $x=3\cos 24^\circ\approx 2.74$ cm.

EXAMPLE 1b

Find $x$ where the side $38$ mm is adjacent to a $69^\circ$ angle, and $x$ mm is the opposite side. Round to 2 d.p.

$x$ is opposite, $38$ is adjacent → use $\tan$.
$\tan 69^\circ=\dfrac{x}{38}$
$x=38\times\tan 69^\circ=\boxed{98.99\text{ mm}}$ (2 d.p.)

Now you try: Adj 21 m, angle 50°, find opposite $x$. Answer: $x=21\tan 50^\circ\approx 25.03$ m.

Practice 2.1 — find $x$ (unknown is the numerator), 2 d.p.

$\sin 35^\circ=\dfrac{x}{12}$
$\cos 42^\circ=\dfrac{x}{9}$
$\tan 28^\circ=\dfrac{x}{5}$
$\cos 61^\circ=\dfrac{x}{8.4}$
$\sin 70^\circ=\dfrac{x}{15}$
$\tan 54^\circ=\dfrac{x}{4.6}$

a) $x=12\sin 35^\circ\approx 6.88$
b) $x=9\cos 42^\circ\approx 6.69$
c) $x=5\tan 28^\circ\approx 2.66$
d) $x=8.4\cos 61^\circ\approx 4.07$
e) $x=15\sin 70^\circ\approx 14.10$
f) $x=4.6\tan 54^\circ\approx 6.33$

Part 3 — Solving for the denominator (Example 2, ~10 min)

If the unknown is on the bottom of the fraction, you need two steps: cross-multiply first, then divide.

EXAMPLE 2a

$\sin 33^\circ=\dfrac{21}{x}$ (opposite 21 m, hypotenuse $x$ m). Find $x$.

$\sin 33^\circ=\dfrac{21}{x}$
Multiply both sides by $x$: $x\times\sin 33^\circ=21$
Divide both sides by $\sin 33^\circ$: $x=\dfrac{21}{\sin 33^\circ}=\boxed{38.56\text{ m}}$ (2 d.p.)

Now you try: $\sin 31^\circ=\dfrac{6}{x}$. Answer: $x=\dfrac{6}{\sin 31^\circ}\approx 11.65$ m.

EXAMPLE 2b

$\tan 53^\circ=\dfrac{71.3}{x}$. Find $x$, 2 d.p.

$\tan 53^\circ=\dfrac{71.3}{x}$
$x\times\tan 53^\circ=71.3$
$x=\dfrac{71.3}{\tan 53^\circ}=\boxed{53.73}$ (2 d.p.)

Now you try: $\tan 46^\circ=\dfrac{27.2}{x}$. Answer: $x\approx 26.27$.

Practice 3.1 — find $x$ (unknown is the denominator), 2 d.p.

$\sin 40^\circ=\dfrac{14}{x}$
$\cos 38^\circ=\dfrac{25}{x}$
$\tan 52^\circ=\dfrac{9}{x}$
$\sin 72^\circ=\dfrac{30}{x}$
$\cos 45^\circ=\dfrac{18}{x}$
$\tan 67^\circ=\dfrac{12.5}{x}$

a) $x=14/\sin 40^\circ\approx 21.78$
b) $x=25/\cos 38^\circ\approx 31.73$
c) $x=9/\tan 52^\circ\approx 7.03$
d) $x=30/\sin 72^\circ\approx 31.54$
e) $x=18/\cos 45^\circ\approx 25.46$
f) $x=12.5/\tan 67^\circ\approx 5.31$

Part 4 — Quick quiz (5 min)

Working program — Cambridge Ex 6A

After the quiz, open Cambridge Chapter 6 and complete the following:

Section	Foundation	Standard (set work)	Advanced
Fluency 1–3	1–2 (½)	1–2 (½)	1–2 (⅓), 3
Problem-solving 4–9	4, 5	5–7	6–9
Reasoning 10–11	10	10	10, 11
Enrichment 12	—	—	12

If you finish early, start Ex 6B — Finding unknown angles using the other lesson page.

Exit ticket — write in your book

Before you pack up, in your exercise book write one sentence each:

Which side of a right triangle is the hypotenuse?
What does the C and the H in CAH stand for?
In a right triangle you know the hypotenuse and an angle. Which ratio finds the side opposite that angle?