Year 10 Core — Finding Unknown Angles (6B)

Today's lesson

In 6A you used sin, cos and tan to find unknown sides. In 6B you use the inverse trig functions ($\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$) to find unknown angles. By the end of today you should be able to:

Learning intentions

Know that inverse trig functions reverse sin / cos / tan to give you the angle
Use $\sin^{-1}$, $\cos^{-1}$ and $\tan^{-1}$ on the calculator (often the SHIFT or 2nd key)
Find an unknown angle in a right-angled triangle given any two side lengths
Apply this to short worded problems (ramps, ladders, tunnels)

Part 1 — The inverse idea (warm up, ~5 min)

If $\sin 30^\circ=\tfrac{1}{2}$, then the inverse is $\sin^{-1}\!\left(\tfrac{1}{2}\right)=30^\circ$. The inverse function takes a ratio and gives you back the angle.

📺 Walkthrough: from $\cos\theta=\tfrac{1.5}{2.5}$ to $\theta=\cos^{-1}(0.6)\approx 53.13^\circ$.

The three inverse rules

If $\sin\theta=k$ then $\theta=\sin^{-1}(k)$
If $\cos\theta=k$ then $\theta=\cos^{-1}(k)$
If $\tan\theta=k$ then $\theta=\tan^{-1}(k)$

For $\sin^{-1}$ and $\cos^{-1}$, the input $k$ must satisfy $-1\le k\le 1$. For $\tan^{-1}$, $k$ can be any real number.

Calculator tip. The inverse functions live above the sin / cos / tan keys, usually as SHIFT then sin. Always check you're in degree mode (DEG, not RAD) — most calculator marks come down to this.

Building understanding — find $\theta$ to 2 d.p. where necessary.

$\sin\theta=0.4$
$\cos\theta=0.5$
$\tan\theta=0.2$
$\sin\theta=0.1$

a) $\theta=\sin^{-1}(0.4)\approx 23.58^\circ$
b) $\theta=\cos^{-1}(0.5)=60^\circ$
c) $\theta=\tan^{-1}(0.2)\approx 11.31^\circ$
d) $\theta=\sin^{-1}(0.1)\approx 5.74^\circ$

Part 2 — Finding an angle in a triangle (Example 3, ~12 min)

Step 1: name the two sides you've been given (opp / adj / hyp). Step 2: pick the matching ratio. Step 3: apply the inverse.

EXAMPLE 3a

Find $\theta$ for this triangle (opp $=1$, hyp $=2$).

Opp $=1$ and hyp $=2$ → use $\sin$.
$\sin\theta=\dfrac{1}{2}$
$\theta=\sin^{-1}\!\left(\dfrac{1}{2}\right)=\boxed{30^\circ}$

Now you try: opp $=3$, hyp $=6$. Answer: $\theta=\sin^{-1}(\tfrac{1}{2})=30^\circ$.

EXAMPLE 3b — Watch the walkthrough first

Find $\theta$ (adj $=1.5$, hyp $=2.5$). Round to 2 d.p.

Adj $=1.5$ and hyp $=2.5$ → use $\cos$.
$\cos\theta=\dfrac{1.5}{2.5}=0.6$
$\theta=\cos^{-1}(0.6)=\boxed{53.13^\circ}$ (2 d.p.)

Now you try: adj $=7$, hyp $=9$. Answer: $\theta=\cos^{-1}(\tfrac{7}{9})\approx 38.94^\circ$.

Practice 2.1 — find $\theta$ to 2 d.p.

$\sin\theta=\dfrac{3}{5}$
$\cos\theta=\dfrac{8}{17}$
$\tan\theta=\dfrac{7}{24}$
$\sin\theta=\dfrac{9}{15}$
$\cos\theta=\dfrac{2.4}{6}$
$\tan\theta=\dfrac{5}{12}$

a) $\theta\approx 36.87^\circ$
b) $\theta\approx 61.93^\circ$
c) $\theta\approx 16.26^\circ$
d) $\theta\approx 36.87^\circ$
e) $\theta\approx 66.42^\circ$
f) $\theta\approx 22.62^\circ$

Part 3 — A worded application (Example 4, ~10 min)

When you get a worded problem, draw the triangle first with all the lengths and the angle you want to find. Then pick the ratio.

📺 Walkthrough: drawing a right triangle for a mine-tunnel problem, then using $\tan^{-1}$ to find the angle.

EXAMPLE 4 — Mine tunnel

A long, straight mine tunnel is dug into the ground. Its final depth is $120$ m and the end of the tunnel is $100$ m horizontally from the entrance. Find the angle $\theta$ the tunnel makes with the horizontal, to 1 d.p.

Draw a right triangle: horizontal leg $=100$, vertical leg $=120$, angle $\theta$ between the tunnel and the horizontal at the entrance.
Opp $=120$, adj $=100$ → use $\tan$.
$\tan\theta=\dfrac{120}{100}$
$\theta=\tan^{-1}\!\left(\dfrac{120}{100}\right)=\boxed{50.2^\circ}$ (1 d.p.)

Now you try: a rabbit burrow has depth $4$ m and horizontal run $5$ m. Answer: $\theta=\tan^{-1}(\tfrac{4}{5})\approx 38.66^\circ$ ($\approx 38.7^\circ$ to 1 d.p.).

Practice 3.1 — short applications, answer to 1 d.p.

A ladder of length $4$ m has its base $1.2$ m from a wall. What angle does the ladder make with the ground?
A skateboard ramp rises $0.8$ m over a horizontal run of $5$ m. What is the angle of the ramp?
A right triangle has opposite side $9$ and hypotenuse $11$. What is the angle?

a) $\cos\theta=\tfrac{1.2}{4}\Rightarrow\theta\approx 72.5^\circ$
b) $\tan\theta=\tfrac{0.8}{5}\Rightarrow\theta\approx 9.1^\circ$
c) $\sin\theta=\tfrac{9}{11}\Rightarrow\theta\approx 54.9^\circ$

Part 4 — Quick quiz (5 min)

Working program — Cambridge Ex 6B

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

Section	Foundation	Standard (set work)	Advanced
Fluency 1–3	1–2 (½)	1–3 (½)	1–3 (⅓)
Problem-solving 4–7	4, 5	5, 6	6, 7
Reasoning 8–10	8	8, 9	9, 10
Enrichment 11	—	—	11

If you finish early, get started on Ex 6C — Applications in two dimensions (angle of elevation / depression). We'll cover that one on Wednesday.

Exit ticket — write in your book

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

What does $\sin^{-1}$ do in plain English?
If your calculator is in radians, what will $\cos^{-1}(0.5)$ give you instead of $60^\circ$?
If you know the opposite and the adjacent sides, which inverse function finds the angle?