Solutions — Trig 6C Worksheet
Year 10 Mathematics Core · Cambridge Ch 6, §6C · Mr Wong
ANSWER KEY
Warm-up
Q1. Choose the ratio (think: which two sides of the right triangle do you have/want?)
a $\tan$ (opp 800, adj $x$)
b $\tan$ (opp 50, adj $x$)
c $\sin$ (hyp 60, opp $h$)
d $\tan$ (opp 13, adj 40)
Q2. Angle of depression from $B$ to $A$ = $35°$ (equal alternate angles in parallel horizontals — the horizontals at top and bottom are parallel, and the line of sight is a transversal).
Part A — Angle of elevation
3.
$\tan 35^\circ=\dfrac{250}{x}\;\Rightarrow\;x=\dfrac{250}{\tan 35^\circ}\approx$ $357.04$ m horizontal distance
4.
$\sin 62^\circ=\dfrac{h}{80}\;\Rightarrow\;h=80\sin 62^\circ\approx$ $70.64$ m high
5.
$\sin 60^\circ=\dfrac{h}{45}\;\Rightarrow\;h=45\sin 60^\circ\approx$ $38.97$ m tall mast
Part B — Angle of depression
6.
$\Delta h=237-158=79$; $\tan\theta=\dfrac{79}{57}\;\Rightarrow\;\theta\approx$ $54.19°$
7.
Convert $1.5$ km = $1500$ m; $\tan\theta=\dfrac{112}{1500}\;\Rightarrow\;\theta\approx$ $4.27°$
8.
The 25 m wall is opposite the 62° depression. Slant = hyp: $\sin 62^\circ=\dfrac{25}{d}\;\Rightarrow\;d=\dfrac{25}{\sin 62^\circ}\approx$ $28.31$ m
Part C — Mixed application
9.
$\tan 18^\circ=\dfrac{60}{x}\;\Rightarrow\;x=\dfrac{60}{\tan 18^\circ}\approx$ $184.66$ m away
10a
$\sin 11^\circ=\dfrac{h}{5.4}\;\Rightarrow\;h=5.4\sin 11^\circ\approx$ $1.03$ m high
10b
$\cos 11^\circ=\dfrac{x}{5.4}\;\Rightarrow\;x=5.4\cos 11^\circ\approx$ $5.30$ m horizontal
11.
tower height = $30\tan 54^\circ\approx 41.29$; flag-top height = $30\tan 58^\circ\approx 48.01$; flag = $\approx 6.72$ m