Solutions — Trig 6E Worksheet
Year 10 Mathematics Core · Cambridge Ch 6, §6E · Mr Wong
ANSWER KEY
Warm-up — cube of side 2 m
a $2\sqrt{2}$ m
b $2\sqrt{3}$ m
c $35.3°$
d $45°$
Working: $AC=\sqrt{2^2+2^2}=2\sqrt{2}$; $AD=\sqrt{(2\sqrt{2})^2+2^2}=2\sqrt{3}$; $\tan(\angle DAC)=\dfrac{2}{2\sqrt{2}}=\dfrac{1}{\sqrt{2}}$ so $\angle DAC\approx 35.3°$. Triangle $ABC$ is isoceles right, so $\angle CAB=45°$.
Part A — Mast and cables
2a
$\sin 48^\circ=\dfrac{h}{36}\;\Rightarrow\;h=36\sin 48^\circ\approx$ $26.753$ m
2b
$\tan\theta=\dfrac{26.753}{24}\;\Rightarrow\;\theta\approx$ $48.11°$
3a
cable $=\sqrt{8^2+6^2}=$ $10$ m
3b
$\tan\theta=\dfrac{8}{6}\;\Rightarrow\;\theta\approx$ $53.13°$
Part B — Boxes and prisms
4a
base diag $=\sqrt{3^2+4^2}=$ $5$ cm
4b
space diag $=\sqrt{3^2+4^2+5^2}=\sqrt{50}=$ $5\sqrt{2}\approx 7.07$ cm
4c
$\tan\theta=\dfrac{5}{5}=1\;\Rightarrow\;\theta=$ $45.00°$
5a
space diag $=\sqrt{2^2+3^2+6^2}=\sqrt{49}=$ $7.00$ m
5b
base diag $=\sqrt{13}$; $\tan\theta=\dfrac{6}{\sqrt{13}}\;\Rightarrow\;\theta\approx$ $59.00°$
Part C — Pyramids & tents · Challenge
6a
slant height 5, half-base 3 ⇒ $h=\sqrt{5^2-3^2}=$ $4$ m
6b
$\tan\theta=\dfrac{4}{3}\;\Rightarrow\;\theta\approx$ $53.13°$
7.
$\tan\theta=\dfrac{2.5}{1.8}\;\Rightarrow\;\theta\approx$ $54.25°$
8a
half-diagonal of base $=5\sqrt{2}$; $h=\sqrt{12^2-(5\sqrt{2})^2}=\sqrt{94}\approx$ $9.70$ cm
8b
$\cos\theta=\dfrac{5\sqrt{2}}{12}\;\Rightarrow\;\theta\approx$ $53.90°$
9.
Total height $=40\tan 46.40°\approx 42.00$ m; building 30 m so flagpole $\approx$ $12.00$ m
10.
foot-to-wall corner $=\sqrt{1.2^2+1.5^2}\approx 1.92$ m; ladder $=\sqrt{1.92^2+3^2}\approx$ $3.56$ m