Chapter 9: Some Applications of Trigonometry (Heights and Distances)

In right △ABC, $\sin 30°=\frac{AB}{AC}\Rightarrow AB=20\times \frac{1}{2}=10$ m.

The pole is 10 m high.

In right △ABC, $\tan 30°=\frac{AB}{AC}\Rightarrow AB=\frac{8}{\sqrt{3}}$ m, and $\cos 30°=\frac{AC}{BC}\Rightarrow BC=\frac{8}{\sqrt{3}/2}=\frac{16}{\sqrt{3}}$ m.

Height of the tree $=AB+BC=\frac{24}{\sqrt{3}}=8\sqrt{3}$ m.

8√3 m (≈ 13.86 m)

The slide is the hypotenuse, so length $=\frac{\text{height}}{\sin \theta }$:

• Smaller slide: $\frac{1.5}{\sin 30°}=\frac{1.5}{1/2}=3$ m
• Steeper slide: $\frac{3}{\sin 60°}=\frac{3}{\sqrt{3}/2}=\frac{6}{\sqrt{3}}=2\sqrt{3}$ m

3 m and 2√3 m (≈ 3.46 m)

$\tan 30°=\frac{h}{30}\Rightarrow h=\frac{30}{\sqrt{3}}=10\sqrt{3}$ m

10√3 m (≈ 17.32 m)

$\sin 60°=\frac{60}{l}\Rightarrow l=\frac{60}{\sqrt{3}/2}=\frac{120}{\sqrt{3}}=40\sqrt{3}$ m

40√3 m (≈ 69.28 m)

The top is $30-1.5=28.5$ m above his eyes.

• From P: $\tan 30°=\frac{28.5}{\text{distance}}\Rightarrow$ distance $=28.5\sqrt{3}$ m
• From Q: $\tan 60°=\frac{28.5}{\text{distance}}\Rightarrow$ distance $=\frac{28.5}{\sqrt{3}}$ m

Distance walked $=28.5\sqrt{3}-\frac{28.5}{\sqrt{3}}=28.5\times \frac{2}{\sqrt{3}}=\frac{57}{\sqrt{3}}=19\sqrt{3}$ m

19√3 m (≈ 32.9 m)

In △ABC: $\tan 45°=\frac{20}{AB}\Rightarrow AB=20$ m.

In △ABD: $\tan 60°=\frac{20+h}{20}\Rightarrow 20+h=20\sqrt{3}\Rightarrow h=20(\sqrt{3}-1)$ m.

20(√3 – 1) m (≈ 14.64 m)

Let the pedestal be $h$ m high and the point be $x$ m from its foot.

• $\tan 45°=\frac{h}{x}\Rightarrow x=h$
• $\tan 60°=\frac{h+1.6}{x}\Rightarrow h+1.6=h\sqrt{3}$

So $h=\frac{1.6}{\sqrt{3}-1}=\frac{1.6(\sqrt{3}+1)}{2}=0.8(\sqrt{3}+1)$ m.

0.8(√3 + 1) m (≈ 2.19 m)

Let the distance between them be $d$.

• Tower: $\tan 60°=\frac{50}{d}\Rightarrow d=\frac{50}{\sqrt{3}}$
• Building: $\tan 30°=\frac{h}{d}\Rightarrow h=\frac{d}{\sqrt{3}}=\frac{50}{3}$

50/3 m = 16⅔ m

Let P be $x$ m from pole AB and $80-x$ m from pole CD, with height $h$.

• $\tan 60°=\frac{h}{x}\Rightarrow h=\sqrt{3}x$
• $\tan 30°=\frac{h}{80-x}\Rightarrow h=\frac{80-x}{\sqrt{3}}$

So $3x=80-x\Rightarrow x=20$ and $h=20\sqrt{3}$.

Each pole is 20√3 m (≈ 34.64 m) high; the point is 20 m from one pole and 60 m from the other.

Let the width be $BC=x$ and the height $AB=h$.

• $\tan 60°=\frac{h}{x}\Rightarrow h=\sqrt{3}x$
• $\tan 30°=\frac{h}{x+20}\Rightarrow h=\frac{x+20}{\sqrt{3}}$

So $3x=x+20\Rightarrow x=10$ and $h=10\sqrt{3}$.

The tower is 10√3 m (≈ 17.32 m) high and the canal is 10 m wide.

• Depression of D is 45°, so $BD=AB=7$ m, and $AE=7$ m.
• In △AEC: $\tan 60°=\frac{CE}{AE}\Rightarrow CE=7\sqrt{3}$ m.

Height $CD=CE+ED=7\sqrt{3}+7=7(\sqrt{3}+1)$ m.

7(√3 + 1) m (≈ 19.12 m)

The angle of depression equals the angle of elevation from the ship (alternate angles).

• Ship C (45°): $BC=\frac{75}{\tan 45°}=75$ m
• Ship D (30°): $BD=\frac{75}{\tan 30°}=75\sqrt{3}$ m

Distance $CD=75\sqrt{3}-75=75(\sqrt{3}-1)$ m.

75(√3 – 1) m (≈ 54.9 m)

The balloon is $88.2-1.2=87$ m above her eyes.

• At 60°: $AM=\frac{87}{\sqrt{3}}=29\sqrt{3}$ m
• At 30°: $AN=87\sqrt{3}$ m

Distance $PQ=MN=87\sqrt{3}-29\sqrt{3}=58\sqrt{3}$ m.

58√3 m (≈ 100.5 m)

Let the tower be $h$ m high.

• At 30°: $BD=h\sqrt{3}$
• At 60°: $BC=\frac{h}{\sqrt{3}}$

In 6 s the car covers $DC=h\sqrt{3}-\frac{h}{\sqrt{3}}=\frac{2h}{\sqrt{3}}$. The remaining distance $BC=\frac{h}{\sqrt{3}}$ is half of that, and the speed is uniform.

3 seconds