Step 1: Use the given measurements to make a figure.

FROM THE FIGURE: Height of the tower = AB

Width of the road = BD = (BC + CD)

Distance between the two points D and C = 10 cm

Angle of elevation at D = 30\degree

Angle of elevation at C = 60\degree

Step 2: Find the common height of the two triangles by using the trigonometric ratios definitions.

EXAMPLE: We have to find the adjacent, and opposite sides so

take \tan \theta = \frac{opposite}{adjacent}

From \triangle ABD

\tan 30\degree = \frac{AB}{BD}

\frac{1}{\sqrt{3}} = \frac{AB}{BC + CD} ( \because \tan 30\degree = \frac{1}{\sqrt{3}})

AB = \frac{BC + CD}{\sqrt{3}}..............................(1)

From \triangle ABC

\tan 60\degree = \frac{AB}{BC}

AB = BC\sqrt{3} ( \because \tan 60\degree = \sqrt{3} ).............(2)

Step 3: Express BC in terms of CD by solving equation (1) & (2)

Equations:

AB = \frac{BC + CD}{\sqrt{3}}

AB = BC\sqrt{3}

Equate the R.H.S

\sqrt{3}BC = \frac{BC + CD}{\sqrt{3}}

BC = \frac{BC + CD}{(\sqrt{3})^2}

BC + CD = 3BC

CD = 2BC

BC = \frac{CD}{2} ................................(3)

BC = \frac{10}{2} = 5

Step 4: Calculate the width of the road and height of the tower.

Width of the road = BD = BC + CD

BD = 5 + 10 = 15 m

Height of the tower AB = BC \sqrt{3} [ \because equation (2)]

AB = 5 \sqrt{3} m