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