a) 17.3 m
b) 21.9 m
c) 27.3 m
d) 30 m

Answer: c
Explanation:

Let AB be the tower and C and D be the points of observation.
Then, ∠ACB = 30°, ∠ADB = 45° and CD = 20m
Let AB = h then,
$$\eqalign{ & \frac{{AB}}{{AC}} = \tan 30^\circ = \frac{1}{{\sqrt 3 }} \cr & \Rightarrow AC = AB \times \sqrt 3 = h\sqrt 3 {\kern 1pt} {\text{And,}} \cr & \Rightarrow \frac{{AB}}{{AD}} = \tan 45^\circ = 1 \cr & \Rightarrow AD = AB = h \cr & \, \, \, \, \, CD = 20 \cr & \Rightarrow \left( {AC – AD} \right) = 20 \cr & \Rightarrow h\sqrt 3 – h = 20 \cr & h = \frac{{20}}{{\left( {\sqrt 3 – 1} \right)}} \times \frac{{\left( {\sqrt 3 + 1} \right)}}{{\left( {\sqrt 3 + 1} \right)}} \cr & = 10\left( {\sqrt 3 + 1} \right){\text{m}} \cr & = \left( {10 \times 2.73} \right){\text{m}} \cr & = 27.3 {\text{m}} \cr} $$