Answer & Solution
Answer: Option C
Solution:

Let AB be the tower and C and D be the two positions of the car.
Then, \[\angle ACB = {45^ \circ },\] \[\angle ADB = {30^ \circ }\]
Let, AB = h, CD = x and AC = y
$$\eqalign{
& \frac{{AB}}{{AC}} = \tan {45^ \circ } = 1 \cr
& \Rightarrow \frac{h}{y} = 1 \cr
& \Rightarrow y = h \cr
& \frac{{AB}}{{AD}} = \tan {30^ \circ } = \frac{1}{{\sqrt 3 }} \cr
& \Rightarrow \frac{h}{{x + y}} = \frac{1}{{\sqrt 3 }} \cr
& \Rightarrow x + y = \sqrt 3 h \cr
& \therefore x = \left( {x + y - y} \right) \cr
& \,\,\,\,\,\,\, = \sqrt 3 h - h \cr
& \,\,\,\,\,\,\, = h\left( {\sqrt 3 - 1} \right) \cr} $$
Now, $$h\left( {\sqrt 3 - 1} \right)$$ is covered in 12 min.
So, h will be covered in→
$$\eqalign{
& \left[ {\frac{{12}}{{h\left( {\sqrt 3 - 1} \right)}} \times h} \right] \cr
& = \frac{{12}}{{\left( {\sqrt 3 - 1} \right)}}\min \cr
& = \left( {\frac{{1200}}{{73}}} \right)\min \cr
& = 16\min ,\,\,23\sec \cr} $$