Question
A vertical tower standing on a levelled field is mounted with a vertical flag staff of length $$3\,m.$$ From a point on the field, the angles of elevation of the bottom and tip of the flag staff are $${30^ \circ }$$ and $${45^ \circ }$$ respectively. Which one of the following gives the best approximation to the height of the tower ?
A.
$$3.90\,m$$
B.
$$4.00\,m$$
C.
$$4.10\,m$$
D.
$$4.25\,m$$
Answer :
$$4.10\,m$$
Solution :
$${\text{as }}\angle CPA = {45^ \circ }$$
$$\eqalign{ & {\text{so, }}AC = AP = x + 3 \cr & \tan {30^ \circ } = \frac{{AB}}{{AP}} = \frac{x}{{x + 3}} \cr & \frac{1}{{\sqrt 3 }} = \frac{x}{{x + 3}} \cr & x + 3 = \sqrt 3 \,x \cr & x = \frac{3}{{\sqrt 3 - 1}} \times \frac{{\left( {\sqrt 3 + 1} \right)}}{{\left( {\sqrt 3 + 1} \right)}} \cr & x = \frac{{3 \times 2.73}}{2} = \frac{{8.19}}{2} = 4.095\,m \approx 4.1\,m \cr} $$
$${\text{as }}\angle CPA = {45^ \circ }$$
$$\eqalign{ & {\text{so, }}AC = AP = x + 3 \cr & \tan {30^ \circ } = \frac{{AB}}{{AP}} = \frac{x}{{x + 3}} \cr & \frac{1}{{\sqrt 3 }} = \frac{x}{{x + 3}} \cr & x + 3 = \sqrt 3 \,x \cr & x = \frac{3}{{\sqrt 3 - 1}} \times \frac{{\left( {\sqrt 3 + 1} \right)}}{{\left( {\sqrt 3 + 1} \right)}} \cr & x = \frac{{3 \times 2.73}}{2} = \frac{{8.19}}{2} = 4.095\,m \approx 4.1\,m \cr} $$