Question

If the angles of elevation of the top of a tower from three collinear points A, B and C on a line leading to the foot of the tower, are \[{30^ \circ }\], \[{45^ \circ }\] and \[{60^ \circ }\] respectively, then the ratio AB : BC, is:
A) \[\sqrt 3 :1\]
B) \[\sqrt 3 :\sqrt 2 \]
C) \[1 :\sqrt 3 \]
D) \[2:3\]

Answer 1

Hint: Here we will assume the height of the tower as h and then find \[\tan {45^ \circ },\tan {60^ \circ },\tan {30^ \circ }\]in terms of ratio of height and base of the triangle and then finally find the ratio of AB: BC.
\[\tan \theta = \dfrac{{perpendicular}}{{base}}\]

Complete step-by-step answer:
Let us assume the height of the tower as h

We know that:
\[\tan \theta = \dfrac{{perpendicular}}{{base}}\]
Now in\[\Delta OAD\],
Evaluating \[\tan {30^ \circ }\] we get:-
\[\tan {30^ \circ } = \dfrac{{OD}}{{AD}}\]
We know that:
\[\tan {30^ \circ } = \dfrac{1}{{\sqrt 3 }}\]
\[OD = h\]
Hence putting in the values we get:-
\[\dfrac{1}{{\sqrt 3 }} = \dfrac{h}{{AD}}\]
\[ \Rightarrow AD = \sqrt 3 h\]……………………………..(1)
In\[\Delta OBD\],
Evaluating \[\tan {45^ \circ }\] we get:-
\[\tan {45^ \circ } = \dfrac{{OD}}{{BD}}\]
We know that:
\[\tan {45^ \circ } = 1\]
\[OD = h\]
Hence putting in the values we get:-
\[1 = \dfrac{h}{{BD}}\]
\[ \Rightarrow BD = h\]…………………………….(2)
In\[\Delta OCD\],
Evaluating \[\tan {60^ \circ }\] we get:-
\[\tan {60^ \circ } = \dfrac{{OD}}{{CD}}\]
We know that:
\[\tan {60^ \circ } = \sqrt 3 \]
\[OD = h\]
Hence putting in the values we get:-
\[\sqrt 3 = \dfrac{h}{{CD}}\]
\[ \Rightarrow CD = \dfrac{h}{{\sqrt 3 }}\]…………………………….(3)
Now we know that:-
\[AB = AD - BD\]
Hence putting in the respective value from equation 1 and 2 we get:-
\[AB = \sqrt 3 h - h\]
Taking h common we get:-
\[ \Rightarrow AB = h\left( {\sqrt 3 - 1} \right)\]………………………(4)
Now we know that:-
\[BC = BD - CD\]
Hence putting in the respective value from equation 2 and 3 we get:-
\[BC = h - \dfrac{h}{{\sqrt 3 }}\]
Solving it further we get:-
\[BC = \dfrac{{\sqrt 3 h - h}}{{\sqrt 3 }}\]
Taking h common we get:-
\[ \Rightarrow BC = \dfrac{{h\left( {\sqrt 3 - 1} \right)}}{{\sqrt 3 }}\]………………………(5)
Now we have to find the ratio AB: BC
Hence dividing (4) by (5) we get:-
\[\dfrac{{AB}}{{BC}} = \dfrac{{h\left( {\sqrt 3 - 1} \right)}}{{\dfrac{{h\left( {\sqrt 3 - 1} \right)}}{{\sqrt 3 }}}}\]
Solving it further we get:-
\[\dfrac{{AB}}{{BC}} = h\left( {\sqrt 3 - 1} \right) \times \dfrac{{\sqrt 3 }}{{h\left( {\sqrt 3 - 1} \right)}}\]
Cancelling the terms we get:-
\[\dfrac{{AB}}{{BC}} = \dfrac{{\sqrt 3 }}{1}\]
Hence, \[AB:BC = \sqrt 3 :1\]

Therefore, option A is correct.

Note: Students might note in making the diagram, hence they should note that as we move towards the tower the angle of elevation increases so the angle of elevation at point C which is nearest to the tower will be \[{60^ \circ }\] and at point A it would be \[{30^ \circ }\].