Question

A balloon is moving up in air vertically above a point A on the ground. When it is at a height \[ h_1\], a girl standing at a distance d(point B) from A (see figure) sees it at an angle \[{45^ \circ }\] with respect to the vertical. When the balloon climbs up a further height \[ h_2\], it is seen at an angle \[{60^ \circ }\] with respect to the vertical if the girl moves further by a distance 2.464 d (point C). Then the height \[ h_2\] is (given tan \[{30^ \circ }\]° = 0.5774):

Answer 1

Hint: First draw the given diagram with the given information. Then Consider the triangle ABD and obtain the value of the tangent. Then consider the triangle ACE and here also write the tangent value and calculate to obtain the required result.

Formula Used: The tangent of a triangle is
\[\tan \theta = \dfrac{a}{b}\] , where a is the height and b is the base.
\[\tan {45^ \circ } = 1\]
\[\tan {30^ \circ } = 0.5774\]

Complete step by step solution: As the angle A is right angle and the external angle is \[{45^ \circ }\] hence the internal angle of B is also \[{45^ \circ }\].
It is given that the external angle of the point C is \[{60^ \circ }\] therefore the internal angle must be \[{30^ \circ }\] as the line upward from the point C is perpendicular to the line AC.
Hence, the image of the given problem is,




 
Now, from the given image we have,
\[\tan {45^ \circ } = \dfrac{{{h_1}}}{d}\]
\[ \Rightarrow 1 = \dfrac{{{h_1}}}{d}\]
\[ \Rightarrow {h_1} = d\]
So, from the triangle ACE we have,
\[\tan {30^ \circ } = \dfrac{{{h_1} + {h_2}}}{{d + 2.464d}}\]
Substitute \[{h_1} = d\] and \[\tan {30^ \circ } = 0.5774\] in the equation \[\tan {30^ \circ } = \dfrac{{{h_1} + {h_2}}}{{d + 2.464d}}\]for further calculation.
\[0.5774 = \dfrac{{d + {h_2}}}{{3.464d}}\]
\[2d = d + {h_2}\]
\[{h_2} = d\]

Option ‘D’ is correct

Note: To solve this type of problem students must have the knowledge of angles, without the knowledge student cannot be able to draw the required image.