Question

A chimney 20 meters high standing on top of a building subtends an angle whose tangent is \[\dfrac{1}{6}\] at a distance 70m from the foot of the building. Find the height of the building.

Answer 1

Hint: Try to draw a figure using the given properties. You will get a right-angled triangle with two of its sides given as perpendicular and base and one of the angles is also given and use the tangent function to get an equation.
Complete step by step answer:
First draw the figure of the triangle using the given sides and angles
The figure will somehow look like this

Where AD is the length of the chimney DB is the length of the house BC is the distance between the foot of the building and the point of elevation and \[\theta \] is the angle of elevation.
It is already given that \[AD = 20m\& BC = 70m\] . Let us take the height of the building be h meters.
Now in \[\vartriangle ABC\] Which is a right angled triangle,
Therefore, it is already given that \[\tan \theta = \dfrac{1}{6}\] and we also know that \[\tan \theta = \dfrac{p}{b}\] where p stands for perpendicular and b stands for base. In this question
\[\begin{array}{l}
p = AB = 20 + h\\
b = BC = 70
\end{array}\]
Now from these two values we can get the equation
\[\tan \theta = \dfrac{p}{b} = \dfrac{{AB}}{{BC}} = \dfrac{{20 + h}}{{70}} = \dfrac{1}{6}\]
By solving this we will get it as
\[\begin{array}{l}
 \Rightarrow \dfrac{{20 + h}}{{70}} = \dfrac{1}{6}\\
 \Rightarrow 70 = 6(20 + h)\\
 \Rightarrow 70 = 120 + 6h\\
 \Rightarrow 6h = 50\\
 \Rightarrow h = \dfrac{{50}}{6} = \dfrac{{25}}{3}
\end{array}\]
Therefore the height of the building is \[\dfrac{{25}}{3}\] metres.

Note: Sketching figures is a must in these height and distance problems. It gives you a better understanding of what values you have and what you need to find out. 20m is the height of the chimney given students often make the mistake to consider it as a height of AB i.e., height of the building + chimney.