A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is \[{{60}^{\circ }}\] . From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is \[{{30}^{\circ }}\]. Find the height of the tower and the width of the canal.
VerifiedHint: Consider the maximum height as AB and draw the angle of depression at two different points and apply \[\tan \theta \] to the two right angled triangles and we will get two equations and then we have to compute the maximum height from which he falls.
Complete step-by-step answer:
Let \[ AB=h \]
\[ CA=x \]
\[ CD=20 \]
From right \[\Delta BAC\]
\[\cot {{60}^{\circ }}={x}{h}\]
\[{1}{\sqrt{3}}={x}{h}\]
\[x={h}{\sqrt{3}}\]
From right \[\Delta BAD\]
\[\cot {{30}^{\circ }}={AD}{AB}\]
\[\sqrt{3}={\left( x+20 \right)}{h}\]
\[x=h\sqrt{3}-20\]
From (1) and (2) we get
\[{h}{\sqrt{3}}=h\sqrt{3}-20\]
\[h\sqrt{3}-{h}{\sqrt{3}}=20\]
\[{3h-h}{\sqrt{3}}=20\]
\[h=10\sqrt{3}\].
From (1) \[x={10\sqrt{3}}{\sqrt{3}}\]
x \[=10\]
Therefore the width of the canal is 10m and height of the canal is \[10\sqrt{3}\].
Note: If the object observed by the observer is below the level of the observer, then the angle formed between the horizontal line and the observer’s line of sight is called the angle of depression. As the person moves from one point to another angle of depression varies.
