Question

A flag pole 18m high casts a shadow 9.6 m long. Find the distance of the top of the pole from the far end of the shadow.\[\]
A.20m.\[\]
B.20.04m.\[\]
C.20.4m.\[\]
D.24m.\[\]

Answer 1

Hint: Draw a diagram of the flagpole and the shadow as lines and then use the properties of the right angled triangle. Use trigonometry in the right angle triangle.

Complete step by step answer:

 We can see from the figure that we have presented BC as the flagpole whose top is at C and the bottom at the ground is at B. The flagpole casts a shadow up to A, which is the far end of the shadow. A,B and C form a right angled triangle named $\Delta ABC$ .\[\]
 We know from the property of right angled triangles that in a right angled triangle there is an angle known as an right angle with measurement ${{90}^{\circ }}$. The side opposite to the right- angle is called hypotenuse and denoted as $h$. The side which is vertical to the ground is called perpendicular and denoted as $p$ .The side which is horizontal or parallel to the ground is called base and denoted as $b$.\[\]
We see in the triangle in the figure the perpendicular is $p=BC$ whose length is given as height in question as 18m. The base is $b=AB$ whose length is given as the distance from bottom of pole to the far end of shadow as 9.6m . We are asked to find out the distance from the top of flag pole to the far end of the shadow that is the hypotenuse $h=AC$.\[\]
The Pythagoras theorem states that “ The square of hypotenuse is the sum of square of base and square of perpendicular. ” We express it in symbols and put the known and unknown values.
\[\begin{align}
  & {{h}^{2}}={{p}^{2}}+{{b}^{2}} \\
 & \Rightarrow h=\sqrt{{{p}^{2}}+{{b}^{2}}} \\
 & \Rightarrow AC=\sqrt{B{{C}^{2}}+A{{B}^{2}}}=\sqrt{{{18}^{2}}+{{\left( 9.6 \right)}^{2}}}=\sqrt{416.16}=20.4 \\
\end{align}\]

Alternatively,
We know from the trigonometric ratio that the tangent of any angle is the ratio of perpendicular to base. So in $\Delta ABC$
\[\begin{align}
  & \tan \left( \angle BAC \right)=\dfrac{BC}{AB}=\dfrac{18}{9.6}=1.785 \\
 & \Rightarrow \angle BAC={{\tan }^{-1}}\left( 1.785 \right) \\
\end{align}\]
We also the sine of the angle is the ratio of perpendicular to hypotenuse. So in $\Delta ABC$
\[\begin{align}
  & \sin \left( \angle BAC \right)=\dfrac{BC}{AC}=\dfrac{18}{h} \\
 & \Rightarrow h=\dfrac{18}{\sin \left( {{\tan }^{-1}}\left( 1.785 \right) \right)}=20.4 \\
\end{align}\]

So, the correct answer is “Option C”.

Note: The question tests your knowledge of Pythagoras theorem . Careful solving of simultaneous equation, substitution will lead us to the correct result. The question can also be framed to ask the angle elevation from the far end.