Question

At the foot of the mountain the elevation of its summit is found to be ${45^0}$.  After ascending 1000 m towards the mountain up a slope of inclination ${30^0}$ inclination, the elevation is found to be ${60^0}$ . The height of the mountain is
\[
  A.{\text{ }}\left( {\dfrac{{\sqrt 3 + 1}}{2}} \right)m \\
  B.{\text{ }}\left( {\dfrac{{\sqrt 3 - 1}}{2}} \right)m \\
  C.{\text{ }}\left( {\dfrac{{\sqrt 3 + 1}}{{2\sqrt 3 }}} \right)m \\
  D.{\text{ None of these}} \\
 \]

Answer 1

Hint- In order to solve the problem, first of all draw a pictorial representation of the problem statement. Then use the concept of heights and distance along with trigonometric identities and the values for trigonometric terms at some particular angle in order to proceed.

Complete step-by-step answer:

In the given figure AB represents the mountain and h is the height of the mountain, O was the initial viewing point and after ascending 1km or 1000 mt up the viewing point changes to C.
Let us now solve the problem geometrically rather than practically.
$
  {\text{In }}\Delta OCE \\
  \sin \angle COE = \dfrac{{CE}}{{OC}} \\
  \sin {30^0} = \dfrac{{CE}}{{1000}} \\
 $
Let us now substitute the values known
\[
   \Rightarrow \dfrac{1}{2} = \dfrac{{CE}}{{1000}}\;\;{\text{ }}\left[ {\because \sin {{30}^0} = \dfrac{1}{2}} \right] \\
   \Rightarrow CE = \dfrac{1}{2} \times 1000mt = 500mt \\
 \]
Also we have for the same triangle
$
  {\text{In }}\Delta OCE \\
  \cos \angle COE = \dfrac{{OE}}{{OC}} \\
  \cos {30^0} = \dfrac{{OE}}{{1000}} \\
 $
Let us now substitute the values known
\[
   \Rightarrow \dfrac{{\sqrt 3 }}{2} = \dfrac{{OE}}{{1000}}\;\;{\text{ }}\left[ {\because \cos {{30}^0} = \dfrac{{\sqrt 3 }}{2}} \right] \\
   \Rightarrow OE = \dfrac{{\sqrt 3 }}{2} \times 1000mt = 500\sqrt 3 mt \\
 \]
Now let us consider triangle AOB
$
  {\text{In }}\Delta AOB \\
  \tan \angle AOB = \dfrac{{AB}}{{OA}} \\
  \tan {45^0} = \dfrac{{AB}}{{OA}} \\
 $
Let us now substitute the values known
\[
   \Rightarrow 1 = \dfrac{{AB}}{{OA}}\;\;{\text{ }}\left[ {\because \tan {{45}^0} = 1} \right] \\
   \Rightarrow OA = AB \\
 \]
Let us find sides of upper triangle
$
  CD = EA = OA - OE \\
  CD = h - 500\sqrt 3 \\
  BD = BA - DA = BA - CE \\
  BD = h - 500 \\
 $
Now let us consider triangle BCD
$
  {\text{In }}\Delta BCD \\
  \tan \angle BCD = \dfrac{{BD}}{{CD}} \\
  \tan {60^0} = \dfrac{{BD}}{{CD}} \\
 $
Now, let us substitute all the values in above equation
$
  \tan {60^0} = \dfrac{{BD}}{{CD}} \\
   \Rightarrow \sqrt 3 = \dfrac{{h - 500}}{{h - 500\sqrt 3 }}{\text{ }}\left[ {\because \tan {{60}^0} = \sqrt 3 } \right] \\
 $
Now let us solve the above equation for the value of h
\[
   \Rightarrow \left( {h - 500\sqrt 3 } \right)\sqrt 3 = h - 500 \\
   \Rightarrow \sqrt 3 h - 1500 = h - 500 \\
   \Rightarrow \sqrt 3 h - h = 1500 - 500 \\
   \Rightarrow h\left( {\sqrt 3 - 1} \right) = 1000 \\
   \Rightarrow h = \dfrac{{1000}}{{\left( {\sqrt 3 - 1} \right)}} \\
 \]
Now, rationalizing the RHS in order to find the value of h
\[
   \Rightarrow h = \dfrac{{1000}}{{\left( {\sqrt 3 - 1} \right)}} \times \dfrac{{\left( {\sqrt 3 + 1} \right)}}{{\left( {\sqrt 3 + 1} \right)}} \\
   \Rightarrow h = \dfrac{{1000\left( {\sqrt 3 + 1} \right)}}{{{{\left( {\sqrt 3 } \right)}^2} - {{\left( 1 \right)}^2}}} \\
   \Rightarrow h = \dfrac{{1000\left( {\sqrt 3 + 1} \right)}}{{3 - 1}} \\
   \Rightarrow h = \dfrac{{1000\left( {\sqrt 3 + 1} \right)}}{2} \\
   \Rightarrow h = 500\left( {\sqrt 3 + 1} \right) \\
 \]
Hence, the height of the mountain is \[500\left( {\sqrt 3 + 1} \right)\] .
So, option D is the correct option.

Note- In order to solve such questions related to the concept of heights and distances the figure plays the most important role so the problem must be started with the figure. Also the students must remember the values of trigonometric functions of some important angles as they are used several times.