Q19. 
Answer key shows: height of tower = 118.25 m 
But my answer comes out to be 118.49 m. 
Will my marks be deducted in boards if my answer is just 0.24 m wrong?

19. The angle of depression of the top and bottom of a 50 m high building from the top of a tower are 45^o and 60^o respectively. Find the height of the tower and the horizontal distance between the tower and the building. (Use $\sqrt{3}$=1.73

Dear Student,

Please find below the solution to the asked query:

From given information we form our diagram , As :

Here , Height of building = AD  = 50 m  , From diagram we get :  AD =  BE = 50 m

And

Height of tower =  BC

And , Horizontal distance between tower and building = AB = DE 

We know : $\tan \mathrm{\theta } = \frac{\mathrm{Opposite}}{\mathrm{Adjacent}}$ , So

In triangle DEC we get :

$$\begin{gathered} \tan \left(\angle \mathrm{EDC}\right) = \frac{\mathrm{EC}}{\mathrm{DE}} \\ \Rightarrow \tan 45° = \frac{\mathrm{EC}}{\mathrm{AB}} \\ \mathrm{We} \mathrm{know} : \mathrm{We} \mathrm{know} : \tan 45° =1 \\ \mathbf{\Rightarrow }\mathbf{AB}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{EC}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{ }\mathbf{(}\mathbf{\hspace{0.17em}}\mathbf{1}\mathbf{\hspace{0.17em}}\mathbf{)} \end{gathered}$$


And in triangle ABC we get :

$$\begin{gathered} \tan \left(\angle \mathrm{BAC}\right) = \frac{\mathrm{BC}}{\mathrm{AB}} \\ \Rightarrow \tan 60° = \frac{\mathrm{BE} + \mathrm{EC}}{\mathrm{AB}} \\ \mathrm{We} \mathrm{know} : \tan 60° =\sqrt{3} \mathrm{and} \mathrm{from} \mathrm{equation} 1 \mathrm{we} \mathrm{get} : \\ \Rightarrow \sqrt{3} = \frac{50 +\hspace{0.17em}\mathrm{AB}}{\mathrm{AB}} \\ \Rightarrow \mathrm{AB}\sqrt{3} = 50 +\hspace{0.17em}\mathrm{AB} \\ \Rightarrow \mathrm{AB} \sqrt{3} - \mathrm{AB} = 50 \\ \Rightarrow \mathrm{AB} \left(\sqrt{3} -1 \right)= 50 \\ \mathrm{Given} : \sqrt{3} = 1.73 , \mathrm{So} \\ \Rightarrow \mathrm{AB} \left(1.73 -1 \right)= 50 \\ \Rightarrow \mathrm{AB} \left(0.73\right)= 50 \\ \Rightarrow \mathrm{AB} = \frac{50}{ 0.73} \\ \mathbf{\Rightarrow }\mathbf{AB}\mathbf{ }\mathbf{=}\mathbf{68}\mathbf{.}\mathbf{49} \end{gathered}$$
So,
Horizontal distance between tower and building = 68.49 m

And

Height of tower = BC =  BE + EC  =  BE + AB  =  50 + 68.49 = 118.49 m

So, answer provided by you is absolutely correct for given query , May be there is printing mistake on the answer key .



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