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Height and Distance

**Correct Marks**
2

**Incorrectly Marks**
-0.5

Q 1. At a point on a horizontal line through the base of a monument the angle of elevation of the top of the monument is found to be such that its tangent is {tex} \frac{1}{5} {/tex}. On walking 138 metres towards the monument the secant of the angle of elevation is found to be {tex} \frac{\sqrt{193}}{12} {/tex}. The height of the monument (in metre) is

42

49

35

56

{tex} \textbf{Shortcut approach} {/tex}

{tex} \textbf{Ist Case:} {/tex}

{tex} tan \theta =\frac{AB}{BC}=\frac{Perpendicular}{Base}=\frac{1}{5} {/tex}

{tex} \textbf{IInd Case:} {/tex}

{tex} Sec \alpha =\frac{AD}{BD}=\frac{Hypo}{Base} {/tex}

{tex} \frac{\sqrt{193}}{12} {/tex}

{tex} In\ \triangle ABD {/tex}

{tex} Hypo=\sqrt{193} {/tex}

{tex} Base=12 {/tex}

Then perpendicular = 7

(By pythagoras theorem)

In both case perpendicular will be equal

{tex} tan \theta =\frac{1 \times 7}{5 \times 7}=\frac{7\leftarrow Perpen}{35\leftarrow Base} {/tex}

{tex} AB = 42 m {/tex}

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