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Question

Determine the height of a mountain if the elevation of its top at an unknown distance from the base is $$30^{\circ}$$ and at a distance 10 km further off from the mountain, along the same line , the angle of elevation is $$15^{\circ}$$ . (Use tan$$15^{\circ}$$ = 0.27) 
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Solution

Let AB be the mountain of height h kilometer . Let C be point at a distance of x km . from the base of the mountain such that the angle of elevation of the top at C is $$30^{\circ}$$ . Let D be a point at a distance of 10 km from C such that the angle of elevation at D is of$$15^{\circ}$$ 
In $$\triangle $$ CAB , we have 
$$ tan 30^{\circ} \, = \, \dfrac {AB} {AC} $$ 
$$ \Rightarrow \, \, \, \, \dfrac {1}{\sqrt{3}} \, = \, \dfrac {h}{x}$$
$$ \Rightarrow \, \, \, \, x \, = \, \sqrt{3h} $$
In $$ \triangle $$ DAB  we have 
$$ tan15^{\circ} \, = \, \dfrac {AB}{AD} $$ 
$$ \Rightarrow \, \, \, 0.27 \, = \, \dfrac {h}{x \, + \, 10 } $$ 
$$ \Rightarrow $$ (0.27) (x + 10) = h 
substituting x = $$ \sqrt{3h} $$ obtained from equation (i) in equation (ii)  we get 
0.27 ( $$ \sqrt{3h} $$  + 10)  = h 
$$ \Rightarrow  \, 0 . 27  \times \,   10 \,  = \, h \, - \,  0.27  \times \, \, \sqrt{3h} $$ 
$$ \Rightarrow $$ h (1 - 0.27  $$ \times \, \, \sqrt{3})$$  = 2.7
$$ \Rightarrow $$ h (1 - 0.46 )  = 2 . 7 
$$ \Rightarrow \, \, h \, = \, \dfrac {2.7}{0.54} $$ = 5 
Hence  , the height  of the mountains is 5 km 
1017559_1009806_ans_da416b372d944a10a64190ef0ab13e8a.png

Mathematics

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