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)

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 Mathematics

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