1. A tower standing on a horizontal plane subtends a certain angle at a point 160 m apart from the foot of the tower. On advancing 100 m towards it, the tower is found to subtend an angle twice as before. The height of the tower is
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- By: anil on 05 May 2019 02.23 am
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Given : CD is the tower, AD = 160 m and AB = 100 m => BD = 160 - 100 = 60 m To find : CD = $$h$$ = ? Solution : $$angle$$ DBC = $$2 heta$$ and $$angle$$ DAC = $$ heta$$ In $$ riangle$$ ACD, => $$tan( heta)=frac{CD}{DA}$$ => $$tan( heta)=frac{h}{160}$$ -----------(i) Similarly, in $$ riangle$$ BCD, => $$tan(2 heta)=frac{CD}{DB}$$ => $$tan(2 heta)=frac{h}{60}$$ => $$frac{2tan heta}{1-tan^2 heta}=frac{h}{60}$$ Substituting value from equation (i), we get : => $$2 imesfrac{h}{160}=[1-(frac{h}{160})^2] imes(frac{h}{60})$$ => $$frac{60}{80}=1-(frac{h}{160})^2$$
=> $$(frac{h}{160})^2=1-frac{3}{4}=frac{1}{4}$$ => $$frac{h}{160}=sqrt{frac{1}{4}}=frac{1}{2}$$ => $$h=frac{160}{2}=80$$ $$ herefore$$ Height of tower is 80 m => Ans - (A)