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CBSE Class 10 Answered

atower  in a  city  is  100m   high   and  multistoried  hotel  in  the   city  market  is  20m   high  building  h m  high  is   situated   on  the  road  connecting    the  tower   with  city   market  at   a  distance  of  1  km  from  the  tower  the  angle  of  elevation  of  the  top   of  the  tower   at  the  top  of  the  hotel  is  theta   wher  tan  theta  =o.o629,.  find  the  height   h  of   the  building if  the  top  of the  tower  ,  top  of  the  building  and  the  top  of  the   hotel  are  in  the  staight  line  .  also  find   the  distance  of  the  tower   from  the  city  market
Asked by vinay110 | 10 Jan, 2016, 06:29: PM
answered-by-expert Expert Answer
R e f e r space t o space t h e space f i g u r e space a b o v e. A C space i s space t h e space t o w e r. F D space i s space t h e space b u i l d i n g space w h o s e space h e i g h t space i s space apostrophe h apostrophe. G H space i s space t h e space h o t e l space i n space t h e space m a r k e t space p l a c e space w h o s e space h e i g h t space i s space 20 space m. A n g l e space o f space e l e v a t i o n space o f space apostrophe A apostrophe space f r o m space apostrophe G apostrophe space i s space theta. angle F G E equals angle A F B I n space t r i a n g l e space A F B comma space tan theta equals fraction numerator A B over denominator B F end fraction rightwards double arrow 0.0629 equals fraction numerator A B over denominator C D end fraction rightwards double arrow A B equals 0.0629 cross times 1000 equals 62.9 space m B C equals A C minus A B equals 100 minus 62.9 equals 37.1 I n space t h e space f i g u r e space F D equals B C equals 37.1 space left parenthesis h e i g h t space o f space t h e space b u i l d i n g right parenthesis.. left parenthesis 1 right parenthesis I n space t r i a n g l e space F E G comma space F E equals F D minus E D equals F D minus G H equals 37.1 minus 20 equals 17.1 space m I n space triangle F E G comma space tan theta equals fraction numerator F E over denominator E G end fraction rightwards double arrow E G equals fraction numerator 17.1 over denominator 0.0629 end fraction equals 271.86 space m C H equals C D plus D H equals 1000 plus 271.86 equals 1271.86 space m H e n c e comma space d i s tan c e space o f space t h e space t o w e r space f r o m space t h e space c i t y space m a r k e t space i s space 1271.86 space m
Answered by satyajit samal | 11 Jan, 2016, 01:19: PM

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CBSE 10 - Maths
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CBSE 10 - Maths
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