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CBSE Class 10 Answered

Two stations due South of a leaning tower which leans the north are at distance a and b metres from it's foot .If alpha and beta be the angles of elevations of the top of the tower from these stations ,prove that it's inclination theta to the horizontal is given by cot theta=bcotalpha - acotbeta÷b-a  
Asked by tombilongjam | 01 Mar, 2019, 11:19: AM
answered-by-expert Expert Answer
As shown in figure, Let AD is leaning tower inclined at an angle θ with horizontal. B and C are stations
which are at distances b and a respectively from foot of tower.

Let α and β are the angle of elevation of top of tower observed from C and B respectively.
 
Let OD = h and OA = x;
 
we have,
begin mathsize 12px style tan space theta space equals space h over x space semicolon space space c o t space theta space equals space x over h space tan space alpha space equals space fraction numerator h over denominator x plus a end fraction space space space semicolon space space space space c o t space alpha space equals space x over h plus a over h space space space semicolon space space c o t space alpha space equals space c o t space theta space plus space a over h space space..................... left parenthesis 1 right parenthesis tan space beta space equals space fraction numerator h over denominator x plus b end fraction space space semicolon space space space c o t space beta space equals space x over h plus b over h semicolon space space space c o t space beta space equals space c o t space theta space plus space b over h space space......................... left parenthesis 2 right parenthesis M u l t i p l y space e q n. left parenthesis 1 right parenthesis space b y space b comma space space space b space c o t space alpha space equals space b space c o t space theta space plus space fraction numerator b a over denominator h end fraction space.............. left parenthesis 3 right parenthesis M u l t i p l y space e q n. left parenthesis 2 right parenthesis space b y space a comma space space a space c o t space beta space equals space a space c o t space theta space plus space fraction numerator a b over denominator h end fraction space space............... left parenthesis 4 right parenthesis s u b t r a c t space left parenthesis 4 right parenthesis space f r o m space left parenthesis 3 right parenthesis comma space space b space c o t space alpha space minus space a space c o t space beta space equals space left parenthesis b minus a right parenthesis space c o t space theta h e n c e space c o t space theta space equals space fraction numerator b space c o t space alpha space minus space a space c o t space beta over denominator left parenthesis space b minus a space right parenthesis end fraction end style
Answered by Thiyagarajan K | 02 Mar, 2019, 03:08: PM

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