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Circles

Question
CBSEENMA10008335

Two stations due south of a leaning tower which leans towards the north are at distances a and b from its foot. If, α, β, are the elevations of the top of the tower from these stations,

prove that its inclination ө to the horizontal is given by cot  straight theta space equals space fraction numerator straight b space cot space straight phi space minus space straight a space cot space straight beta over denominator straight b space minus straight a space end fraction

Solution

Let CE be the leaning tower. Let A and B be two given stations at distances a and b respectively from the foot of the tower.
Let CD = x and DE = h
In right triangle CDE, we have
tan space straight theta space equals space DE over CD
rightwards double arrow space space tan space straight theta space space equals space straight h over straight x

rightwards double arrow space space straight x space equals space fraction numerator straight h over denominator tan space straight theta end fraction equals straight h space cot space straight theta space space space space space space space space space space space space space space space space space... left parenthesis straight i right parenthesis
In right triangle BDE, we have
tan space straight alpha space space equals space DE over BD
rightwards double arrow space space tan space straight alpha space equals space fraction numerator DE over denominator BC plus CD end fraction
rightwards double arrow space space tan space straight alpha space equals space space fraction numerator straight h over denominator straight a plus straight x end fraction
rightwards double arrow space space straight a space plus space straight x space equals space fraction numerator straight h over denominator tan space straight alpha end fraction
rightwards double arrow space space space straight a space plus space straight x space equals space straight h space cot space straight alpha
rightwards double arrow space space space space straight x space equals space straight h space cot space straight alpha space minus space straight a space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis ii right parenthesis
In right triangle ADE, we have
tan space straight beta space equals space DE over AD
rightwards double arrow space tan space straight beta space equals space fraction numerator DE over denominator AC plus CD end fraction
rightwards double arrow space space tan space straight beta space equals space fraction numerator straight h over denominator straight b plus straight x end fraction
rightwards double arrow space space straight b plus straight x equals fraction numerator straight h over denominator tan space straight beta end fraction
rightwards double arrow space space straight b plus straight x equals space straight h space cot space straight beta
rightwards double arrow space space straight x space equals space straight h space cot space straight beta space minus space straight b space space space space space space space space space space space... left parenthesis iii right parenthesis
Comparing (i) and (ii), we get
straight h space cot space straight theta space equals space straight h space cot space straight alpha space minus space straight a
rightwards double arrow space straight h space cot space straight alpha space minus space straight h space cot space straight theta space equals space straight a
rightwards double arrow space straight h space left parenthesis cot space straight alpha space minus space cot space straight theta right parenthesis space equals space straight a
straight h space equals space fraction numerator straight a over denominator cot space straight alpha space minus space cot space straight theta end fraction space space space space space space space space space space space space space space space space space space space space space... left parenthesis iv right parenthesis
Comparing (i) and (iii), we get
straight h space cot space straight theta space equals space straight h space cot space straight beta space minus space straight b
rightwards double arrow space straight h space cot space straight beta space minus space straight h space cot space straight theta space equals space straight b
rightwards double arrow space straight h space left parenthesis cot space straight beta space minus space cot space straight theta right parenthesis space equals space straight b
straight h space equals space fraction numerator straight b over denominator cot space straight beta space minus space cot space straight theta end fraction space space space space space space space space space space space space space space space space space space... left parenthesis straight v right parenthesis
Comparing (iv) and (v), we get
fraction numerator straight a over denominator cot space straight alpha space minus space cot space straight theta end fraction space equals space fraction numerator straight b over denominator cot space straight beta minus cot space straight theta end fraction
rightwards double arrow space space straight a left parenthesis cot space straight beta space minus space cot space straight theta right parenthesis space equals space straight b left parenthesis cot space straight alpha space minus space cot space straight theta right parenthesis
rightwards double arrow space space straight a left parenthesis cot space straight beta space minus space straight a space cot space straight theta right parenthesis space equals space straight b space cot space straight alpha space minus space straight b space cot space straight theta
rightwards double arrow space space straight b space cot space straight theta space minus straight a space cot space straight theta space equals straight b space cot space straight alpha space minus space straight a space cot space straight beta
rightwards double arrow space cot space straight theta space left parenthesis straight b minus straight a right parenthesis space equals space straight b space cot space straight alpha minus space straight a space cot space straight beta
rightwards double arrow space space space space space space space cot space straight theta space equals space fraction numerator straight b space cot space straight alpha space minus space straight a space cot space straight beta over denominator straight b minus straight a end fraction
Hence, inclination ө to the horizontal is given by cot straight theta space equals space fraction numerator straight b space cot space straight alpha space minus space straight a space cot space straight beta over denominator straight b minus straight a end fraction

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