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Home / Doubts and Solutions/JEE FOUNDATION/Class 10/Mathematics

ABC is a Triangular Park with ab is equal to AC is equal 200 m a clock tower situated at the midpoint of BC.The angles of elevation of the top of the tower at A and B are cot inverse 3.2 and cosec inverse 2.6 respectively. The height of the tower is

Asked by mokshithreddy125 30th July 2018, 7:33 PM
Answered by Expert
Answer:
begin mathsize 16px style AB space equals AC space equals 200 space straight m space in space increment BX apostrophe straight X angle XBX apostrophe space equals space cosec to the power of negative 1 end exponent open parentheses 2.6 close parentheses cosec open parentheses angle XBX apostrophe close parentheses equals 2.6 cot open parentheses angle XBX apostrophe close parentheses equals square root of cosec squared open parentheses angle XBX apostrophe close parentheses minus 1 end root cot open parentheses angle XBX apostrophe close parentheses equals 2.4 fraction numerator BX apostrophe over denominator XX apostrophe end fraction equals 2.4 in space increment AX apostrophe straight X angle XAX apostrophe space equals space cot to the power of negative 1 end exponent open parentheses 3.2 close parentheses cot open parentheses angle XAX apostrophe close parentheses equals 3.2 fraction numerator AX apostrophe over denominator XX apostrophe end fraction equals 3.2 fraction numerator BX apostrophe over denominator 2.4 end fraction equals fraction numerator AX apostrophe over denominator 3.2 end fraction...... left parenthesis 1 right parenthesis increment AX apostrophe straight B apostrophe space forms space straight a space right space angled space triangle AB squared space equals space AX apostrophe squared plus BX apostrophe squared...... left parenthesis 2 right parenthesis from space left parenthesis 1 space and space 2 right parenthesis find space AX apostrophe comma space BX apostrophe space then space XX apostrophe end style
Answered by Expert 31st July 2018, 11:39 AM
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