PQRS is a square and it’s diagonals meets at O. A and B are the point on PQ such that PO = PB and QA = QO. If ∠AOB = θ, then what is the value of tanθ?
Answer
Correct Answer : a ) \(1\)
Explanation :According to the question
Let ‘c’ be the mid-point of PQ and perpendicular to PQ and length of the side of Square be ‘a’
We know that
If the Length of side of Square is ‘a’ unit then the length of it’s diagonal is \({\sqrt 2}{a}\).
And OC = \(a \over 2\)
⇒SQ =\({\sqrt 2}{a}\).
⇒OQ=\(SQ \over 2\)
⇒ OQ = \({a} \over {\sqrt 2} \)
⇒ AQ = OQ = \({a} \over {\sqrt 2} \)………(1)
C is the mid-point of PQ
⇒CQ=\(a \over 2\)
⇒ AC = AQ – CQ
⇒AC=\({a \over {\sqrt 2}}-{a \over 2}\)
OC is the angle bisector of ∠AOB
In △ AOC
⇒∠AOC = \(θ \over 2\)
tan\(θ \over 2\)=\(AC \over CO\)
Substituting the values
tan\(θ \over 2\)=\({{a \over {\sqrt 2}}-{a \over 2}} \over {a \over 2}\)
⇒ \(\sqrt 2-1\)
We know that
tanθ=\({2\ tan{θ \over 2}} \over { 1-tan^2{θ \over 2}}\)
Substituting the values
tanθ=\({2({ \sqrt 2-1})} \over {1-({{ \sqrt 2}-1)}^2}\)
⇒tanθ=\({2{ \sqrt 2-2}} \over {2{{ \sqrt 2}-}2}\)
⇒ tan θ = 1
Hence, (a) is the correct answer.
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