Question of The Day25-10-2021

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.

0
COMMENTS

Comments

Share QOTD