A vertical tower OP stands at the centre O of a square ABCD. Let h and b denote the lengths OP and AB respectively. Suppose ∠APB = 60°, then the relationship between h and b can be expressed as :	Select the correct option:  A) 2b^2 = h^2  B) 2h^2 = b^2  C) 3b^2 = 2h^2  D) 3h^2 = 2b^2  E) None of these
Explanation :   As OP is at centre of square so clearly PA = PB.   so, angle ∠PAB = ∠ PBA.   and as ∠APB = 60,so we can see that in ∆ APB, ∠ PAB = ∠ PBA = 60.   Hence, its an equilateral triangle thus making AP = PB = AB = b.   Now lets see ∆ AOB ,   O is centre of square so angle AOB=900​.   giving us AO = OB = AB/√2.   so AO= b/√2   Now in triangle AOP right angled at O,   =&gt; AO2 + OP2 = PA2 .   =&gt; b2/2 + h2 = b2 .   =&gt; h2 = b2/2   Answer : 2h^2 = b^2
<result>B</result>
A vertical tower OP stands at the centre O of a square ABCD. Let h and b denote the lengths OP and AB respectively. Suppose ∠APB = 60°, then the relationship between h and b can be expressed as :