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# A vertical tower OP stands at the center O of a square ABCD. Let h and b denote the length OP and AB respectively. Suppose ÐAPB = 60° then the relationship between h and b can be expressed asa. 2b2 = h2 b. 2h2 = b2 c. 3b2 = 2h2 d. 3h2 = 2b2Please can you explain in detail.

7 years ago
angle APB =60
as op is at centre of square so clearly PA=PB
so angle PAB=anglePBA
and as angle APB=60 so we can see that in triangle APB angle PAB=angle PBA=60

hence its an equilateral triangle thus making AP=PB=AB=b

now lets see triangle AOB
O is centre of square so angle AOB=90
giving us AO=OB=AB/root(2)
so AO= b/root(2)

now in triangle AOP right angled at O
AO2+OP2=PA2
b2/2+ h2= b2
h2=b2/2
`Thanks & RegardsSaurabh Singh,askIITians FacultyB.Tech.IIT Kanpur`