A vertical tower OP stands at the centre O of a square ABCD Let h and b denote the length - Maths -

Question

A vertical tower OP stands at the centre O of a square ABCD Let h
and b denote the length OP and AB respectively Suppose angle APB=60
then the relationship between h and b can be expressed as:
options-a)2b2=h2
b)2h2=b2
c)3b​2=2h2
d)3h2=2b2

Gursheen kaur. · Accepted Answer

Given: â€“âˆ APB = 60Â° and AB = b âˆ´PQ= $\frac{b}{2} \sqrt{3}$ Next, $\frac{b}{2}$ , h and PQ form a right angle triangle âˆ´ By Pythagoras theorem $\Rightarrow {(\frac{b}{2})}^{2} + {(A B)}^{2} = {(P Q)}^{2} \Rightarrow \frac{b^{2}}{4} + h^{2} = {(\frac{\sqrt{3} b}{2})}^{2} \Rightarrow \frac{b^{2}}{4} + h^{2} = \frac{3 b^{2}}{4} \Rightarrow 2 h^{2} = b^{2}$

A vertical tower OP stands at the centre O of a square ABCD.Let h and b denote the length OP and AB respectively.Suppose angle APB=60 then the relationship between h and b can be expressed as: options-a)2b2=h2 b)2h2=b2 c)3b​2=2h2 d)3h2=2b2

A vertical tower OP stands at the centre O of a square ABCD.Let h and b denote the length OP and AB respectively.Suppose angle APB=60 then the relationship between h and b can be expressed as:
options-a)2b²=h² b)2h²=b²
c)3b²=2h²d)3h²=2b²