Question

hellobegin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3} }{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }& 1 & \sqrt{3} & \rm Not \: De fined \\ \\ \rm cosec A & \rm​

Answer 1

Answer:

Let one of the odd positive integer be x 

then the other odd positive integer is x+2

their sum of squares

=x2+(x+2)2

             

=x2+x2+4x+4                          

=2x2+4x+4

Given that their sum of squares = 290

2x2+4x+4=290 

2x2+4x=286

2x2+4x−286=0

x2+2x−143=0

x2+13x−11x−143=0

x(x+13)−11(x+13)=0

(x−11)=0,(x+13)=0

Therfore ,x=11or−13

We always take positive value of x 

So , x=11 and (x+2)=11+2=13

 

Therefore , the odd positive integers are 11 and 13 

Answer 2

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