The distance between the points (a cos 9 + b sin 9, 0)and (0, a sin 9-b cos 9) is...?

"( '9' - theta)"

Question

The distance between the points (a cos 9 + b sin 9, 0)and (0, a
sin 9-b cos 9) is ?
"( '9' - theta)"

Manmeet Singh · Accepted Answer

Let point A = $((a \cos θ + b \sin θ), 0)$

and point B = $(0, (a \sin θ - b \cos θ))$

The distance between A and B can be calculated by using distance formula,

$d i s \tan c e = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} \Rightarrow \sqrt{(0 - (a \cos θ + b \sin θ))^{2} + ((a \sin θ - b \cos θ) - 0)^{2}} \Rightarrow \sqrt{(a^{2} \cos^{2} θ + b^{2} \sin^{2} θ + 2 a b \cos θ \sin θ) + (a^{2} \sin^{2} θ + b^{2} \cos^{2} θ - 2 a b \sin θ \cos θ)} \Rightarrow \sqrt{(a^{2} + b^{2}) \cos^{2} θ + (a^{2} + b^{2}) \sin^{2} θ} = \sqrt{(a^{2} + b^{2})} (\sin c e \cos^{2} θ + \sin^{2} θ = 1)$

The distance between the points (a cos 9 + b sin 9, 0)and (0, a sin 9-b cos 9) is...? "( '9' - theta)"

The distance between the points (a cos 9 + b sin 9, 0)and (0, a sin 9-b cos 9) is...?

"( '9' - theta)"