Question

The distance between the points (a cos 0 + b sin 0, 0) and (0, a sin 0-b cos 0),​

Answer 1

Answer:

$\sqrt{ {a}^{2} + {b}^{2} }$

Step-by-step explanation:

By using distance formula,

$distance = \sqrt{ {(x2 - x1)}^{2} + {(y2 - y1)}^{2}}$

$\sqrt{ (0 - acos0 - bsin0)^{2} + {( a\sin0 - b \cos0 - 0)}^{2} } \\ = \sqrt{ {a}^{2} { \cos}^{2}0 + {b}^{2} { \sin}^{2}0 + 2sin0cos0 + { {a}^{2} { \sin}^{2}0 + {b}^{2} { \cos}^{2}0 - 2sin0cos0}} \\ = \sqrt{{a}^{2} { \cos}^{2}0 + {b}^{2} { \sin}^{2}0 + { {a}^{2} { \sin}^{2}0 + {b}^{2} { \cos}^{2}0 }} \\ = \sqrt{{a}^{2} { \cos}^{2}0 + {a}^{2} { \sin}^{2}0 + {b}^{2} { \sin}^{2}0 { + {b}^{2} { \cos}^{2}0 }} \\ = \sqrt{ {a}^{2} ( { \sin }^{2}0 + { \cos }^{2} ) + {b}^{2} ( { \sin }^{2}0 + { \cos }^{2}0 )} \\ = \sqrt{ {a}^{2} + {b}^{2} }$