Question

A sin 30 0)(0 sin60) find distance between two points

Answer 1

${\bold{\huge{\underline{\green{Answer!!}}}}}$

$\textsf{distance between points A(sin30°,0),B(0,sin60°)}$

$\large\sf\implies\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$\large\sf\implies\sqrt{(0-sin30°)^2+(sin60°-0)^2}$

$\large\sf\implies\sqrt{(sin^2 30° )+(sin^2 60°)}$

$\large\sf\implies\sqrt{\dfrac{1}{4}+\dfrac{3}{4}}$

$\large\sf\implies\sqrt\dfrac{\cancel{4}}{\cancel{4}}$

$<b><u>distance between two points is 1$

Answer 2

Answer:

Distance between 2 points A(sin30 A,0) B(0,sin60A)

$\sqrt{(x_{2} -x_{1})^{2}+y_{2} -y_{1})^{2}$

=>   $\sqrt{(0-sin30A)^{2} + (0-sin60A)^{2} }$

=>  $\sqrt{(sin30A)^{2} + (sin60A)^{2} }$

=>  $\sqrt{(\frac{1}{2} )^{2} + (\frac{\sqrt{3} }{2})^{2} }$

=>  $\sqrt{\frac{1}{4}+\frac{3}{4} }$

=> $\sqrt{\frac{4}{4} }$ = $\sqrt{1}$

We know that $cos^{2} + sin^{2}$ = 1

=>  $\sqrt{cos^{2}+sin^{2} }$

So when we take it out of root, it becomes

= cos+sin A