Question

The distance between the points $(a\cos {20}^{\circ }+b\sin {20}^{\circ },0)$ and $(0,a\sin {20}^{\circ }-b\cos {20}^{\circ })$ is:

• A. $(a+b)$
• B. $(a-b)$
• C. $\sqrt{a^2-b^2}$
• D. $\sqrt{a^2+b^2}$

Answer 1

The correct option is D $\sqrt{a^2+b^2}$

Given points are $\left(a\cos {20}^0+b\sin {20}^0,0\right)$ and $(0,a\sin {20}^0-b\cos {20}^0)$
Distance $=\sqrt{{\left(a\cos {20}^0+b\sin {20}^0-0\right)}^2+{(0-a\sin {20}^0-b\cos {20}^0)}^2}$
$=\sqrt{a^2{\cos }^2{20}^0+b^2{\sin }^2{20}^0+2ab\sin {20}^0\cos {20}^0+a^2{\sin }^2{20}^0+b^2{\cos }^2{20}^0-2ab\sin {20}^0\cos {20}^0}$
$=\sqrt{a^2[{\cos }^2{20}^0+{\sin }^2{20}^0]+b^2[{\cos }^2{20}^0+{\sin }^2{20}^0]}$
$=\sqrt{a^2+b^2}$