- Home
- Standard 11
- Physics
3-1.Vectors
hard
The resultant of two vectors at an angle $150^{\circ}$ is $10$ units and is perpendicular to one vector. The magnitude of the smaller vector is ....... units
A
$10$
B
$10 \sqrt{3}$
C
$10 \sqrt{2}$
D
$5 \sqrt{3}$
Solution
(b)
$\Rightarrow R^2+A^2=B^2 \ldots (1)$
$R=10$
Also $\tan 30^{\circ}=\frac{\text { Perpendicular }}{\text { Base }}$
$\frac{1}{\sqrt{3}}=\frac{R}{A}$
From equation $(1)$ $A=10 \sqrt{3}$
$(10)^2+(10 \sqrt{3})^2=B^2$
$B=20$
Standard 11
Physics
