Question

If vectors $\overrightarrow{A},\overrightarrow{B}$ and $\overrightarrow{C}$ have magnitudes $5,12$ and $13$ units and $\overrightarrow{A}+\overrightarrow{B}=\overrightarrow{C},$ then the angle between $\overrightarrow{B}$ and $\overrightarrow{C}$ is

• A. $si{n}^{-1}\frac{5}{13}$
• B. $co{s}^{-1}\frac{12}{13}$
• C. $si{n}^{-1}\frac{12}{13}$
• D. $co{s}^{-1}\frac{5}{13}$

Answer 1

Answer: (A), (B)

The correct options are
A $si{n}^{-1}\frac{5}{13}$
B $co{s}^{-1}\frac{12}{13}$

As $\overrightarrow{A}+\overrightarrow{B}=\overrightarrow{C}$
$\therefore \overrightarrow{A}=\overrightarrow{C}-\overrightarrow{B}$
$\Rightarrow |\overrightarrow{A}{|}^2=|\overrightarrow{C}-\overrightarrow{B}{|}^2$
$\Rightarrow |\overrightarrow{A}{|}^2=|\overrightarrow{C}{|}^2+|\overrightarrow{B}{|}^2-2|\overrightarrow{C}||\overrightarrow{B}|cos\theta$
$\Rightarrow cos\theta =\frac{{13}^2+{12}^2-5^2}{2\times 13\times 12}=\frac{288}{312}=\frac{12}{13}$
$\Rightarrow \theta =co{s}^{-1}\frac{12}{13}$
Also, $\sin \theta =\sqrt{1-{\cos }^2\theta }=\sqrt{1-(\frac{12}{13}{)}^2}=\frac{5}{13}$
$\Rightarrow \theta ={\sin }^{-1}\frac{5}{13}$