Question

If the forces of $12$, $5$ and $13$ unit-wt balance at a point, two of them are inclined at

• A. $30°$
• B. $45°$
• C. $90°$
• D. $60°$

Answer 1

The correct option is C

$90°$

Step 1 Given data:

The forces of $12$, $5$ and $13$ unit-wt balance at a point.

Step 2 Formula used:

$|\mathrm{A}+\mathrm{B}{|}^2=|\mathrm{A}{|}^2+|\mathrm{B}{|}^2+2\mathrm{ABcos}\left(\mathrm{\theta }\right)$

Step 3 Angle

Let;

$$\begin{gathered} \left|A\right|=12 \\ \left|B\right|=5 \\ \left|C\right|=13 \end{gathered}$$

so,

$$\begin{gathered} |\mathrm{A}+\mathrm{B}{|}^2=|\mathrm{A}{|}^2+|\mathrm{B}{|}^2+2\mathrm{ABcos}\left(\mathrm{\theta }\right) \\ {13}^2={12}^2+5^2+2\times 12\times 5\cos \left(\mathrm{\theta }\right) \\ 169=144+25+120\cos \left(\mathrm{\theta }\right) \\ 169=169+120\cos \left(\mathrm{\theta }\right) \\ 169-169=120\cos \left(\mathrm{\theta }\right) \\ 0=120\cos \left(\mathrm{\theta }\right) \\ 0=\cos \left(\mathrm{\theta }\right) \end{gathered}$$

That is,

$$\begin{gathered} \cos \left(\mathrm{\theta }\right)=0 \\ \theta ={\cos }^{-1}0 \\ \theta =90° \end{gathered}$$

Therefore, if the forces of $12$, $5$ and $13$ unit-wt balance at a point, then two of them are inclined at $90°$. Hence, option C is correct.