Forces 5 N, 12 N and 13 N are in equilibrium. If ( sin\text{... | Physics | SolveeIt | Solveeit

Today, August 23

Question

Forces 5 N, 12 N and 13 N are in equilibrium. If $sin\text{ }23{}^\circ =\frac{5}{13},$ the angle between 5 N force and 13 N force is

A

$23{}^\circ$

B

$67{}^\circ$

C

$90{}^\circ$

D

$113{}^\circ$

Answer

$67{}^\circ$

Explanation

Solution

As ${{5}^{2}}+{{12}^{2}}={{13}^{2}},$ so these forces will be acting along the sides of a right angled triangle. Angle between 5 N and 13 N $=180-(90+23)$ $=180-113={{67}^{o}}$