## 26 KN is the resultant of the two forces one of which is shown in the figure. Determine the other force.

**Solution:**

Let Q be the unknown force which makes an angle θ with horizontal

Given R = 26KN

\begin{equation*}

\theta_1=\tan^{-1}(\frac{12}{5})=67.38^{\circ}\\

\theta_2=\tan^{-1}(\frac{3}{4})=36.87^{\circ}\\

\end{equation*}

\begin{equation*}

\sum Fx = Rx\\

\end{equation*}

10cos36.87 + Qcosθ = 26cos67.38

Qcosθ = 26cos67.38 – 10cos36.87

Qcosθ = 1.99

\begin{equation*}

\sum Fy = Ry

\end{equation*}

10sin36.87 + Qsinθ = 26sin67.38

Qsinθ = 26sin67.38 – 10sin36.87

Qsinθ = 17.99

By Adding,

Qcosθ + Qsinθ = 1.99 + 17.99

Squaring both side

Q = 18.196KN

So the other force will be Q = 18.196KN