Explanation:
Here \(\alpha = 90^\circ ,Q = 5,R = 2P\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)\)
\(\;\tan \alpha = \frac{{Q\sin \theta }}{{P + Q\cos \theta }}\)
\(\therefore \infty = \frac{{5\sin \theta }}{{P + 5\cos \theta }}\,\,\,\,\,\,\,\,\,\,\,\,(\because \tan 90^\circ = \infty )\)
\( \Rightarrow P + 5\cos \theta = 0\)
\(\therefore \cos \theta = \frac{{ - P}}{5}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)\)
Since,
\({R^2} = {P^2} + {Q^2} + 2PQ\cos \theta \)
\(\therefore \quad 4{P^2} = {P^2} + {5^2} + 2P \times 5 \times \left( {\frac{{ - P}}{5}} \right)\)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,[{\rm{From}}\,(1)\,{\rm{and}}\,(2)]\)
\(\therefore \quad 4{P^2} = 25 - {P^2}\)
\(\therefore \quad 5{P^2} = 25\)
\({P^2} = 5\) units.
