A) \[\text{12}0{}^\circ \]
B) \[\text{15}0{}^\circ \]
C) \[\text{135}{}^\circ \]
D) None of these
Correct Answer: B
Solution :
\[\frac{B}{2}=\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\ \cos \theta }\] ?(i) \ \[\tan 90{}^\circ =\frac{B\sin \theta }{A+B\cos \theta }\Rightarrow A+B\cos \theta =0\] \ \[\cos \theta =-\frac{A}{B}\] Hence, from (i) \[\frac{{{B}^{2}}}{4}={{A}^{2}}+{{B}^{2}}-2{{A}^{2}}\Rightarrow A=\sqrt{3}\frac{B}{2}\] Þ \[\cos \theta =-\frac{A}{B}=-\frac{\sqrt{3}}{2}\]\ \[\theta =150{}^\circ \]
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