A) obtuse angled
B) acute angled
C) isosceles
D) right angled
Correct Answer: D
Solution :
Given, \[A-B={{60}^{o}}\] Day gream By sine rule, \[\frac{2a}{\sin \,A}=\frac{a}{\sin \,B}\] \[\Rightarrow \] \[\sin A-2\sin B=0\] \[\Rightarrow \] \[\sin ({{60}^{o}}+B)-2\sin B=0\] \[\Rightarrow \] \[\frac{\sqrt{3}}{2}\cos \,B+\frac{1}{2}\sin B-2\sin B=0\] \[\Rightarrow \] \[\frac{\sqrt{3}}{2}\cos B-\frac{3}{2}\sin B=0\] \[\Rightarrow \] \[\sqrt{3}\left( \frac{1}{2}\cos B-\frac{\sqrt{3}}{2}\,\sin B \right)=0\] \[\Rightarrow \] \[\sqrt{3}\left( \frac{1}{2}\,\cos B-\frac{\sqrt{3}}{2}\,\sin B \right)=0\] \[\Rightarrow \] \[\sqrt{3}[\cos \,({{60}^{o}}+B)]=0\] \[\Rightarrow \] \[{{60}^{o}}+B={{90}^{o}}\] \[\Rightarrow \] \[B={{30}^{o}}\] \[\Rightarrow \] \[A={{90}^{o}}\] Hence, it is right angled triangle,
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