A) 60
B) 55
C) 60
D) 40
Correct Answer: D
Solution :
Perimeter \[=b+a+10=x+y\cos \,z\] |
Using sine rule |
\[\frac{10\times 2}{\sqrt{3}}=\frac{a}{\sin 50{}^\circ }=\frac{b}{\sin 70{}^\circ }\] |
\[a=\frac{20}{\sqrt{3}}\sin 50{}^\circ ,\,\,b=\frac{20}{\sqrt{3}}\,\,\sin 70{}^\circ \] |
\[a+b=\frac{20}{\sqrt{3}}(\sin 50{}^\circ +\sin 70{}^\circ )\]\[=\frac{20}{\sqrt{3}}\,\,2\cos 10{}^\circ \,\,\sin 60{}^\circ \]=\[20\,\,\cos \,\,10{}^\circ \] |
Perimeter \[10+a+b=10+20\,\cos \,10{}^\circ \]\[=x+y\,\cos \,z\] |
On comparing we get |
\[x+y+z=10+20+10=40\] |
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