KVPY Sample Paper KVPY Stream-SX Model Paper-19

A triangle has base 10 cm long and the base angles of \[50{}^\circ \] and \[70{}^\circ .\] If the perimeter of the triangle is \[x+y\text{ }cos\text{ z}{}^\circ ,\] where \[z\in (0,90{}^\circ ),\] then the value of \[x\text{ }+\text{ }y\text{ }+\text{ }z\] equals -

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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