In a \triangle {\ABC}(a, b, c are the lengths of the opposite sides of angles A,B,C respectively), the perimeter of a \triangle {\ABC} is 6 times the arithmetic mean of the sine values of its angles. If the side a=1, then the angle {\A} is:

NIMCET 2023 Question Paper with solutions

Question : 47 of 120

Marks: +1, -0

In a

△ {\ABC}(a, b, c

are the lengths of the opposite sides of angles A,B,C respectively), the perimeter of a

△ {\ABC}

is 6 times the arithmetic mean of the sine values of its angles. If the side

a=1

, then the angle

{\A}

is:

Solution:

Let the sides of the triangle are

a, b, c

It is given that the perimeter of a triangle

{\ABC}

is 6 times the Arithmetic Mean of the sines of its angles.

∴ a+b+c\;=6 (\;{ sin \; {\A}+ sin \; {\B}+ sin \; {\C}}/{3})

a+b+c\;=2( sin \; {\A}+ sin \; {\B}+ sin \; {\C})....(i)

From the law of

\ sin

\;\;{a}/{ sin \; {\A}}=\;{b}/{ sin \; {\B}}=\;{c}/{ sin \; {\C}}=k

\;⇒ \;\; a=k sin \; {\A}

\;⇒ \;\; b=k sin \; {\B}

\;⇒ \;\; c=k sin \; {\C}

∴ a+b+c=2( sin \; {\A}+ sin \; {\B}+ sin \; {\C})

....(ii)

\;\text " Hence, "\;\;k\;=2

⇒\;a\;=2 sin \; {\A}

⇒ 1\;=2 sin \; {\A}

⇒ sin \; {\A}\;=\;{1}/{2}

{\A}\;=\;{π}/{6} .

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