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If Triangle ABC,if a=2, b=3 and sin A=2/3, then​

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Answer:

The VALUE of ANGLE B is  \dfrac{\pi }{4}

Step-by-step explanation:

Given as :

For any TRIANGLE ABC

a = 2  

b = 3

Sin A = \dfrac{2}{3}

Let The value of ∠B = x°

According to question

By Applying Sin Law , we get

\dfrac{a}{SinA}  = \dfrac{b}{SinB} = \dfrac{c}{SinC}

Or,   \dfrac{a}{SinA}   =  \dfrac{b}{SinB}

Or, \dfrac{2}{\dfrac{2}{3} }  =  \dfrac{3}{SinB}

Or, \dfrac{6}{2}  = \dfrac{3}{SinB}

or, \dfrac{3}{SinB} = 3

∴  Sin B = \dfrac{3}{3}

i.e Sin B = 1

Or, B = Sin^{-1} 1

So, B = \dfrac{\pi }{4}

So, The value of angle B = \dfrac{\pi }{4}

Hence , The value of angle B is  \dfrac{\pi }{4}   Answer


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