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9.     In triangle ABC, right-angled at B, if  \tan A =\frac{1}{\sqrt{3}},find the value of:

            (i) \sin A\: \cos C + \cos A\: \sin C
            (ii) \cos A\: \cos C + \sin A\: \sin C

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Given a triangle ABC, right-angled at  B and \tan A =\frac{1}{\sqrt{3}}   \Rightarrow A=30^0

According to question,
\tan A =\frac{1}{\sqrt{3}} = \frac{BC}{AB}
By using Pythagoras theorem,
\\AC^2 = AB^2+BC^2\\ AC = \sqrt{1+3} =\sqrt{4}
AC = 2 
Now, 
\\\sin A = \frac{BC}{AC} = \frac{1}{2}\\ \sin C =\frac{AB}{AC} = \frac{\sqrt{3}}{2}\\ \cos A = \frac{AB}{AC} = \frac{\sqrt{3}}{2}\\ \cos C = \frac{BC}{AC} = \frac{1}{2}

Therefore,
  
(i) \sin A\: \cos C + \cos A\: \sin C
\\\Rightarrow \frac{1}{2}\times\frac{1}{2}+\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}\\ \Rightarrow1/4 +3/4\\ \Rightarrow4/4 = 1

(ii) \cos A\: \cos C + \sin A\: \sin C
\\\Rightarrow \frac{\sqrt{3}}{2}\times \frac{1}{2}+\frac{1}{2}\times\frac{\sqrt{3}}{2}\\ \Rightarrow \frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}\\ \Rightarrow \frac{\sqrt{3}}{2}

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manish

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