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  • Given 15 cot A = 8, find sin A and sec A.

4. Given 15 \: \cot A=8, find \sin A and \sec A.

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We have,
15 \: \cot A=8, \Rightarrow \cot A =8/15 
It implies that In the triangle ABC in which \angle B =90^0. The length of AB be 8 units and the length of BC  = 15 units

Now, by using Pythagoras theorem,
AC = \sqrt{64 +225} = \sqrt{289}
\Rightarrow AC =17 units

So,                  \sin A = \frac{BC}{AC} = \frac{15}{17}
and                 \sec A = \frac{AC}{AB} = \frac{17}{8} 

Posted by

manish

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