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Question
Given 15 cot A = 8. Find sin A and sec A.
Solution
Consider a right-angled triangle, right-angled at B.
It is given that
cot A = `8/15`
`("AB")/("BC")=8/15`
Let AB be 8k.
Therefore, BC will be 15k, where k is a positive integer.
Applying Pythagoras theorem in ΔABC, we obtain
AC2 = AB2 + BC2
= (8k)2 + (15k)2
= 64k2 + 225k2
= 289k2
AC = 17k
sin A = `(15k)/(17k)`
= `15/17`
sec A = `("AC")/("AB")`
= `17/8`
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