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Given 15 cot A = 8. Find sin A and sec A

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#### Solution

Consider a right-angled triangle, right-angled at B.

cot A =`(AB)/(BC)`

`= (AB)/(BC)`

it is given that

cot A = 8/15

`(AB)/(BC)=8/15`

Let AB be 8*k*.Therefore, BC will be 15*k*, where *k* is a positive integer.

Applying Pythagoras theorem in ΔABC, we obtain

AC^{2} = AB^{2} + BC^{2}

= (8*k*)^{2} + (15*k*)^{2}

= 64*k*^{2} + 225*k*^{2}

= 289*k*^{2}

AC = 17*k*

sin A `= (15k)/(17k) = 15/17`

sec A ` = (AC)/(AB) = 17/8`

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