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Given 15 cot A = 8, find sin A and sec A. find it asap and provide it to me

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one month ago Anand Kumar Pandey
1677 Points
```							Dear StudentLet us assume a right angled triangle ABC, right angled at BGiven: 15 cot A = 8So, Cot A = 8/15We know that, cot function is the equal to the ratio of length of the adjacent side to the opposite side.Therefore, cot A =Adjacent side/Opposite side = AB/BC = 8/15Let AB be 8k and BC will be 15kWhere, k is a positive real number.According to the Pythagoras theorem,the squares of the hypotenuse side is equal to the sum of the squares of the other two sides of a right angle triangle and we get,AC^2=AB^2+ BC^2Substitute the value of AB and BCAC^2= (8k)^2+ (15k)^2AC^2= 64k^2+ 225k^2AC^2= 289k^2Therefore, AC = 17kNow, we have to find the value of sin A and sec A We know that,Sin (A) =Opposite side /HypotenuseSubstitute the value of BC and AC and cancel the constant k in both numerator and denominator, we get Sin A = BC/AC = 15k/17k = 15/17Therefore, sin A = 15/17Since secant or sec function is the reciprocal of the cos function which is equal to the ratio of the length of the hypotenuse side to the adjacent side.Sec (A) = Hypotenuse/Adjacent sideSubstitute the Value of BC and AB and cancel the constant k in both numerator and denominator,we get,AC/AB = 17k/8k = 17/8Thereforesec (A) = 17/8Thanks
```
one month ago
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