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

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

Grade:12

1 Answers

Harshit Singh
askIITians Faculty 5963 Points
3 years ago
Dear Student

Let us assume a right angled triangle ABC, right angled at B
Given: 15 cot A = 8
So, Cot A = 8/15
We 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 = 8/15

Let AB be 8k and BC will be 15k
Where, 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^2
Substitute the value of AB and BC
AC^2= (8k)^2+ (15k)^2
AC^2= 64k^2+ 225k^2
AC^2= 289k^2
Therefore, AC = 17k
Now, we have to find the value of sin A and sec A We know that,
Sin (A) =Opposite side /Hypotenuse

Substitute the value of BC and AC and cancel the constant k in both numerator and denominator, we get Sin A = BC/AC = 15k/17k = 15/17

Therefore, sin A = 15/17
Since 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 side
Substitute the Value of BC and AB and cancel the constant k in both numerator and denominator,
we get,
AC/AB = 17k/8k = 17/8
Therefore
sec (A) = 17/8

Thanks

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