Question

Given: in a right triangle, cot0 = 28/45
1. what is csc0
2. what is sec0

Answer 1

i)
`cosec0 = (hypotenuse)/(opposite side) = (53)/(45)`

ii)
`sec0 = (hypotenuse)/(adjacent side) = (53)/(28)`

Explanation:

Step(i):-

Given that

`cot 0 = (28)/(45) = (adjacentside)/(opposite side)`

Hypotenuse AC² = AB²+BC²

⇒ AC² = 45²+28² = 2809

⇒ AC = √2809

⇒ AC = 53

Hypotenuse AC =53

Step(ii):-

i)
`cosec0 = (hypotenuse)/(opposite side) = (53)/(45)`

ii)
`sec0 = (hypotenuse)/(adjacent side) = (53)/(28)`