Given sec θ = 13/12, caluculate all other trigonometric ratios.
Answers
Answered by
9
given=
secΘ = 13/12
so, sec= hypotenuse/base
let the lengths be 13x and 12x.
so, perpendicular = √h² - b²
= √ (13x)² -(12x)²
= √169x² - 144x²
=√ 25x
= 5x
so, sin Θ=p/h = 5/13
cos Θ = b/h = 12/13
tan Θ =p/b = 5/12
cot = b/p = 12/5
cosecΘ = h/p =13/12
secΘ = 13/12
so, sec= hypotenuse/base
let the lengths be 13x and 12x.
so, perpendicular = √h² - b²
= √ (13x)² -(12x)²
= √169x² - 144x²
=√ 25x
= 5x
so, sin Θ=p/h = 5/13
cos Θ = b/h = 12/13
tan Θ =p/b = 5/12
cot = b/p = 12/5
cosecΘ = h/p =13/12
Answered by
7
Sec θ =
Sec θ =
So Hypotenuse = 13 and Base = 12
To find Perpendicular
H² = P² + B²
13² = P² + 12²
169 - 144 = P²
25 = P²
P = 5
Now for other trigonometric terms.
Sin θ = =
Cos θ = =
Tan θ = =
Cosec θ = =
Cot θ = =
Sec θ =
So Hypotenuse = 13 and Base = 12
To find Perpendicular
H² = P² + B²
13² = P² + 12²
169 - 144 = P²
25 = P²
P = 5
Now for other trigonometric terms.
Sin θ = =
Cos θ = =
Tan θ = =
Cosec θ = =
Cot θ = =
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