Question

If sin thita=11/15 find the value of other trigonometric ratio thita ​

Answer 1

Answer:

$\sin(Ѳ) = \frac{11}{15}$

$it \: is \: a \: right \: angled \: triangle \: so \: other \: side$

$= \sqrt{ {15}^{2} - {11}^{2} }$

$since \: 15 \: is \: hypotenuse \:$

$other \: side = \sqrt{225 - 121} = \sqrt{104}$

$so \: \cos(Ѳ) = \frac{ \sqrt{104} }{15}$

$\tan(Ѳ) = \frac{11}{ \sqrt{104} }$

$\cot(Ѳ) = \frac{ \sqrt{104} }{11}$

$\sec(Ѳ) = \frac{15}{ \sqrt{104} }$

$\csc(Ѳ) = \frac{15}{11}$

Answer 2

Answer:

Step-by-step explanation:

We have, sin θ = 11/15 ………. (1)

By definition,

sin θ = Perpendicular/ Hypotenuse …. (2)

On Comparing eq. (1) and (2), we get;

Perpendicular = 11 and Hypotenuse= 15

Now, using Pythagoras theorem in Δ ABC

AC2 = AB2 + BC2

Putting the value of perpendicular (BC) and hypotenuse (AC) to get the base (AB), we have

152 = AB2 +112

AB2 = 152 – 112

AB2 = 225 – 121

AB2 = 104

AB = √104

AB= √ (2×2×2×13)

AB= 2√(2×13)

AB= 2√26

Hence, Base = 2√26

By definition,

cos θ = Base/Hypotenuse

∴ cosθ = 2√26/ 15

And, cosec θ = 1/sin θ

∴ cosec θ = 15/11

And, secθ = Hypotenuse/Base

∴ secθ =15/ 2√26

And, tan θ = Perpendicular/Base

∴ tanθ =11/ 2√26

And, cot θ = Base/Perpendicular

∴ cotθ =2√26/ 11