Question

In each of the following, one of the six trigonometric ratios is given. Find the values of the other trigonometric ratios.
(iv)$sin\Theta=\frac{11}{15}$
(v)$tan\alpha =\frac{5}{12}$
(vi)$sin\Theta=\frac{\sqrt{3} }{2}$

Answer 1

SOLUTION  

(iv) Given : sinθ = 11/15

sin θ = 11/15  =  Perpendicular/ Hypotenuse  

In right ∆ ABC ,  

Perpendicular side (BC) = 11 and

Hypotenuse (AC) = 15

By using Pythagoras theorem,

AC² = AB² + BC²

15² = AB² + 11²

AB² = 15² – 11²

AB² = 225 – 121

AB² = 104

AB= √104 = √2×2×2×13 = √4 × 2 ×13

Hence, Base (AB) = 2√26

Now, cos θ=Base/Hypotenuse

cos θ =2√26 / 15

cosec θ = Hypotenuse/Perpendicular

cosec θ = 15/11

sec θ = Hypotenuse / Base

sec θ = 15/2√26

tan θ = Perpendicular /Base

tan  θ  =  11/2√26

cot  θ  = Base/ Perpendicular

cot  θ  = 2√26/11

(v) Given : tan α = 5/12

tan  α = 5/12  =  Perpendicular /Base  

In right ∆ ABC ,  

Perpendicular (BC) = 5

Base (AB) = 12

By using Pythagoras theorem,

AC² = AB² + BC²

AC²  = 12² + 5²

AC² = 144+ 25

AC² = 169

AC = √169

AC = 13

Hence, Hypotenuse (AC) = 13  

Now, sin α = Perpendicular/ Hypotenuse

sin α = 5/13

cos α = base/Hypotenuse

cos α = 12/13

cosec α = Hypotenuse/Perpendicular

cosec α =13/5

sec α = Hypotenuse / Base

sec α = 13/12

cot α = Base/ Perpendicular

cot α = 12/5

(vi) Given : sin θ = √3/2

sin θ = √3/2  =  Perpendicular/ Hypotenuse  

In right ∆ ABC ,  

Perpendicular side (BC) = √3 and

Hypotenuse (AC) = 2

By using Pythagoras theorem,

AC² = AB² + BC²

2² = AB² + (√3)²

AB² = 2² –  (√3)²

AB² = 4 - 3

AB² = 1

AB= √1 = 1

AB = 1

Hence, Base (AB) = 1

Now, cos θ = Base/Hypotenuse

cos θ = 1/2

cosec θ = Hypotenuse/Perpendicular

cosec θ = 2/√3

sec θ = Hypotenuse / Base

sec θ = 2/1

tan θ = Perpendicular /Base

tan  θ  = √3/1

cot  θ  = Base/ Perpendicular

cot  θ  = 1/√3

HOPE THIS ANSWER WILL HELP YOU.

Answer 2

Step-by-step explanation:

cot teta = 2 root 26/11

IF IT HELPS YOU

MARK AS BRAINLIEST PLEASE