Question

In each of the following, one of the six trigonometric ratios is given. Find the values of the other trigonometric ratios.
(i)$sin A=\frac{2}{3}$
(ii)$cos A=\frac{4}{5}$
(iii)tan θ = 11

Answer 1

SOLUTION :

(i) Given : sinA = 2/3

sinA = 2/3  =  Perpendicular/ Hypotenuse  

In right ∆ ABC ,  

Perpendicular side (BC) = 2 and

Hypotenuse (AC) = 3

By using Pythagoras theorem,

AC² = AB² + BC²

3² = AB² + 2²

AB² = 3² – 2²

AB² = 9 – 4

AB² = 5

AB= √5

Hence, Base (AB) = √5

Now, cos A=Base/Hypotenuse

cosA =√ 5/3

cosec A = Hypotenuse/Perpendicular

cosec A = 3/2

sec A = Hypotenuse / Base

sec A = 3/√5

tan A = Perpendicular /Base

tan A =  2/√5

cot A = Base/ Perpendicular

cot A = √5/2

(ii) Given : CosA = 4/5

CosA = 4/5  =  Base/Hypotenuse  

In right ∆ ABC ,  

Base (AB) = 4

Hypotenuse (AC) = 5

By using Pythagoras theorem,

AC² = AB² + BC²

5² = 4² + BC²

25 = 16 + BC²

BC² = 25 - 16

BC² = 9

BC = √9

BC = 3  

Hence, Perpendicular side (BC)  = 3

Now, sin A = Perpendicular/ Hypotenuse

sin A = 3/5

cosec A = Hypotenuse/Perpendicular

cosec A = 5/ 3

sec A = Hypotenuse / Base

sec A = 5/4

tan A = Perpendicular /Base

tan A =  3/4

cot A = Base/ Perpendicular

cot A = 4/3

(iii) Given : tan θ = 11

tan θ = 11/1  =  Perpendicular /Base  

In right ∆ ABC ,  

Perpendicular (BC) = 11

Base (AB) = 1

By using Pythagoras theorem,

AC² = AB² + BC²

AC²  = 1² + 11²

AC² = 1 + 121

AC² = 122

AC = √122

Hence, Hypotenuse (AC) = √122

Now, sin θ = Perpendicular/ Hypotenuse

sin θ = 11/√122

cos θ = base/Hypotenuse

cos θ = 1/√122

cosec θ = Hypotenuse/Perpendicular

cosec A =√122/11

sec θ = Hypotenuse / Base

sec θ = √122/1

cot θ = Base/ Perpendicular

cot θ = 1/11

HOPE THIS ANSWER WILL HELP YOU...

Answer 2

Step-by-step explanation:

I hope it is helpful to u