Question

In each of the following, one of the six trigonometric ratios is given. Find the values of the other trigonometric ratios.
(x)$sec\Theta =\frac{13}{5}$
(xi)$cosec\Theta=\sqrt{10}$
(xii)$cos\Theta =\frac{12}{15}$

Answer 1

SOLUTION

(x) Given : sec θ = 13/5

sec θ = 13/5  =  Hypotenuse /base

In right ∆ ABC ,  

Base (AB) = 5

Hypotenuse (AC) = 13

By using Pythagoras theorem,

AC² = AB² + BC²

13² = 5² + BC²

169 = 25 + BC²

BC² = 169 - 25

BC² = 144

BC = √144

BC = 12  

Hence, Perpendicular side (BC)  = 12

Now, sin θ = Perpendicular/ Hypotenuse

sin θ = 12/13

cos θ = Base/Hypotenuse

cos θ = 5/13

cosec θ = Hypotenuse/Perpendicular

cosec θ = 13/12

sec θ  = Hypotenuse / Base

sec θ = 13/5

tan θ = Perpendicular /Base

tan θ = 12/5

cot θ = Base/ Perpendicular

cot  θ= 5/12

(xi) Given : cosec θ = √10

cosec θ = √10/1 = Hypotenuse / Perpendicular

In right ∆ ABC ,  

Perpendicular side (BC) = 1 and

Hypotenuse (AC) = √10

By using Pythagoras theorem,

AC² = AB² + BC²

(√10)² = AB² + (1)²

AB² = (√10)² –  (1)²

AB² = 10 - 1

AB² = 9

AB= √9 = 3

AB = 3

Hence, Base (AB) = 3

sin θ = Perpendicular/Hypotenuse

sin θ = 1/√10

cos θ = Base/Hypotenuse

cos θ = 3/√10

sec θ = Hypotenuse / Base

sec θ = √10/3

tan θ = Perpendicular /Base

tan  θ  = 1/3

cot  θ  = Base/ Perpendicular

cot  θ  = 3/1= 3

(xii) Given : Cos θ = 12/15

Cos θ = 12/15  =  Base/Hypotenuse  

In right ∆ ABC ,  

Base (AB) = 12

Hypotenuse (AC) = 15

By using Pythagoras theorem,

AC² = AB² + BC²

15² = 12² + BC²

225 = 144 + BC²

BC² = 225 - 144

BC² = 81

BC = √81

BC = 9

Hence, Perpendicular side (BC)  = 9

Now, sin θ = Perpendicular/ Hypotenuse

sin A = 9/15

cosec θ = Hypotenuse/Perpendicular

cosec θ = 15/9

sec θ  = Hypotenuse / Base

sec θ = 15/12

tan θ = Perpendicular /Base

tan θ = 9/12

cot θ = Base/ Perpendicular

cot  θ= 12/9

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