Question

In each of the following, one of the six trigonometric ratios is given. Find the values of the other trigonometric ratios.
(vii)$cos\Theta=\frac{7}{25}$
(viii)$tan\Theta =\frac{8}{15}$
(ix)$cot\Theta =\frac{12}{5}$

Answer 1

SOLUTION

(vii) Given : Cos θ = 7/25

Cos θ = 7/25  =  Base/Hypotenuse  

In right ∆ ABC ,  

Base (AB) = 7

Hypotenuse (AC) = 25

By using Pythagoras theorem,

AC² = AB² + BC²

25² = 7² + BC²

625 = 49 + BC²

BC² = 625 - 49

BC² = 576

BC = √576

BC = 24  

Hence, Perpendicular side (BC)  = 24

Now, sin θ = Perpendicular/ Hypotenuse

sin A = 24/25

cosec θ = Hypotenuse/Perpendicular

cosec θ = 25/24

sec θ  = Hypotenuse / Base

sec θ = 25/7

tan θ = Perpendicular /Base

tan θ = 24/7

cot θ = Base/ Perpendicular

cot  θ= 7/24

(viii)

Given : tan θ = 8/15

tan θ = 8/15  =  Perpendicular /Base  

In right ∆ ABC ,  

Perpendicular (BC) = 8

Base (AB) = 15

By using Pythagoras theorem,

AC² = AB² + BC²

AC²  = 15² + 8²

AC² = 225 + 64

AC² = 289

AC = √289 = 17

AC = 17  

Hence, Hypotenuse (AC) = 17

Now, sin θ = Perpendicular/ Hypotenuse

sin θ = 8/17

cos θ = base/Hypotenuse

cos θ = 15/17

cosec θ = Hypotenuse/Perpendicular

cosec θ = 17/8

sec θ = Hypotenuse / Base

sec θ = 17/15

cot θ = Base/ Perpendicular

cot θ = 15/8

(ix) Given : cot θ = 12/5

cot θ = 12/5 = Base/ Perpendicular

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 θ = Perpendicular/ Hypotenuse

sin θ = 5/13

cos θ = base/Hypotenuse

cos θ = 12/13

cosec θ = Hypotenuse/Perpendicular

cosec θ = 13/5

sec θ = Hypotenuse / Base

sec θ = 13/12

tan θ = perpendicular/base  

tan θ =  5/12

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