In each of the following one of the six trigonometric ratios is given fund the other trigonometric ratios cos theta is equal to 7/25
Answers
Answer:
cos=7/25
sin=24/25
tan=24/7
cosec=25/24
sec=25/7
cot=7/24
Answer:
Step-by-step explanation:
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
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