Grade 1210 grade maths
If cosec θ = √10, the find the values of all T-ratios of θ.
If cosec θ = √10, the find the values of all T-ratios of θ.
2 Answers
Harshit Singh
Dear Student
Given function: cosecθ = √10
Draw a right ∆ABC,∠C = 90 degrees and∠A = θ
We know that, cosecθ = hypotenuse/perpendicular
= (k√10)/k
where k is a positive integer
By Pythagoras Theorem:
AC^2 = AB^2 + BC^2
= 10k^2 + k^2 = 9k^2
AC = 3k
Find other T-rations using their definitions:
Sin𝜃= BC/AB = 1/√10
cos𝜃= AC/AB = (3k)/(k√10) = 3/√10
tan θ = BC/AC = sinθ /cosθ = 1/3
secθ = AB/AC = 1/cosθ = √10/3
cotθ = AC/BC = 1/tanθ = 3
Thanks
Given function: cosecθ = √10
Draw a right ∆ABC,∠C = 90 degrees and∠A = θ
We know that, cosecθ = hypotenuse/perpendicular
= (k√10)/k
where k is a positive integer
By Pythagoras Theorem:
AC^2 = AB^2 + BC^2
= 10k^2 + k^2 = 9k^2
AC = 3k
Find other T-rations using their definitions:
Sin𝜃= BC/AB = 1/√10
cos𝜃= AC/AB = (3k)/(k√10) = 3/√10
tan θ = BC/AC = sinθ /cosθ = 1/3
secθ = AB/AC = 1/cosθ = √10/3
cotθ = AC/BC = 1/tanθ = 3
Thanks
Ram Kushwah
From the fig we know that
cosecθ =Hypotenuse/Height =AC/AB=√10/1=√10
Using pythagoras theorem in the Δ ABC
BC²=AC²-AB²
=(√10)²-1²=10-1=9
so BC=√9 = 3
Thus the other T-ratios can be written as :
sin θ =AB/AC = 1/√10
cos θ=BC/AC= 3/√10
tanθ=AB/BC= 1/3
cotθ=BC/AB= 3/1= 3
secθ=AC/BC=√10/3
Thanks
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