Cotθ=2 how do i work it out? thank you
cota = 2
But we know that:
cota = cosa/sina = 2
==> cosa = 2sina.........(1)
Also, we know that:
sin^2 a + cos^2 a = 1
==> cosa = sqrt(1-sin^2 a).........(2)
Now substitute (2) in (1):
==> sqrt(1-sin^2 a) = 2sina
Squre both sides:
==> 1-sin^2 a = 4sin^2 a
==> sin^2 a + 4sin^2 a = 1
==> 5sin^2 a = 1
==> sin^2 a = 1/5
==> sina = 1/sqrt5
==> cosa = 2/sqrt5
==> a= 26.6 (approx.)
@hala718 - I read your answer and I was wondering:
sin^2 a = 1/5, then sin a is either 1/sqrt 5 or -1/sqrt 5, depending on the position of the angle on the trigonometric circle.
This is what I've been taught, at least. Am I wrong? :-s
cotx = 2
To solve for x.
Let ABC be a right angled triangle, With angle at A equal to theta.
And angle at B =90 degree.
Then by definition cot(theta) = AB/BC = 2.
Therefore , construct a triangle with AB = 10 cm, BC = 5cm. Then AC = sqrt(AB^2+BC^2) = (10^2+5^2)^(1/2) = 11.18cm
Now measure angle A = theta = arc cot (2) = arc tan (1/2) = 26 .565.. degree nearly.
But in construction we may get only approximate values because of our practical errors and judgement.