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.

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