# Knowing that sinx = m/n, find cosx .Solve by using right angled triangles .

hala718 | Certified Educator

sinx= m/n  find cos.

Let us assume that m, and n are sides of a right angle

triangle

==> m is the opposite side

and n is the hypotenuse side

==? the adjacent side is sqrt(n^2-m^2)

==> cos x= adjacent/ hypotenuse = sqrt(n^2-m^2)/n

neela | Student

sinx = m/n To find cosx.

Solution:

We  can have  right angled triangle ABC with right angle at B .

Let angle A = x , angle B = right angle, and C = 90-x.

We construct this triangle such that BC = m and AC = n.

Then by definition, sin A = sinx = BC/AC = m/n. Or

sinx = m/n.

By Pythagorus theorem, AB^2+BC^2 = AC^2. Or

AB^2 = AC^2-BC^2 = n^2-m^2.

AB^2 = n^2-m^2, Dividing both sides by AC^2, we get:

AB^2 = (n^2-m^2)/n^2. Or

AB/AC =  sqrt {(n^2-m^2)/n^2] Or

By definition cosx = AB/AC = (1/n)sqrt(n^2-m^2) .

giorgiana1976 | Student

Based on the fact the sine function, in a right angle triangle, is a ratio the has at numerator the opposite cathetus and at denominator, the hypotenuse.

In this case, is given the sin x = m/n, so we'll conclude that one cathetus is m and the hypotenuse  is n.

cos x = joined cathetus/hypotenuse = joined cathetus/n

Also, in a right angle triangle,by applying Pythagorean theorem:

(hypotenuse)^2 = (cathetus)^2 + (cathetus)^2

(n)^2 = m^2 + joined cathetus^2

joined cathetus^2 = (n)^2 - (m)^2

joined cathetus = sqrt[(n)^2 - (m)^2]

cos x =[sqrt(n^2 - m^2)]/n