If the sides of a right angle triangle at B ABC are AB=6 and BC=7, find the value of cos A and cos B
Given the right angle triangle at B is ABC such that:
AB = 6
Then the hypotenuse is AC
Now we will calculate the length of the hypotenuse.
==> AC^2 = AB^2 + BC^2
==> AC^2 = 6^2 + 7^2 = 36+49= 85
==> AC = sqrt85
Now let us calculate cosA and cosB
We know that cosA= adjacent/ hypotenuse.
==> cosA= AB/AC= 6/sqrt85
==> cosA= 6sqrt85 / 85
Now we know B is the right angle.
==> B= pi/2
==> cosB = cos(pi/2) = 0
Then the answer is:
cosA= 6*sqrt85 / 85
cosB = 0
The sides of a right triangle ABC which is right angled at B are AB=6 and BC=7.
Using the Pythagorean Theorem
AC^2 = AB^2 + BC^2
=> AC^2 = 36 + 49
=> AC^2 = 85
=> AC = sqrt 85
cos A = 6 / sqrt 85
cos B = cos 90 = 0
The values of cos A = 6/ sqrt 85 and cos B = 0