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

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

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

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

Given the right angle triangle at B is ABC such that:

AB = 6

BC= 7

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

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