# ABC is a right angle triangle. AB = 10 A = 27 degree. Find BC if AC is the hypotenuse.

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We are given that ABC is a right triangle. AB is equal to 10 and the Angle A is 27 degree. Also, AC is the hypotenuse.

Now cos A = adjacent side / hypotenuse

=> cos A = AB/ AC

=> cos 27 = 10 / AC

=> 0.891 = 10 / AC

=> AC = 10/ .891

=> AB = 11.22 ( approximately)

For BC we can use the Pythagorean Theorem. We have

AC^2 = AB^2 + BC^2

=> 11.22^2 = (10)^2 + BC^2

=> BC^2 = 11.22^2 - 10^2

=> BC^2 = 125.96 - 100

=> BC^2 = 25.96

=> BC = 5.09 ( approximately)

**Therefore the required value of BC is 5.09**

Given the right angle triangle ABC such that:

AB = 10

angle A = 27 degree

AC = hypotenuse.

Then we will use the trigonometric identities to find the other sides.

We know that :

cos A = adjacent / hypotenuse

==> cos A = AB/ AC

==> cos 27 = 10 / AC

==> 0.891 = 10/ AC

**==> AC = 11.22**

Now we will calculate the side BC.

We know that.

AC^2 = AB^2 + BC^2

==> 11.22^2 = 10^2 + BC^2

==> 125.96 = 100 + BC^2

==> BC^2 = 25.96

**==> BC = 5.09**

ABC is a right angled triangle with angle A = 27 degree.

Since AC is hypotenuse, the angle B opposite to hypotenuse AC is the right angle.

Therefore, tan27 = opposite side of angle 27

degree/adjacent side of angle 27 degree = BC/AB .

Or

tan27 = BC/AB.

BC = AB*tan27.

Substitute the given value of AB = 10 in BC = AB*tan27 and we get:

BC = 10 tan27 = 5.0953 nearly.