# A right angle triangle has a 40 degree angle and hypotenuse = 10. Find the other angles and sides.

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Let ABC be a right angle triangle such that B is the right angle.

==> AC is the hypotenues = 10

==> A = 40 degrees.

==> Then, we conclude that the angle C = 180-A-B = 180 - 90 - 40 = 50 degrees.

==> C = 50 degrees.

Now we will use trigonometric identities to find the sides.

We know that:

cos A = adjacent / hypotenuse.

==> cos 40 = AB/ AC

==> cos 40 = AB/ 10

==>** AB = 10*cos 40 = 7.66**

**==> BC = 10*cos 50 = 6.43**

Let ABC be a right angled triangle with B as right angle. Then the hypotenuse AC = 10. Let angle A = 40 degree.

By trigonometry,

sinA = Opposite sine of angle A/ hypotenuse = BC/AC = BC/10.

So BC = 10 sinA = 10 sin40deg = 6.4279.

Similarly cosA = Adjacent side ofangle A /hpotenuse = AB/AC = AB/10.

So AB = 10 cosA = 10cos40 deg = 7.6604.

Angle B = 90 degree( implied by data0. angle A = 40 deg (given).

Since sum of all 3 angles is 180 degrees, so angle C = 180-(B+A).

=> C = 180-(90+40) = 50 degree.

Angles ,A = 40 deg, B = 90 deg, C = 50 deg.

AC = 10 , BC = 6.4279 and AB = 7.6604.

Let's suppose that the measure of the angle is A = 90.

That means that B+C = 40 + C = 90

C = 90 - 40

**C = 50 and B = 40**

To determine the lengths of the other sides, we'll apply the Pythagorean theorem and the sine function.

The sine function is a ratio between the opposite cathetus and the hypothenuse. We'll note the cathetus as x and y.

sin B = x/10

sin 40 = x/10

x = 10*sin 40

x = 10*0.64

**x = 6.4 units**

We'll apply Pythagorean theorem in a right angle triangle:

10^2 = x^2 + y^2

100 = 40.96 + y^2

y^2 = 100 - 40.96

y^2 = 59.04

**y = 7.6 units**

We'll accept only the positive value for y, since it is a length of a triangle and it cannot be negative.