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

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.