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.
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.
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