In a triangle ABC, angle B is 90⁰ and AB is 4cm and BC is 3cm and AC is 5cm. The magnitude of angle BAC will be
Tan A = 3/4 = 0.75
so
angle A = tan^-1 ( 0.75 ) = 36.9
degrees (rounded to one decimal place).
or sin A = 3/5 = 0.6
A = 36.9
or cos A = 4/5 = 0.8
A = 36.9
Yes, that's correct. Since the angles in a triangle always add up to 180 degrees, you can find the other two angles by subtracting angle A from 180 and dividing by 2. In this case:
B = (180 - 36.9) / 2 = 71.55 degrees
C = (180 - 36.9) / 2 = 71.55 degrees
So the three angles of the triangle are approximately 36.9 degrees, 71.55 degrees, and 71.55 degrees.
Nonsense !
I apologize if my response was incorrect or unclear. Can you please clarify what part you think is nonsense so that I can correct it or explain it further?
the sides are 3 and 4 and the hypotenuse is 5
angle A is 36.9 deg
angle B is 90 deg (right angle)
angle C is 90 - 36.9 = 53.1 deg
You are correct, and I am sorry for my mistake earlier. Thank you for pointing it out and providing the correct information.
To summarize, for the right triangle with sides of 3 and 4 and hypotenuse of 5, we can find the angles as follows:
- Angle A is the angle opposite the side of length 3, so we can use sin A = opposite/hypotenuse to find that sin A = 3/5. Taking the inverse sine of both sides, we get A ≈ 36.87 degrees.
- Angle B is the right angle between the sides of length 3 and 4, so it is 90 degrees.
- Angle C is the angle opposite the hypotenuse of length 5, so we can use cos C = adjacent/hypotenuse to find that cos C = 4/5. Taking the inverse cosine of both sides, we get C ≈ 36.87 degrees. Since the angles of a triangle add up to 180 degrees, we can subtract angles A and B from 180 to find that angle C is approximately 53.13 degrees.
Thank you again for correcting me and I apologize for any confusion I may have caused.