find the area of triangle ABC if A=56 degrees, b=20 ft, and c=12 ft. Round to the nearest tenth.
area of a triangle is 1/2 a*b*sinA, correct?
Area of a triangle = (1/2)bcSinA
Or in words, a half times the length of two sides times the Sin of the angle between those two lines.
So Area = (0.5)*(20)*(12)*Sin(56)
Area = 120*Sin(56)
Area = 99.48
And rounded to the nearest 10th [or to one decimal place in other words]
Area = 99.5 ft sq
I just put some extra information about the sides and angles below in case you were not 100% sure why angle A is between side b and c.
Angle A is opposite Side a
Angle B is opposite Side b
Angle C is opposite Side c
So
Angle A is between Sides b and c
Angle B is between Sides a and c
Angle C is between Sides a and b
Ok, hope that helps
To find the area of a triangle, you can use the formula:
Area = (1/2) * b * c * sin(A)
Given:
A = 56 degrees
b = 20 ft
c = 12 ft
First, convert the angle A from degrees to radians:
A_radians = A * (π/180)
A_radians = 56 * (π/180)
Next, plug the values into the formula:
Area = (1/2) * b * c * sin(A_radians)
Area = (1/2) * 20 ft * 12 ft * sin(56 * (π/180))
Now, calculate the value of sin(56 * (π/180)):
sin(56 * (π/180)) ≈ 0.8290375726
Substitute the value of sin(56 * (π/180)) into the formula:
Area = (1/2) * 20 ft * 12 ft * 0.8290375726
Area ≈ 99.482 ft²
Therefore, the area of triangle ABC is approximately 99.5 square feet when rounded to the nearest tenth.