Two triangles each have adjacent sides of length 120 feet and 180 feet. The first triangle has an angle between the two sides of 40 degrees, while the second triangle has an angle between the two sides of 60 degrees. What is the approximate difference between the areas of the two triangles?
To find the area of a triangle, we can use the formula:
Area = (1/2) * base * height
Since we have the lengths of the adjacent sides of the triangles, we can use the law of sines to find the altitude or height of each triangle.
The law of sines states:
sin(A) / a = sin(B) / b = sin(C) / c
Let's apply the law of sines to the first triangle (triangle A):
sin(A) / a = sin(B) / b
sin(40°) / 120 = sin(90°) / height_A
Since sin(90°) is equal to 1:
sin(40°) / 120 = 1 / height_A
Rearranging the equation to isolate height_A:
height_A = 120 / sin(40°)
Calculating that gives us:
height_A ≈ 184.35 feet
Now, let's find the height of the second triangle (triangle B):
sin(A) / a = sin(B) / b
sin(60°) / 180 = sin(90 - 60) / height_B
Since sin(90 - 60) is equal to sin(30):
sin(60°) / 180 = sin(30°) / height_B
Rearranging the equation to isolate height_B:
height_B = 180 / sin(60°)
Calculating that gives us:
height_B ≈ 207.84 feet
Now, let's find the areas of the two triangles:
Area_A = (1/2) * base_A * height_A
= (1/2) * 120 * 184.35
Area_B = (1/2) * base_B * height_B
= (1/2) * 180 * 207.84
Calculating the areas gives us:
Area_A ≈ 11,040 sq. feet
Area_B ≈ 18,764.8 sq. feet
The difference between the areas of the two triangles is:
Approximate difference = Area_B - Area_A
≈ 18,764.8 - 11,040
≈ 7,724.8 sq. feet
Therefore, the approximate difference between the areas of the two triangles is approximately 7,724.8 square feet.