Question

Two triangles each have adjacent sides of length 120 ft and 180 ft. The first triangle has an angle between the two sides of 40°, while the second triangle has an angle between the two sides of 60°.
What is the approximate difference between the areas of the two triangles?

Answer 1

Area of a triangle = 1/2absin C

Difference between the areas of the two triangles = 1/2(120)(180) (sin 60 - sin 40) = 10800(0.2232) = 2411 square feet.

Answer 2

Answer:

The difference is around 2410.96 square feet.

Step-by-step explanation:

To solve this problem, we need to use the definiton of a triangular area but including angles, like the following

$A=\frac{1}{2}\times a \times b \times sin(C\°)$

(This formula is used in  triangles, where you know two sides and the angle formed).

Where $C\°$ is the angle of the vertex C, which is equivalent to the difference between the other angles.

So, let's find out the area of each triangle:

$A=\frac{1}{2}\times 120 \times 180 \times sin(40\°) \approx 6942.11 ft^{2}$

$A=\frac{1}{2}\times 120 \times 180 \times sin(60\°) \approx 9353.07 ft^{2}$

The difference between areas would be

$d=9353.07 - 6942.11 = 2410.96 ft^{2}$

Therefore, the difference is around 2410.96 square feet.