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What is the area of triangle ABC if angle A is 60 degrees, side b is 40 feet and side c is 20 feet?

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Answer:the area of triangle ABC is

346.4 feet^2

Explanation:

The given triangle ABC is shown in the attached photo.

The formula for determining the area of a triangle is expressed as

Area = 1/2(abSinC) = 1/2(bcSinA)

A = 60 degrees

b = 40 feet

c = 20 feet

Therefore, area of the triangle is

1/2(bcSinA) = 1/2(40 × 20 × Sin60)

= 1/2 × 40 × 20 × 0.8660

= 346.4 feet^2

Alternatively, we can apply Heron's formula. We would determine the length of a by applying cosine rule,

a^2 = b^2 + c^2 -2bcCosA

Then, Heron's formula is expressed as

Area = √S(S - a)(S- b)(S - c)

Where S = (a + b + c)/2

What is the area of triangle ABC if angle A is 60 degrees, side b is 40 feet and side-example-1
User Steve Ganem
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