^{2}sint

Suppose you have an isosceles triangle, and each of the equal sides has a length of foot. Suppose the angle formed by those two sides is 45 degrees . The area of the triangle is 0.3536 square feet.

Suppose you have an isosceles triangle, and each of the equal sides has a length of r feet. Suppose the angle formed by those two sides is t .

what is the area of the triangle in square feet?

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Area of the triangle, A = (1/2)base x height = (1/2) 2rsin(t/2) rcos(t/2)

Answer: A = (1/2)r^{2}sint

Check: r = 1, t = 45, A = (1/2)sin(45) = .3536

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Attn: You can extend it to two unequal sides a and b, and the angle t included by the two sides.

A = (1/2)absint

Area of a triangle is 1/2*base*altitude. For a triangle with sides a and b and angle θ between them, altitude to side a is b*sinθ. Therefore, area of a triangle is:

S=½a*b*sin(θ).

For isosceles triangle:

S=½a^{2}*sin(θ). If the side is 1 foot and θ=45°, then S=0.5*1^{2}*sin(45)=(√2)/4≈0.3536 ft^{2}

For the isosceles triangle with the side r and angle t, area is given by:

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