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Example of finding the area of an obtuse triangle
Question:
ABC is a triangle, BC = 6 feet and AC = 7 square root of 3 feet. If the angle C = 120o. Find the area of the triangle ABC. Solution:
Area of a triangle = (1/2)base × height extended BC, drawing a line from point A and perpendicular to the extended BC line, these two lines intesect at point D
AD is the height of the obtuse triangle when base is BC
since AD is perpendicular to CD, so triangle ACD is a right triangle
since BCD is a straight line
so the angle ACB + angle ACD = 180o
angle ACD = 180o - angle ACB = 180o - 120o = 60o
since ACD is a right triangle
so sin(angle ACD) = opposite side/hypotenuse
that is, sin(angle ACD) = AD/AC
so AD = AC sin(angle ACD)
= (7 square root of 3)(sin60o)
= (7 square root of 3)(square root of 3)/2
= 7 × 3 ÷ 2
= 21/2
let A be the area of triangle ABC
A = (1/2) base × height
= (1/2) BC × AD
= (1/2) × 6 × 21/2
= 6 × 21 ÷ 4
= 31.5
so the area of the triangle ABC is 31.5 square feet.

The area of a obtuse triangle is the base multiply its altitude. altitude is perpendicular to the base. In a triangle, if an angle is a right angle, then the triangle is a right triangle. In a right triangle, the Sine of an angle is equal to the length of its opposite side divide by the length of the hypotenuse. In other word, the length of the opposite side of an angle is equal to the length of the hypotenuse multiply the Sine of the angle.