- 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 = 120
^{o}. 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 = 180
^{o} - angle ACD = 180
^{o}- angle ACB = 180^{o}- 120^{o}= 60^{o} - 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)(sin60
^{o}) - = (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.