Question

A triangle has a side with length 6 feet and another side with length 8 feet. The angle between the sides measures 73 degrees. Find the area of the triangle. Please help! 
A. 1752.0 
B. 7.0 
C. 45.9 
D. 23.0

Answer 1

In a triangle, a = 6 ft, b = 8 ft and C = 73°.

Law of cosines: c² = a² + b² - 2ab Cos C

Substituting known values we get,

c² = 6² + 8² [2(6)(8) Cos 73]

c² = 36 + 64 - [96(0.2924)]

c² = 100 - 28.1

c² = 71.93

c = √71.93

c = 8.48

s = (a + b + c) / 2

Substitute a = 6, b =8 and c = 8.48 we get,

s = (6 + 8 + 8.48) / 2

s = 22.48 / 2

s = 11.24

A = √s(s - a)(s - b)(s - c)

Substitute a = 6, b =8, c = 8.48 and s = 11.24 we get,

A = √[11.24(11.24 - 6)(11.24 - 8)(11.24 - 8.48)]

A = √[11.24(5.24)(3.24)(2.76)]

A = √[11.24(46.86)]

A = √526.69

A = 22.95 ≅ 23.0

Therefore the answer is D.

Answer 2

Area of an Oblique Triangle :

The area of any triangle is one - half the product of the lengths of two sides times the sine of their included angle. That is,

Area = 1/2 bc sin A = 1/2 ab sin C = 1/2 ac sin B.

The lengths of the triangle are 6 feet and 8 feet, and the angle between these sides is 73 degrees.

Area = 1/2 bc sin A.

Area = 1/2 (6 feet)(8 feet) sin (73^o)

Area ≅ (3)(8) (0.9563) feet²

Area ≅ (24)(0.9563) feet²

Area ≅ 22.951314 feet²

Area ≅ 23 feet²

The option D is correct choice.