Find area of triagle

Question

Find area of triangle ABC where side a = 19, base = 112 and side c = 113

Answer 1

Use Heron's formula if the triangle is not a right triangle

Area = sqr(p(p-a)(p-b)(p-c)) where p= 1/2 perimeter

p = (112+113 + 19)/2 = 244/2 = 122

Area = sqr(122)(122-112)(122-113)(122-19))

=sqr(122(10)(9)(103)) = 1063.46

or you could use laws of sines or cosines to calculate the area

sides A, B, C, with opposite angles, a, b, c

a is the angle opposite side A

A^2 = B^2 + C^2 -2BCcos(a)

113^2 = 112^2 + 19^2 - 2(112)(19)cosa

12769 = 12544+361 -4256cosa

cosa = 136/4256 = .032

a = cos^-1(.032) = 88.17, almost a right triangle

If it had been 90 degrees, then A would be the hypotenuse

and base times the other side/ 2 = area = 112x19/2 = 1064

but it's only 88.17 not 90 degrees

area = absinC/2 = 112(19)sin88.17/2 = 2128/2 =1064

the law of sines says ratios of sides to sines of opposite sides are equal

sina/A =sinb/B = sinc/C

sin88.17/113 = sinb/112=sinc/19

use those equations to solve for b or c

for a 3rd route to the solution = about 1064

or just go on line to a website that calculates it for you

when you have 2 sides and one angle