a) We can use the law of cosines to find the angles.
c2 = a2 + b2 - 2ab cos C , where C is the angle opposite side c .
Let a = 19 , b = 20 , c = 21 and A , B , and C be the angles opposite their respective sides.
212 = 192 + 202 - 2(19)(20)cos C
441 = 361 + 400 - 760 cos C
441 = 761 - 760 cos C
Subtract 761 from both sides of the equation.
-320 = -760 cos C
Divide both sides by -760 .
8/19 = cos C
Take the inverse cosine of both sides.
C = acos( 8/19 ) ≈ 65.099°
Now let's use the law of cosines again to find the angle opposite the 20 cm side.
202 = 192 + 212 - 2(19)(21)cos B
400 = 361 + 441 - 798 cos B
400 = 802 - 798 cos B
-402 = -798 cos B
67/133 = cos B
B = acos( 67/133 ) ≈ 59.751°
Since there are 180° in every triangle, A + B + C = 180° and A = 180° - B - C
A ≈ 180° - 59.751° - 65.099° ≈ 55.15°
b)
Let the base be the 19 cm side, and the height be 20 sin 65.099°
area ≈ (1/2)(19)(20 sin 65.099°) ≈ 172.337 sq cm