using trig to solve problems
1. a triangle has sides of lengths 19cm, 20cm and 21cm
a) calculate the size of each angle of the triangle
b) calculate the area of the triangle
+9466 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
Acute scalene triangle.
Sides: a = 19 b = 20 c = 21
Area: T = 172.3376879396
Perimeter: p = 60
Semiperimeter: s = 30
Angle ∠ A = α = 55.1550095421° = 55°9' = 0.96325507479 rad
Angle ∠ B = β = 59.75109669494° = 59°45'3″ = 1.04328511045 rad
Angle ∠ C = γ = 65.09989376296° = 65°5'56″ = 1.13661908012 rad
Height: ha = 18.1410724147
Height: hb = 17.23436879396
Height: hc = 16.4133036133
Median: ma = 18.17327818454
Median: mb = 17.34993515729
Median: mc = 16.43992822228
Inradius: r = 5.74545626465
Circumradius: R = 11.57661641211
Vertex coordinates: A[21; 0] B[0; 0] C[9.57114285714; 16.4133036133]
Centroid: CG[10.19904761905; 5.47110120443]
Coordinates of the circumcenter: U[10.5; 4.87441743668]
https://www.triangle-calculator.com/?what=sss&a=19&b=20&c=21&submit=Solve
+9466 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
