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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

Jan 24, 2018

#2
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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

Jan 24, 2018

#1
0

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

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

Jan 24, 2018
#2
+2

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

hectictar Jan 24, 2018