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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 

Guest Jan 24, 2018

Best Answer 

 #2
avatar+7324 
+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
 #1
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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

Guest Jan 24, 2018
 #2
avatar+7324 
+2
Best Answer

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

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