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If a=3 cm b=4 cm C=5 cm what is the magnitude of angle B    

triangle ABC in which a= 5 cm B= 9 cm C= 12 cm determine the largest and the smallest angle 

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Guest Sep 7, 2017
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We can use the law of cosines for these.

The law of cosines says:     c2  =  a2 + b2  -  2ab cos C  ,   where  C  is the angle opposite side  c .

 

In the first problem, we want to know the measure of the angle across from the side that is  4  cm.

 

42  =  32 + 52 - 2(3)(5)cos B

                                                    Simplify.

16  =  9 + 25 - 30 cos B

16  =  34 - 30 cos B

                                                    Subtract  34  from both sides.

-18  =  -30 cos B

                                                    Divide both sides by  -30 .

0.6  =  cos B

                                                    Take the inverse cosine of both sides.

acos( 0.6 )  =  B

 

      53.13°  ≈  B

 

 

For the second problem,  note that the smallest angle is opposite the smallest side, and the largest angle is opposite the largest side.   ( You can look at this to see why. )

 

So...the angle across from the side that is  5 cm  will be the smallest.

We can find the measure of this angle,  " A " , with the law of cosines again.

 

52  =  92 + 122  -  2(9)(12) cos A

 

25  =  225 - 216 cos A

 

200/216  =  cos A

 

acos(200/216)  =  A

 

           22.19º   ≈   A

 

And the largest angle is across from the  12 cm  side... " C " .

 

122  =  52 + 92  -  2(5)(9) cos C

 

144  =  106 - 90 cos C

 

38/ - 90  =  cos C

 

114.98º  ≈  C

hectictar  Sep 8, 2017

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