remember the law of sines
your largest angle will be opposite your longest side
and you have a total of 180 degrees to work with
If the sides of a triangle are 4p, 5p and 6p, calculate the size of the lagest angle.
6 answers
jolly am still lost, pls direct me sir.
forget the p. It is just a scale factor.
using the law of cosines,
6^2 = 4^2 + 5^2 - 2*4*5*cosA
A = 82.82°
Now do a similar equation for B, and then C is easy, since A+B+C=180
using the law of cosines,
6^2 = 4^2 + 5^2 - 2*4*5*cosA
A = 82.82°
Now do a similar equation for B, and then C is easy, since A+B+C=180
sorry Somay - Law of Cosines:
c^2 = a^2 + b^2 – 2ab cosC
where a, b and c are the lengths of the three sides opposite the three angles A, B and C (respectively)
a is the side opposite angle A, etc.
Now let a = 4p and b = 5p and c = 6p
Then you have cosC = – (c^2 – a^2 – b^2)/2ab = – (36 – 16 – 25) / 40 = 1/8 = 0.125
cosC = 82.82 degrees
pls check my algebra in case i made a mistake
c^2 = a^2 + b^2 – 2ab cosC
where a, b and c are the lengths of the three sides opposite the three angles A, B and C (respectively)
a is the side opposite angle A, etc.
Now let a = 4p and b = 5p and c = 6p
Then you have cosC = – (c^2 – a^2 – b^2)/2ab = – (36 – 16 – 25) / 40 = 1/8 = 0.125
cosC = 82.82 degrees
pls check my algebra in case i made a mistake
okay that's great. Steve thanks a lot
oh okay. Thank you jolly, that's great.
1 answer