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
Show working.
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