Question

How do you find the measure of each of the angles of a triangle given the measurements of the sides are 32, 40, 38?

Answer 1

The angles are `/_A=48.36^0, ./_B=69.09^0,/_C= 62.55^0`

Explanation:

The sides of triangle are `a=32 , b=40 ,c =38`
Let the angles opposite to sides `a,b,c` are `/_A ,/_B,/_C` respetively.
`cosA= (b^2+c^2-a^2)/(2bc)=(40^2+38^2-32^2)/(2*40*38)=0.6644 :./_A= cos^-1 (0.6644)=48.36^0(2dp)`

`cosB= (c^2+a^2-b^2)/(2ca)=(38^2+32^2-40^2)/(2*38*32)=0.3569 :./_B= cos^-1 (0.3569)=69.09^0(2dp)`

`cosC= (a^2+b^2-c^2)/(2ab)=(32^2+40^2-38^2)/(2*32*40)=0.4609 :./_C= cos^-1 (0.4609)=62.55^0(2dp)`

The angles are `/_A=48.36^0, ./_B=69.09^0,/_C= 62.55^0` [Ans]