How do you solve the triangle given #a=1.42, b=0.75, c=1.25#?

1 Answer
May 16, 2017

Answer:

A = 86.68
B = 31.82
C = 61.5
(2dp)

Explanation:

Just to clarify, #A# is the angle opposite side #a#, #B# opposite side #b#, and the same for #C#.

1st step : Get your calculator!

2nd step : Let's start by working out the angle #A#. The Law of Cosines states that #cos A = (b^2+c^2-a^2)/(2bc)#
Let's substitute the known values :
#cos A = (0.75^2+1.25^2-1.42^2)/(2*0.75*1.25)#

3rd step : At this point, you will probably need a calculator. Ok, let's simplify this. With my calculator, I got : #181/3125#

4th step : So right now, #cos A = 181/3125#
We need to get #A# on its own. To do that, we take the inverse cosine of #181/3125#.
#cos^-1(181/3125) = 86.67957016......#
Let's just round that to 2 decimal places : 86.68

Next, let's work out side B. Although you can solve this using sine, since you asked us about using the Law of Cosine, I will use the Law of Cosine.

#cos B = (a^2+c^2-b^2)/(2ac)#

Substitute known values :

#cos B = (1.42^2+1.25^2-0.75^2)/(2*1.42*1.25)#

Plug in numbers on your calculator

#cos B = 7541/8875#

Take the inverse cosine of that :

#B = 31.82201662#
Round it to 2 decimal places : 31.82

Ok, so because interior angles add up to 180 in a triangle, let's set this equation for the last angle :

#31.82+86.68+C=180#
#118.5+C=180#
#C=61.5#

I'll leave it for you to check angle C using the Law of Cosines.