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

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.