How do you evaluate cos A = 0.7431?

May 24, 2018

$\angle A = {42}^{\circ}$

Explanation:

$\text{given } \cos A = 0.7431$

$\text{then } \angle A = {\cos}^{-} 1 \left(0.7431\right) = {42}^{\circ}$

May 24, 2018

$A = \pm \textrm{A r c} \textrm{\cos} \left(0.7431\right) + {360}^{\circ} k \quad$integer $k$

$A \approx \pm {42.0038}^{\circ} + {360}^{\circ} k \quad$integer $k$

Explanation:

Cosines, unlike sines, uniquely determine a triangle angle between $0$ and ${180}^{\circ} .$

But we're not given any such constraint here. As far as we know, $A$ can be any angle, any real number of degrees or radians.

What's important to remember is the general solution to $\cos x = \cos a$ is $x = \pm a + {360}^{\circ} k ,$ integer $k .$

The principal value of the inverse cosine gives us a particular solution and we apply the recipe to turn it into the general solution.

$\cos A = 0.7431 = \cos \textrm{A r c} \textrm{\cos} \left(0.7431\right)$

$A = \pm \textrm{A r c} \textrm{\cos} \left(0.7431\right) + {360}^{\circ} k \quad$integer $k$

That's the exact, general answer. It gives all the $A$s whose cosine is $0.7431$.

Now we're supposed to get out our calculator and evaluate the inverse cosine. I don't like that part. I got a nice exact answer here and I have to muck it up.

$A \approx \pm {42.0038}^{\circ} + {360}^{\circ} k \quad$integer $k$