Given $\cos A$ and $\cos B$ , find $\cos C$

Question

In a triangle $ABC$, $\cos A = 3/5$ and $\cos B = 24/25$. Find the value of $\cos C$.

Question: Is this even possible since cosine is adjacent over hypotenuse? Or is there a formula or theorem involving these kind of problems?

Answer 1

In a triangle $A+B+C=\pi$, hence $$\cos(C)=\cos(\pi-(A+B))=-\cos(A+B)=\sin(A)\sin(B)-\cos(A)\cos(B)$$ where $\cos(A)=\frac{3}{5}$ implies $\sin(A)=\frac{4}{5}$ and $\cos(B)=\frac{24}{25}$ implies $\sin(B)=\frac{7}{25}$. It follows that $$ \cos(C) = \frac{4\cdot 7-3\cdot 24 }{5\cdot 25}=\color{red}{-\frac{44}{125}}.$$

Answer 2

From $\cos A$ and $\cos B$ we know angles A and B by $\arccos$ and use $A+B+C=180°$ or as alternative note that from the latter we can use trigonometric identities to find out $\cos C$.

Notably with the forst method we obtain

$$\cos C=\cos \left(\pi - \arccos \frac35 - \arccos \frac{24}{25}\right)=-\frac{44}{125}$$

For the second, as an alternative to the standard way shown in other answers, for example we can use the following

$$\cos ^{2}A +\cos ^{2}B +\cos ^{2}C =-2 \cos A \cos B \cos C +1 \\\implies\cos^2 C+\frac{144}{125}\cos C+\frac{176}{625}=0 \implies \cos C=-\frac{44}{125}$$

Answer 3

Hint: $$A+B+C = \pi \implies C = \pi -(A+B) \implies \cos(C) = -\cos(A+B)$$

Answer 4

it is $$-\cos\left(180^{\circ}-(A+B)\right)=-\cos(A+B)=-\cos(A)\cos(B)+\sin(A)\sin(B)$$