The area of an acute-angled triangle with sides $|AC|=40$ and $|BC|=29$ is $420$. Calculate the length of the radius of the circle inscribed in this triangle.
SOLUTION:
(1) We sketch a triangle
(2) We express the area of a triangle using the formula when two sides and the included angle of a triangle are given: $$ \frac12 |AC||BC| \sin \gamma =420 $$ (3) This enables us to find $\sin \gamma$: $$ \begin{align} \sin \gamma &=\frac{840}{40\cdot 29} \cr \sin \gamma &=\frac{21}{29} \end{align} $$
(4) Now we calculate $\cos \gamma$: $$ \begin{align} \cos \gamma & =\sqrt{1-\sin^2\gamma } \cr \cos \gamma &=\sqrt{1- \frac{441}{841}} \cr \cos \gamma &=\frac{20}{29} \end{align} $$
(5) Now that we know the value of $\cos\gamma$ , we can determine the length of the side $c$ using the Law of Cosines: $$ \begin{align} c^2&=|AC|^2+|BC|^2-2|AC||BC| \cos \gamma \cr c^2&=1600+841-2\cdot 40\cdot 29\cdot \frac{20}{29} \cr c^2&=841 \cr c&=29 \end{align} $$ (6) Finally, we apply the formula for the radius $r$ of an inscribed circle in a triangle: $$ r=\frac{P}{s} $$ where $P$ is the area of the triangle and $s=|AB|+|BC|+|AC|$, so $$ \begin{align} r&=\frac{420}{98} \cr r&=\frac{30}7 \end{align} $$
Is the solution all right or is there an error in the solution? Select the correct answer.
The error is in step (6). For $s$ we must substitute $\frac12 (|AB|+|BC|+|AC|)$.
The error is in step (4). It should also have been $$ \cos \gamma =-\sqrt{1-\sin^2\gamma} $$ so this example has two solutions.
An error occurred in step (5) while applying the Law of Cosines. It should have been $$ c^2=|AC|^2+|BC|^2-|AC||BC| \cos \gamma $$
The solution is correct.
Notice that the area of a tringle can be expressed in terms of the lengths of its sides and the length $r$ of the radius of the inscribed circle:
$$ P=\frac12 r(|AB|+|BC|+|AC|) $$
