Approximate the area of the triangle with α = 36°, γ = 50° and b = 78.2, to the nearest tenth.

Step-by-step solution

1. 1

   Find the third angle. The angles of a triangle sum to $180^{\circ}$:

   $\beta=180^{\circ}-36^{\circ}-50^{\circ}=94^{\circ}$

   Note that $b$ is the side opposite $\beta$, so the given side lies between the two given angles — the AAS/ASA configuration.

2. 2

   Choose the right area formula. With two angles and the side between them, the standard result is

   $K=\frac{b^{2}\sin\alpha\,\sin\gamma}{2\sin\beta}$

   It follows from $K=\tfrac12 ac\sin\beta$ together with the law of sines, $a=\tfrac{b\sin\alpha}{\sin\beta}$ and $c=\tfrac{b\sin\gamma}{\sin\beta}$.

3. 3

   Evaluate the trigonometric values (degree mode).

   $\sin 36^{\circ}=0.587785,\qquad \sin 50^{\circ}=0.766044,\qquad \sin 94^{\circ}=0.997564$

4. 4

   Substitute.

   $b^{2}=78.2^{2}=6115.24$

   $K=\frac{6115.24\times 0.587785\times 0.766044}{2\times 0.997564}=\frac{2753.50}{1.995128}=1380.1$

   $\boxed{K\approx 1380.1}$

5. 5

   Cross-check by finding the sides first. $a=\dfrac{78.2\sin 36^{\circ}}{\sin 94^{\circ}}=46.08$ and $c=\dfrac{78.2\sin 50^{\circ}}{\sin 94^{\circ}}=60.05$, so $K=\tfrac12(46.08)(60.05)\sin 94^{\circ}=\tfrac12(2766.97)(0.997564)=1380.1$ — agreeing with the formula ✓.

6. 6

   Sanity-check the size. The triangle fits inside a rectangle of about $60$ by $46$, area $2767$, and $1380$ is almost exactly half of that — which is right, because $\beta=94^{\circ}$ is very close to a right angle, making the triangle nearly half of the rectangle on sides $a$ and $c$.