A triangle has angle alpha = 36 degrees, angle gamma = 50 degrees, and the side b = 78.2 between them. Approximate the area of the triangle to the nearest tenth.

Step-by-step solution

1. 1

   Find the third angle first — nothing else can start without it. The angles of a triangle sum to $180^\circ$:

   $\beta=180^\circ-\alpha-\gamma=180^\circ-36^\circ-50^\circ=94^\circ$

   Because $\beta$ is obtuse, side $b$ (opposite $\beta$) is the longest side of the triangle — a useful reality check on the numbers that follow.

2. 2

   Recognise the configuration: the known side is between the known angles (ASA). Standard labelling puts side $b$ opposite angle $\beta$, so with $\alpha$ and $\gamma$ known, $b$ is the included side. Every angle and one length are now determined, which means the triangle is fixed and there is no ambiguous-case worry (that only arises in SSA problems).

3. 3

   Derive the area formula from the law of sines. The area is $K=\tfrac12ac\sin\beta$, and the law of sines gives the two unknown sides in terms of $b$:

   $a=\frac{b\sin\alpha}{\sin\beta},\qquad c=\frac{b\sin\gamma}{\sin\beta}$

   Substituting,

   $K=\frac12\cdot\frac{b\sin\alpha}{\sin\beta}\cdot\frac{b\sin\gamma}{\sin\beta}\cdot\sin\beta=\frac{b^2\sin\alpha\sin\gamma}{2\sin\beta}$

   Using this single formula avoids rounding the intermediate side lengths, which is where two-step solutions typically lose a tenth.

4. 4

   Substitute the values. With $b^2=78.2^2=6115.24$ and (to seven decimals)

   $\sin36^\circ=0.5877853,\quad \sin50^\circ=0.7660444,\quad \sin94^\circ=0.9975641$

   the numerator and denominator are

   $6115.24\times0.5877853\times0.7660444=2753.5068,\qquad 2\times0.9975641=1.9951281$

5. 5

   Divide and round.

   $K=\frac{2753.5068}{1.9951281}=1380.1153\ldots\approx1380.1$

   Note the digits after the decimal point: the unrounded value is $1380.115$, so it rounds down to $1380.1$. Carrying only four-decimal sines gives $1380.2$ and would produce the wrong rounded answer.

6. 6

   Cross-check with a completely different method. Solving for the sides gives $a=b\sin\alpha/\sin\beta=46.0770$ and $c=b\sin\gamma/\sin\beta=60.0510$. Then $K=\tfrac12ac\sin94^\circ=1380.1153$, and Heron's formula with $s=\tfrac{a+b+c}{2}=92.1640$ also returns $1380.1153$. Three independent routes agreeing to four decimals settles the answer.