Question

Can you help me with this problem? Find the area of the triangle. Round to the nearest tenth. Hint: first use the Law of Sines to solve for c.

Answer 1

Recall that the sine law states that:

`(CA)/(\sin B)=(AB)/(\sin C)=(BC)/(\sin B)\text{.}`

Therefore:

`(c)/(\sin105^(\circ))=(7)/(\sin 35^(\circ))\text{.}`

Solving for c we get:

`\begin{gathered} c=(7*\sin 105^(\circ))/(\sin 35^(\circ)), \\ c\approx11.7883. \end{gathered}`

Now, we know that the interior angles of a triangle add up to 180 degrees, therefore the measure of angle A is:

`\begin{gathered} \measuredangle A=180^(\circ)-35^(\circ)-105^(\circ), \\ \measuredangle A=40^(\circ). \end{gathered}`

Using the sine law we get:

`(CB)/(\sin40^(\circ))=(7)/(\sin 35^(\circ))\text{.}`

Solving the above equation for CB we get:

`\begin{gathered} CB=(7*\sin 40^(\circ))/(\sin 35^(\circ)), \\ CB\approx7.8447. \end{gathered}`

Finally, using Heron's formula we get that the area of the given triangle is:

`\begin{gathered} A=\sqrt[]{13.3165(13.3165-7.8447)(13.3165-11.7883)(13.3165-7)}, \\ A=\sqrt[]{13.3165(5.4718)(1.5282)(6.3165)}, \\ A\approx\sqrt[]{703.3589}, \\ A\approx26.5. \end{gathered}`

Answer:

`A\approx26.5\text{ units}^2.`