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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.

Can you help me with this problem? Find the area of the triangle. Round to the nearest-example-1
User Mohamed AL ANI
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1 Answer

15 votes
15 votes

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.

User Trevorsky
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