Euclidean Geometry - Triangles Problem Solving

This wiki is about problem solving on triangles. You need to be familiar with some (if not all) theorems on triangles.

Triangles \(ABC\) and \(CDE\) are equilateral triangles of the same size. If \(AC=10\) and \(\angle BCD=80\), find the area of triangle \(ABF\).

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Since \(\angle ACB=60^\circ\), \(\angle ACD=60^\circ+80^\circ=140^\circ.\)

Since \(\triangle ACD\) is isosceles, \(\angle CAD=\angle ADC=\dfrac{180^\circ-140^\circ}{2}=20^\circ.\)

It then follows that \(\angle BAF=\angle BAC-\angle DAC=60^\circ-20^\circ=40^\circ.\)

By the exterior angle theorem, \(\angle BFA=20^\circ+60^\circ=80^\circ.\)

By applying sine law on \(\triangle BAF\), we have

\[\dfrac{BF}{\sin 40^\circ}=\dfrac{10}{\sin 80^\circ}\implies BF=\dfrac{\sin 40^\circ}{\sin 80^\circ}(10).\]

Note that the area of a triangle is half the product of two adjacent sides multiplied by the sine of the included angle. We have

\[A=\dfrac{1}{2}(AB)(BF)(\sin 60^\circ)=\dfrac{1}{2}(10)\left(\dfrac{\sin 40^\circ}{\sin 80^\circ}\right)(10)(\sin 60^\circ) \approx 28.263.\ _\square \]

• Triangles
• Pythagorean Theorem
• Isosceles Triangle Theorem