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 theexterior angle theorem, $\angle BFA=20^\circ+60^\circ=80^\circ.$

By applyingsine lawon $\triangle BAF$, 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