Triangles ABC and CDE are equilateral triangles of the same size. If AC=10 and ∠BCD=80, find the area of triangle ABF.
Since ∠ACB=60∘, ∠ACD=60∘+80∘=140∘.
Since △ACD is isosceles, ∠CAD=∠ADC=2180∘−140∘=20∘.
It then follows that ∠BAF=∠BAC−∠DAC=60∘−20∘=40∘.
By the exterior angle theorem, ∠BFA=20∘+60∘=80∘.
By applying sine law on △BAF, we have
sin40∘BF=sin80∘10⟹BF=sin80∘sin40∘(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=21(AB)(BF)(sin60∘)=21(10)(sin80∘sin40∘)(10)(sin60∘)≈28.263. □