Find ∠ A B C \angle ABC

Geometry Level 3

In △ A B C \triangle ABC , point D D on A B AB is such that A C = D B AC=DB , ∠ C A B = 1 2 ∘ \angle CAB=12^\circ , and ∠ A C D = 6 ∘ \angle ACD = 6^\circ . Find ∠ A B C \angle ABC in degrees.


The answer is 30.

Parallax's Den by Parallax's Den , 19, Bangladesh

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1 solution

Let ∣ A C ‾ ∣ = x |\overline {AC}|=x . Then ∣ A D ‾ ∣ = x sin ⁡ 6 ° sin ⁡ 18 ° , ∣ A B ‾ ∣ = x sin ⁡ 18 ° + sin ⁡ 6 ° sin ⁡ 18 ° |\overline {AD}|=\dfrac{x\sin 6\degree}{\sin 18\degree}, |\overline {AB}|=x\dfrac{\sin 18\degree+\sin 6\degree}{\sin 18\degree} . Applying sine rule to △ A B C \triangle {ABC} we get

x sin ⁡ 18 ° + sin ⁡ 6 ° sin ⁡ 18 ° = x sin ⁡ ( α + 12 ° ) sin ⁡ α x\dfrac{\sin 18\degree+\sin 6\degree}{\sin 18\degree}=x\dfrac{\sin (α+12\degree)}{\sin α} or

cot ⁡ α = sin ⁡ 18 ° + sin ⁡ 6 ° − sin ⁡ 18 ° cos ⁡ 12 ° sin ⁡ 18 ° sin ⁡ 12 ° \cot α=\dfrac{\sin 18\degree+\sin 6\degree-\sin 18\degree\cos 12\degree}{\sin 18\degree\sin 12\degree} .

After some simplification work, we get cot ⁡ α = cot ⁡ 30 ° \cot α=\cot 30\degree , and hence α = 30 ° α=\boxed {30\degree}

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