What sort of construction is useful here?

Geometry Level 5

Let A B C ABC be a triangle with ∠ A B C = 4 5 ∘ \angle ABC = 45^{\circ} . Let D D be the point on B C BC such that 2 B D = C D 2BD=CD . If ∠ B A D = 1 5 ∘ \angle BAD = 15^{\circ} , find ∠ A C B \angle ACB in degrees.


The answer is 75.

Sharky Kesa by Sharky Kesa , 19, United Kingdom

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1 solution

Let ∠ A C B = θ \angle ACB = \theta

Then, ∠ D A C = 2 π 3 − θ \angle DAC=\dfrac{2π}{3}-\theta

In △ A B D \triangle ABD :

A B sin ⁡ π 3 = B D sin ⁡ π 12 \dfrac{AB}{\sin \dfrac{π}{3}}=\dfrac{BD}{\sin \dfrac{π}{12}}

In △ A B C \triangle ABC :

A B sin ⁡ θ = 3 B D sin ⁡ ( 3 π 4 − θ ) \dfrac{AB}{\sin \theta}=\dfrac{3BD}{\sin (\dfrac{3π}{4}-\theta)}

Solving the above equations , we get:

θ = 5 π 12 \theta=\dfrac{5π}{12}

3 Helpful 0 Interesting 1 Brilliant 1 Confused

Mind the title, good solve though..

Vishwash Kumar ΓΞΩ - 3 years, 1 month ago

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