Find alpha

Geometry Level 5

In Triangle A B C ABC , A D AD is the angle bisector at A A . If ∠ A B D = 3 0 ∘ , ∠ D B C = 1 ∘ , ∠ A C D = 8 9 ∘ \angle ABD = 30 ^ \circ, \angle DBC = 1 ^ \circ, \angle ACD = 89 ^ \circ , what is the measure of ∠ D C B \angle DCB in degrees?


The answer is 2.

Aly Ahmed by Aly Ahmed , 34, Egypt

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1 solution

A p p l y i n g S i n L a w , Δ A B D , A D ∗ S i n ( A / 2 ) = B D ∗ S i n 30. Δ A C D , A D ∗ S i n ( A / 2 ) = C D ∗ S i n 89. Δ B C D , S i n ( α ) = B D ∗ S i n 1 C D = S i n 1 ∗ S i n 89 S i n 30 = 2. Applying\ Sin\ Law,\\ \Delta\ ABD,\ \ AD*Sin(A/2)=BD*Sin30.\\ \Delta\ ACD,\ \ AD*Sin(A/2)=CD*Sin89.\\ \Delta\ BCD,\ \ Sin(\alpha)=BD*\dfrac{Sin1}{CD}=Sin1*\dfrac{Sin89}{Sin30}= \Large\ \ \color{#D61F06}{2}.\\

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Yeah did the same way

will jain - 4 years, 8 months ago

Nice. I applied the same process in a different angle by using trigonometric form of ceva's theorem.

Swapnil Das - 4 years, 8 months ago

Nyc Approach Sir!

nibedan mukherjee - 4 years, 8 months ago

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Thank you.

Niranjan Khanderia - 4 years, 8 months ago

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