NMTC Practice problem 3
Geometry
Level
3
In triangle ABC , D is the mid point of BC. Angle ADB=45 , Angle ACD=30 . Determine angle BAD.
The answer is 30.
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I n Δ C A D , i n t e r n a l ∠ C A D = 4 5 − 3 0 = 1 5 . I n Δ D A B ∠ A B D = 1 8 0 − 4 5 − D A B = 1 3 5 − D A B . A p p l y i n g S i n L a w t o Δ s A B D A C D , S i n B A D S i n ( 1 3 5 − B A D ) = A D B D = A D D C = S i n 1 5 S i n 3 0 E x p a n d i n g S i n a n d s i m p l i f y i n g , C o t B A D = 2 ∗ 2 ∗ C o s 1 5 − 1 . ∠ B A D = T a n − 1 ( 2 ∗ 2 ∗ C o s 1 5 − 1 1 ) = 3 0 o .