In triangle ABC, D is the midpoint of BC. If angle ADB = 45 degrees and angle ACD = 30 degrees, what is angle BAD?
+128475 Angle ADC = 180 - 45 = 135
So angle DAC = 180 - 30 - 135 = 15
Let angle BAD = x
Using the Law of Sines
sin 30 / AB = sin ( 15 + x) / (2BD) ⇒ (2BD) / AB = sin (15 + x) / sin 30 (1)
Also
sin 45 / AB = sin x / BD
sin 45/ AB = 2 sin x / (2BD) ⇒ (2BD) / AB = 2 sin x / sin 45 (2)
Equate (1) , (2)
sin (15 + x) / sin 30 = 2 sin x / sin 45
sin (x + 15) / ( 2sin 30) = sin x / sin 45 { 2sin 30 = 1 }
sin (x + 15) = sin x / sin 45
sin x cos 15 + cos x sin 15 = sin x / sin 45
sin x ( 1/sin 45 - cos 15) = cosx sin 15
sin x / cos x = sin 15 / ( 1/sin 45 - cos 15)
tan x = sin 15 / ( √2 - cos 15)
tan x = 1/ √3
arctan (1/ √3) = x = 30° = BAC
