Question bank/JEE Main/Mathematics

In a triangle ABC, let D be the midpoint of BC. If angle ADB is
45° and angle ACD is 30° then angle BAD will be equal to ?
f=Hkqt ABC esa ekuk D Hkqtk BC fcUnq s.k ADB dk eku
45º s.k ACD dk eku 30º rc dks.k BAD

A15°B30°C45°D$22\frac{1}{2}^{\circ}$

Show answer and solution▾

Using m-n rule
m-n
$2cot135^{\circ} = cot\alpha - cot15^{\circ}$
$- 2 = cot\alpha - 2 - \sqrt{3}$
$cot\alpha = \sqrt{3}$ ⇒ $\alpha = 30^{\circ}$

Source: Grass-G / M-26 Mathematics Paper 1 15 Jan.docx