Question

3. In AABC, AC = BC. angle BAC is bisected by AD
and AD = AB. Find angle ACB.​

Answer 1

Answer

36°

Explanation

Given AC = BC

→ angle BAC = angle ABC

Let, angle ABC = y

→ angle BAC = y

Nowz angle BAC is bisected by AD

→ angle BAD = (angle BAC)/2 = y/2

Also, given that AD = AB

→ angle ABD = angle ADB

angle ABC = angle ABD (as they both coincide with each other)

→ angle ABD = y

→ angle ADB = y

By angle sum property in ∆ ABD

→ y/2 + y + y = 180°

→ 5y/2 = 180°

→ y = 180/5 × 2

→ y = 36 × 2

→ y = 72°

Now, in ∆ACB,

angle BAC + angle ABC + angle ACB = 180° (by angle sum property)

→ y + y + angle ACB = 180°

→ 72 + 72 + angle ACB = 180°

→ angle ACB = 180 - 144

→ angle ACB = 36°

Answer 2

$\huge{\mathfrak{</p><p><strong>Answer:</strong></p><p>}}$

$\huge{\bf{\underline{Given}}}$

$\large{\sf{AD \: = \: AB}}$

$\large{\sf{AC \:=\:BC}}$

$\large{\sf{{\angle} \:BAC \:=\:angle\:ABC}}$

$\large{\sf{Let \:\: {\angle} \:\:ABC\:\:be \:\:and \:BAC\:\:be\:\:x.}}$

$\large{\sf{Now \: \: {\angle} \:BAC \: is \:\: bisected \:\:by \:\:AD. }}$

$\large{\sf{{\angle} \: \: BAC \: = \: x/2}}$

$\large{\sf{{\angle} \: ABD \: = y}}$

$\large{\sf{{\angle} \: ADB \: = y}}$

$\huge{\bf{By \:angle\:sum\:property}}$

$\large{\sf{x/2 + x + x = 180^{\circ}}}$

$\large{\sf{5x/2 = 180}}$

$\large{\sf{ 5x = 360}}$

$\large{\sf{x = 72^{\circ}}}$

» x = 72°

$\large{\sf{{\angle} \:BAC \:+ \:{\angle}\:ABC \:+\:{\angle} \:ACB \: = \: 180^{\circ}}}$

$\large{\sf{x + x + {\angle} \: ABC = 180^{\circ}}}$

$\large{\sf{{\angle} \: ACB = 36^{\circ}}}$