Question

The side BC of ΔABC is produced to D. If ∠ACD = 114° and ∠ABC= (1/2)∠BAC, then what is the value (in degrees) of ∠BAC?

A) 36 B) 48 C) 76 D) 84

Answer 1

★ Figure refer to attachment

Given

• The side BC of ΔABC is produced to D
• ∠ACD = 114°
• ∠ABC= (1/2)∠BAC

Find out

• Find the value of ∠BAC

Solution

★ Let the ∠ABC be x and ∠BAC be 2x

**According to exterior angle property**

An exterior angle of a triangle is equal to the sum of the opposite interior angles.

→ ∠BAC + ∠ABC = ∠ACD

→ 2x + x = 114°

→ 3x = 114°

→ x = 114/3

→ x = 38°

Hence,

• ∠ABC = x = 38°
• ∠BAC = 2x = 38 × 2 = 76°

Additional Information

★If two parallel lines are cut by a transversal,then

• Pairs of corresponding angles are equal
• pairs of alternate angles are equal
• interior angles on the same side of the transversal are supplementary

Answer 2

Given:-

• ∠ACD = 114°
• ∠ABC= $\sf \dfrac{1}{2}$ ∠BAC

To find:-

• The value (in degrees) of ∠BAC?

Solution:-

Let ∠ABC be x

Then, ∠BAC = 2x

Using Exterior Angle Property:-

An exterior Angle of a triangle is equals to sum of opposite interior angles of a triangle.

$:\implies$ ∠BAC + ∠ABC = ∠ACD

$:\implies$ 2x + x = 144°

$:\implies$ 3x = 144°

$:\implies$ x= $\sf \cancel{ \dfrac{144^\circ}{3}}$

$:\implies$ x = 38°

Then, ∠BAC = 2x

$:\implies$ 2 × 38°

$:\implies$ 76°

Hence, ∠BAC is 76°.

Therefore, Option (C) is correct.

$\rule{200}3$