Side BC of a ΔABC is extended upto D such that CD = AC. If ∠BAD = 118º and ∠ACB = 80º, then the value of ∠ABC is equal to:

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This question was previously asked in

SSC CHSL Exam 2024 Tier-I Official Paper (Held On: 08 Jul, 2024 Shift 4)

Answer 1

Given:

∠BAD = 118º

∠ACB = 80º

CD = AC (so ΔACD is isosceles)

Solution:

Step 1: Use the exterior angle theorem in ΔACD

Since ΔACD is isosceles, we have:

⇒ ∠CAD = ∠CDA

From the exterior angle theorem:

∠ACB = ∠CAD + ∠CDA

Given ∠ACB = 80º, and since ∠CAD = ∠CDA:

⇒ 80º = 2∠CAD

⇒ ∠CAD = 40º

Step 2: Find ∠BAC using the exterior angle theorem

We are given ∠BAD = 118º. From the exterior angle theorem in ΔABC:

∠BAD = ∠BAC + ∠CAD

⇒ 118º = ∠BAC + 40º

⇒ ∠BAC = 118º - 40º = 78º

Step 3: Use the angle sum property in ΔABC

In ΔABC, the sum of interior angles is 180º:

∠BAC + ∠ABC + ∠ACB = 180º

Substitute the known values:

78º + ∠ABC + 80º = 180º

⇒ ∠ABC = 180º - 158º = 22º

∴ The value of ∠ABC is 22º.

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