In the following question, ABCDE and HIJKL are regular pentagons and AEFGHL is a regular hexagon. If ∠LKB = ∠ABK, then, what is the value of ∠ABK?

In the following question, ABCDE and HIJKL are regular pentagons and AEFGHL is a regular hexagon. If ∠LKB = ∠ABK, then, what is the value of ∠ABK? Correct Answer <span lang="EN-IN" style=" line-height: 107%; background-image: initial; background-position: initial; background-size: initial; background-repeat: initial; background-attachment: initial; background-origin: initial; background-clip: initial;">48°</span>

Given:

∠LKB = ∠ABK and ABCDE and HIJKL are regular pentagons and AEFGHL is a regular hexagon.

Formula used:

Sum of the interior angles of the polygon = (n - 2) × 180°

Each Interior angle = / n

Calculation:              

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Each interior angle of pentagon = (5 - 2)× 180° / 5 = 108°

Each interior angle of hexagon = (6 - 2) × 180° / 6 = 120°

Now, ∠BAE + ∠EAL + ∠BAL = 360°

⇒ 108° + 120° + ∠BAL = 360°

⇒ 228° + ∠BAL = 360°

⇒ ∠BAL = 132°

Similarly, ∠KLA = 132°

Suppose, ∠ABK = ∠LKB = x°

⇒ x° + x° + 132° + 132° = 360°

⇒ 2x°+ 264° = 360°

⇒ 2x° = 96°

⇒ x° = 96° / 2

⇒ x° = 48°