In the given diagram, ABCD is a square and ΔBCT is an equilateral tri

1.	In the given diagram, ABCD is a square and ΔBCT is an equilateral triangle. ∠BTD equals(a) 30° (b) 15° (c) 45° (d) 35°
Answer» (c) 45° ∠DCB = 90° (ABCD is a square) ∠TCB = 60° (DCT is an equilateral Δ) ∴ ∠DCT = 90° + 60° = 150° DC = CB (Adj sides of a square) CB = CT (Sides of an equilateral Δ) ⇒ DC = CT ⇒ ∠CTD =∠CDT (isos. Δ property) In ΔDCT, ∠CTD = \(\frac12\) (180° –∠DCT) = \(\frac12\) (180° – 150°) = 15° ∴ ∠BTD =∠BTC – ∠CTD= 60° – 15° = 45°.

In the given diagram, ABCD is a square and ΔBCT is an equilateral triangle. ∠BTD equals(a) 30° (b) 15° (c) 45° (d) 35°

Answer»

(c) 45°

∠DCB = 90° (ABCD is a square)

∠TCB = 60° (DCT is an equilateral Δ)

∴ ∠DCT = 90° + 60° = 150°

DC = CB (Adj sides of a square)

CB = CT (Sides of an equilateral Δ)

⇒ DC = CT ⇒ ∠CTD =∠CDT (isos. Δ property)

In ΔDCT, ∠CTD = \(\frac12\) (180° –∠DCT)

= \(\frac12\) (180° – 150°) = 15°

∴ ∠BTD =∠BTC – ∠CTD= 60° – 15° = 45°.