Question

In the given figure, BP and CP are the bisectors of angles ∠CBD and ∠BCE respectively. If ∠BAC = 30°, then ∠ABP + ∠ACP equals

• A. 215°
• B. 240°
• C. 255°
• D. 225°

Answer 1

The correct option is C 255°


Since, BP bisects ∠CBD.
$\therefore \angle CBP=\angle PBD=\frac{1}{2}\angle CBD$ ...(i)
Similarly, CP bisects ∠BCE.
$\therefore \angle BCP=\angle PCE=\frac{1}{2}\angle BCE$ ....(ii)
Now, ∠CBD + ∠ABC = 180° (Linear pair)
⇒ ∠ABC = 180° – 2∠CBP [From (i)] …..(iii)
In ΔABC,
∠BAC + ∠ABC = ∠BCE (Exterior angle property)
⇒ 30° + 180° – 2∠CBP = 2∠BCP [From (ii) and (iii)]
⇒ 2(∠BCP + ∠CBP) = 210°
⇒ ∠BCP + ∠CBP = 105°
∴ ∠DBP + ∠PCE = 105° [From (i) and (ii)] …..(iv)
Now, ∠ABP + ∠ACP = (180° – ∠DBP) + (180° – ∠PCE)
= 360° – (∠DBP + ∠PCE)
= 360° – 105° [From (iv)]
= 255°
Hence, the correct answer is option (c).