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2. (a) In Figure (1) BP bisects ZABC and AB = AC. Find x.
(b) Find x in Figure (ii).
Given : DA = DB = DC, BD bisects
ZABC and ZADB = 70°.
A
x PP
Answers
Answer:
AB = AC, and BP bisects ∠ABC
AP ∥ BC is drawn
Now ∠PBC = ∠PBA
(∵ PB is the bisector of ∠ABC)
∴ AP||BC
∴ ∠APB = ∠PBC (Alternate angles)
⇒ x = ∠PBC …(i)
In ∆ ABC, ∠A = 60°
and ∠B = ∠C (∵ AB = AC)
But ∠A + ∠B + ∠C = 180°
(Angles of a triangle)
⇒ 60° + ∠B + ∠C = 180°
⇒ 60° + ∠B + ∠B = 180°
⇒ 2∠B = 180° - 60° = 120°
∠B = 120°/2 = 60°
= 1/2∠B = 60°/2 = 30° ⇒ ∠PBC = 30°
∴ From (i)
x= 30°
(b) In the figure (ii),
DA = DB = DC
BD Bisects ∠ABC
and ∠ADB = 70°
but ∠ADB + ∠DAB + ∠DBA = 180
(Angles of triangle)
⇒ 70° + ∠DBA + ∠DBA = 180° (∵ DA = DB)
⇒ 70° + 2∠DBA = 180°
⇒ 2∠DBA = 180° - 70° = 110°
∴ ∠DBA = 110°/2 = 55°
∵ BD is the bisector of ∠ABC,
∴ ∠DBA = ∠DBC, = 55°
But in ∆ DBC,
DB = DC
∴ ∠DCB = ∠DBC
⇒ X = 55°
Answer:
a) 30
b)55
Step-by-step explanation:
AB = AC, and BP bisects ∠ABC
AP ∥ BC is drawn
Now ∠PBC = ∠PBA
(∵ PB is the bisector of ∠ABC)
∴ AP||BC
∴ ∠APB = ∠PBC (Alternate angles)
⇒ x = ∠PBC …(i)
In ∆ ABC, ∠A = 60°
and ∠B = ∠C (∵ AB = AC)
But ∠A + ∠B + ∠C = 180°
(Angles of a triangle)
⇒ 60° + ∠B + ∠C = 180°
⇒ 60° + ∠B + ∠B = 180°
⇒ 2∠B = 180° - 60° = 120°
∠B = 120°/2 = 60°
= 1/2∠B = 60°/2 = 30° ⇒ ∠PBC = 30°
∴ From (i)
x= 30°
(b) In the figure (ii),
DA = DB = DC
BD Bisects ∠ABC
and ∠ADB = 70°
but ∠ADB + ∠DAB + ∠DBA = 180
(Angles of triangle)
⇒ 70° + ∠DBA + ∠DBA = 180° (∵ DA = DB)
⇒ 70° + 2∠DBA = 180°
⇒ 2∠DBA = 180° - 70° = 110°
∴ ∠DBA = 110°/2 = 55°
∵ BD is the bisector of ∠ABC,
∴ ∠DBA = ∠DBC, = 55°
But in ∆ DBC,
DB = DC
∴ ∠DCB = ∠DBC
⇒ X = 55°