* In Triangle ABC , CD bisects AB and the bisector of angle B and angle E meet at E such that D-E-C. If angle A= 30, angle B=30. Find (i) angle ECB (ii)angle BED
Answers
Answer:
angle BDC=angle ADC (CD is bisector)
angle BDC+angle ADC=180
2x=180
x=180/2
x=90°
therefore angle BDC=angle ADC=90°
angle DCB=angle DCA= 1/2×BCA (C is bisected) (Eq. 1)
angle CBE= angle DBE= 1/2×DBC =15°(B is bisected,as 30° is bisected,so made it half) (Eq. 2)
Also
In ∆DBE
angle BDE=90°
angle DBE=15°
angle BED=?
we know
angle BDE+angle DBE+angle BED=180°(sum of all interior angles of a triangle)
90+15+angle BED=180
105+angle BED=180
angle BED=180-105
angle BED=75°
but we've to find angle ECB
we know from (Eq. 1)
angle DCB+angle DCA=angle BCA
angle DCB=angle DCA=1/2×angle BCA
angle ECB=angle DCB (the angle won't change) (Eq. 3)
also
in ∆BDC
angle DBC=30°
angle BDC=90°
angle BCD=?
angle DBC+angle BDC+angle DCB=180°
90+30+angle DCB=180
120+angle DCB=180
angle DCB=180-120
angle DCB=60°
and from (Eq. 3)
angle DCB=angle ECB
angle ECB=60°
hope it helps
Given : Triangle ABC , CD bisects AB and the
bisector of angle B and angle E meet at E such that D-E-C.
If angle A= 30, angle B=30.
To Find : (i) angle ECB (ii )angle BED
Solution:
∠A = ∠B = 30°
=> BC = AC as Sides opposites to equal angles are equal in a triangle
, CD bisects AB
=> BD = AD = AB/2
in ΔBCD and ΔACD
BD = AD
∠B = ∠A
BC = AC
=> ΔBCD ≅ ΔACD Using SAS
∠BDC = ∠ADC
∠BDC + ∠ADC = 180°
=> ∠BDC = ∠ADC = 90°
D-E-C straight line
=> ∠BDE = 90°
BE is bisector of angle B
=> ∠DBE = ∠CBE = ∠B/2 = 30°/2 = 15°
in ΔBDE , sum of angles of a triangle is 180°
=> ∠BDE + ∠DBE + ∠BED = 180°
=> 90° + 15° + ∠BED = 180°
=> ∠BED = 75°
∠BED = ∠CBE + ∠ECB
Exterior angle of triangle = Sum of opposite two interior angles
=> 75° = 15° + ∠ECB
=> ∠ECB = 60°
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