Question

the side B C of AA BC is produced to point D. the bisector of cA meets sides Beinl. If LAB (230- and LACD 215. Eind 2Alcz​

Answer 1

Answer:

opposite interior es^ prime prime we

get,

loge Delta*ABC

angle ACD= angle CAB+ angle CBA

115^ = angle CAB+30^

angle CAB=115^ -30^

angle CAB=85^

Now, as AL is the bisector of angle A

<CAL = 1/2 CAB

ZCAL = 1/2 (85°)

angle CAL=44 1^ 2

Also, angle ACD is the exterior angle of triangle ALC

Thus,

Again, using the property, "exterior angle of a triangle is equal to the sum of the two opposite interior angles", we get,

In*Delta*A*L_{C}

angle ACD= angle CAL+ angle ALC

115=44 1 2 + angle ALC

=115-44 1^ 2 angle ALC=1

angle ALC=72 1^ 2

Thus, angle ALC=72 1^ 2

Answer 2

Answer:

Answer:

opposite interior es^ prime prime we

get,

loge Delta*ABC

angle ACD= angle CAB+ angle CBA

115^ = angle CAB+30^

angle CAB=115^ -30^

angle CAB=85^

Now, as AL is the bisector of angle A

<CAL = 1/2 CAB

ZCAL = 1/2 (85°)

angle CAL=44 1^ 2

Also, angle ACD is the exterior angle of triangle ALC

Thus,

Again, using the property, "exterior angle of a triangle is equal to the sum of the two opposite interior angles", we get,

In*Delta*A*L_{C}

angle ACD= angle CAL+ angle ALC

115=44 1 2 + angle ALC

=115-44 1^ 2 angle ALC=1

angle ALC=72 1^ 2

Thus, angle ALC=72 1^ 2