Question

a) In the following figure, AC=CD, AD = BD and C=38° FIND CAB​

Answer 1

Answer:

where is the figure ?????

Answer 2

Solution :-

In ∆ACD we have,

→ ∠ACD = 38° (given)

→ AC = CD (given)

so,

→ ∠CAD = ∠CDA (Angle opposite to equal sides are equal in measure .

then,

→ ∠ACD + ∠CAD + ∠CDA = 180° (Angle sum property.)

→ 38° + 2*∠CAD = 180°

→ 2*∠CAD = 180° - 38°

→ 2*∠CAD = 142°

→ ∠CAD = 71° --------- Eqn.(1)

now, in ∆ADB ,

→ ∠ADB = ∠CAD + ∠ACD (Exterior angle is equal to sum of two interior opposite angles.)

→ ∠ADB = 71° + 38°

→ ∠ADB = 109°

and,

→ AD = DB (given)

so,

→ ∠DAB = ∠DBA (Angle opposite to equal sides are equal in measure .

then,

→ ∠ADB + ∠DAB + ∠DBA = 180° (Angle sum property.)

→ 109° + 2*∠DAB = 180°

→ 2*∠DAB = 180° - 109°

→ 2*∠DAB = 71°

→ ∠DAB = 35.5° --------- Eqn.(2)

therefore,

→ ∠CAB = ∠CAD + ∠DAB

putting values from Eqn.(1) and Eqn.(2),

→ ∠CAB = 71° + 35.5°

→ ∠CAB = 106.5° (Ans.)

Learn more :-

In ABC, AD is angle bisector,

angle BAC = 111 and AB+BD=AC find the value of angle ACB=?

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