Question

can you please can you please tell me the answer of this question this is emergency I have to submit my assignment ​

Answer 1

Answer:

Step-by-step explanation:

In triangle ACD,

Since AC = CD,

triangle ACD is an isosceles triangle.

∠CAD = ∠CDA

In triangle ACD,

∠CAD + ∠CDA + ∠ACD = 180°

2∠CAD + ∠ACD = 180°

∠ACD = 180° - 2∠CAD

In triangle ABC,

∠CAB + ∠ABC + ∠BCA = 180°

(∠CAD + ∠BAD) + ∠ABC + (180° - 2∠CAD) = 180°

∠CAD + ∠BAD + ∠ABC - 2∠CAD = 180° - 180°

∠BAD + ∠ABC = 2∠CAD - ∠CAD

∠BAD + ∠ABC = ∠CAD          -(i)

∠CAB - ∠ABC = 30°

(∠CAD+∠BAD) - ∠ABC = 30°

∠BAD - ∠ABC = 30° - ∠CAD          -(ii)

Adding equations (i) and (ii), we get-

∠BAD + ∠ABC + ∠BAD - ∠ABC = ∠CAD + 30° - ∠CAD

2∠BAD = 30°

∠BAD = 15°

Hope this helps . . . please comment if further doubt.