Question

solve the following question​

Answer 1

∠A=40°

Step-by-step explanation:

Given In ∆ABC ,

AB=AC and <B=70°

/* In ∆ABC , If AB=AC then <C=<B */

Given \: \angle C=\angle B=70\degreeGiven∠C=∠B=70°

\angle A+\angle B+\angle C=180\degree∠A+∠B+∠C=180°

\implies \angle A+70 \degree+70 \degree=180 \degree⟹∠A+70°+70°=180°

\implies \angle A+140\degree=180 \degree⟹∠A+140°=180°

\implies\angle A=180\degree-140 \degree⟹∠A=180°−140°

\implies \angle A=40 \degree⟹∠A=40°

Therefore,

\angle A=40 \degree∠A=40°

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Answer 2

Answer:

150°

Step-by-step explanation:

here ABC is an isosceles triangle

so angle B = C = 70°

so Angle BAC = 40 ( sum of angle of ∆ is 180)

now given , angle BAD = 80°

= angle BAC + angle ACD = 80°

= 40° + <ACD = 80°

therefore

angle ACD = 40°

now

angle ADE = <DCA + <CAD { external angle = sum of opposite angles }

= ADE = 110° + 40° ...

therefore < ADE = 150°