Math, asked by vratikas3001, 4 months ago


ABC and DBC are both isosceles triangles on a common base BC such that A and Ole on the
came side of BC. Are triangles ADB and ADC congruent? Which condition do you use?
BAC = 40° and Z BDC = 100°, then find angle ADB

CAN ANYONE EXPLAIN THIS QUESTION IN DETAIL​

Answers

Answered by dhruvsharmaind1
0

Answer:

ADB = 130°

Step-by-step explanation:

We are given that, ∆ ABC and ∆ DBC are isosceles triangles and at least two sides of isosceles triangles are equal in length,

∴ BD = DC ….. (i)

And,

AB = AC ….. (ii)

Step 1:

Considering ∆ ADB & ∆ ADC, we have

BD = DC ….. [from (i)]

AB = AC ….. [from (ii)]

AD = AD ….. [common]

∴ By SSS congruence criterion, ∆ ADB ≅ ∆ ADC

∴ ∠BAD = ∠CAD …. [C.P.C.T] ….. (iii)

Also,  

∠BAC =∠BAD + ∠CAD

⇒ 40° = 2 * ∠BAD ….. [∵ ∠BAC = 40° (given) and ∠BAD = ∠CAD from (iii)]

⇒ ∠BAD = 40° / 2 = 20° ….. (iv)

Step 2:

Since from (ii), we have AB = AC

∴ ∠ABC = ∠ACB …. [∵ angles opposite to equal sides are equal] …. (v)

In ∆ABC, applying angle sum property,

∠A + ∠ABC +∠ACB = 180°

⇒ 40° + 2* ∠ABC = 180° …. [from (v) and given ∠A = 40° ]

⇒ ∠ABC = 140°/2 = 70°  

Also, in ∆ DBC, applying angle sum property,

∠BDC + ∠DBC + ∠DCB = 180°

⇒ 100° + 2* ∠DBC = 180° …. [∵ angles opposite to equal sides BD = CD are equal i.e., ∠DBC = ∠DCB]

⇒ ∠DBC = 80°/2 = 40°  

Now,

∠ABD = ∠ABC – ∠DBC

⇒ ∠ABD = 70° – 40° = 30° ….. (vi)

Step 3:

In ∆ ABD, applying angle sum property,

∠BAD + ∠ADB + ∠ABD = 180°

⇒ 20° + ∠ADB + 30° = 180° …. [substituting values from (iv) & (vi)]

⇒ ∠ADB = 180° - 50° = 130°  

HOPE IT HELPS, PLS MARK AS BRAINLIEST!

Answered by bhoomikabhatia
0

Answer:

I hope this will help you

Step-by-step explanation:

please mark it as a brainlist

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