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Calculate: ∠ADC ∠ABC ∠BAC - Mathematics

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प्रश्न

Calculate:

  1. ∠ADC
  2. ∠ABC
  3. ∠BAC

योग

उत्तर

Given: ACE = 130°; AD = BD = CD

Proof:

(i) ∠ACD + ∠ACE = 180°

⇒ ∠ACD = 180° − 130°

⇒ ∠ACD = 50° 

Now, CD = AD

⇒ ∠ACD = ∠DAC = 50°  ..... (i)[ Since angles opposite to equal

sides are equal]

In ΔADC,

∠ACD = ∠DAC = 50°  

∠ACD + ∠DAC + ∠ADC = 180° 

 50°  + 50°  + ∠ADC = 180° 

∠ADC =180° − 100°

∠ADC = 80°

(ii) ∠ADC = ∠ABD + ∠DAB ....[Exterior angle is equal to the sum of opp. interior angle] 

But AD = BD

∴ ∠DAB = ∠ABD

⇒ 80° = ∠ABD + ∠ABD

⇒ 2∠BD = 80°

⇒ ∠ABD = 40° = ∠DAB .....(ii)

(iii) ∠BAC = ∠DAB + ∠DAC

substituting the value from (i) and (ii)

∠BAC = 40° + 50°

⇒ ∠BAC = 90°

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 10: Isosceles Triangles - Exercise 10 (A) [पृष्ठ १३१]

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सेलिना Concise Mathematics [English] Class 9 ICSE
अध्याय 10 Isosceles Triangles
Exercise 10 (A) | Q 2 | पृष्ठ १३१
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