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EleoNora [17]
3 years ago
12

In ∆ABC the angle bisectors drawn from vertices A and B intersect at point D. Find m∠ADB if: m∠C=130°

Mathematics
2 answers:
Scilla [17]3 years ago
7 0

Answer:

∠ADB=155°.

Step-by-step explanation:

In Δ ABC, let A = x°

By angle-sum property.

A+B+C=180°

But, it is given that C=130°

So, x+B+130=180

B=180-130-x

B=50-x

Since AD and BD are internal bisectors of A and B,

∠DAB=x/2 and

∠DBA=

In ΔADB, by angle-sum property,

∠DBA+∠DAB+∠ADB=180°

+∠ADB=180°

25+∠ADB=180°

∠ADB=180-25=155°

Hence, ∠ADB=155°.

Zepler [3.9K]3 years ago
3 0

Answer:

  m∠ADB = 155°

Step-by-step explanation:

  m∠ADB = (m∠C)/2 + 90° = 130°/2 + 90°

  m∠ADB = 155°

_____

Reference brainly.com/question/17443119 for the derivation.

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   11g2h2 + h2 + 13

 ———————————

                h2      

Step-by-step explanation:

Step  1  :

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Simplify   ——

           h2

Equation at the end of step  1  :

               9              4

 ((((4•(g2))+————)+(7•(g2)))+——)+1

             (h2)            h2

Step  2  :

Equation at the end of step  2  :

               9         4

 ((((4•(g2))+————)+7g2)+——)+1

             (h2)       h2

Step  3  :

            9

Simplify   ——

           h2

Equation at the end of step  3  :

              9        4

 ((((4•(g2))+——)+7g2)+——)+1

             h2       h2

Step  4  :

Equation at the end of step  4  :

             9              4    

 (((22g2 +  ——) +  7g2) +  ——) +  1

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Step  5  :

Rewriting the whole as an Equivalent Fraction :

5.1   Adding a fraction to a whole

Rewrite the whole as a fraction using  h2  as the denominator :

            22g2     22g2 • h2

    22g2 =  ————  =  —————————

             1          h2    

Equivalent fraction : The fraction thus generated looks different but has the same value as the whole

Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator

Adding fractions that have a common denominator :

5.2       Adding up the two equivalent fractions

Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

22g2 • h2 + 9     4g2h2 + 9

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Rewriting the whole as an Equivalent Fraction :

6.1   Adding a whole to a fraction

Rewrite the whole as a fraction using  h2  as the denominator :

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   7g2 =  ———  =  ————————

           1         h2  

Adding fractions that have a common denominator :

6.2       Adding up the two equivalent fractions

(4g2h2+9) + 7g2 • h2      11g2h2 + 9

————————————————————  =  ——————————

         h2                  h2    

Equation at the end of step  6  :

  (11g2h2 + 9)     4    

 (———————————— +  ——) +  1

       h2         h2    

Step  7  :

Adding fractions which have a common denominator :

7.1       Adding fractions which have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

(11g2h2+9) + 4     11g2h2 + 13

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      h2               h2    

Equation at the end of step  7  :

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      h2          

Step  8  :

Rewriting the whole as an Equivalent Fraction :

8.1   Adding a whole to a fraction

Rewrite the whole as a fraction using  h2  as the denominator :

        1     1 • h2

   1 =  —  =  ——————

        1       h2  

Adding fractions that have a common denominator :

8.2       Adding up the two equivalent fractions

(11g2h2+13) + h2     11g2h2 + h2 + 13

————————————————  =  ————————————————

       h2                   h2      

Trying to factor a multi variable polynomial :

8.3    Factoring    11g2h2 + h2 + 13

Try to factor this multi-variable trinomial using trial and error

Factorization fails

Final result :

 11g2h2 + h2 + 13

 ————————————————

        h2      

Processing ends successfully

plz mark me as brainliest :)

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