Question

IN triangle abc ab+bc =2ac angle A = C+90 find the value of 100 cos b​

Answer 1

Answer:

ans is 45

Step-by-step explanation:

Answer 2

Concept

Properties of Triangle

Given

In a triangle abc , ab+bc =2ac and angle A = C+90 '

Find

The value of 100 cos b​

Solution

                      a+c = 2b. .. (1)

 We know A+B+C=180 (angle sum property of triangle)

        90+C+ B+C  = 180

                   B+2C = 90

       B = 90–2C, A = 90+C

 Some relations between sides and angles,

•           a/ cosC = b / cos 2C = c / sin C

•         b/a = 2 cos^2C —1 / cos C, c/a = tan C

•         b/a = 2 cos C — sec C, a = c cot C

           Using in 1 i.e. a+c = 2b

                      c cot C + c = cCot C ( 2 cos C — secC)

                         1 + cot C = 2 cos^2 C / sin C — cosec C

      sin C + cos C / sin C = 1/sin C ( cos 2C).

                  sin C + cos C = cos 2C

                  sinC+ cosC=2coscsinC

      sinC+CosC-2sinCcosC=0    

           let sinC=t; then

         sqrt (1-t^2) = 1–2t^2 -t

              1 - 5t^2 = (1-t)^2 —4 (1-t)t^2

               1–5t^2 = 1+t^2 -2t -4t^2 +4t^3

  4t^3 + 2t^2 -2t = 0

         2t^2 +t —1 = 0

                         t = —1 +— sqrt (1 +8) /4

                          t= —1+—3/4

                          t= —1, 1/2

             sinC=-1 or C=-180 or sinC =1/2 or C=30°

C = 30°, A = 120°, B = 30°

cosB=cos30°= √3/2 ,

Hence,

100 cos B = 50√3

The value of 100 cos b​ is equal to 50√3

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