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  • 14-01-2022
  • Mathematics
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Special Right Triangle

Special Right Triangle class=

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Аноним Аноним
  • 14-01-2022

let ABC be a triangle

where,

  • AB= 6cm
  • BC= a
  • AC = b

now

applying tan 30°

tan theta = perpendicular/base

=>tan 30° = 6/a

(since tan 30° = 1/√3)

=>1/√3 = 6/a

=> a = 6√3

applying cos Q for finding b

=>cos 60° = b/h

(since cos 60° = 1/2)

=>1/2= 6/b

=> b= 12

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