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Mekhanik [1.2K]
3 years ago
11

Special Right Triangle

Mathematics
1 answer:
MA_775_DIABLO [31]3 years ago
3 0
<h3>let ABC be a triangle</h3>

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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