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denis23 [38]
3 years ago
15

Help mi plzz.........​

Mathematics
1 answer:
faust18 [17]3 years ago
3 0
Tan3A=tan(2A+A)

We know that , tan(x+y)=(tanx + tany)/(1 - (tanx)(tany))

tan(2A+A)=(tan2A+tanA)/(1 - (tan2A)(tanA))— (1)

We know that , tan2x=2tanx/(1 - tan^2x)

So, by substituting tan2A in (1),we get,

=[2tanA/(1 - tan^2A) + tanA]/1- (2tanA/(1 - tan^2A))(tanA)]

=[2tanA+tanA - tan^3A]/[1 - tan^2A - 2tan^2A]

=[3tanA - tan^3A]/[1-3tan^2A]

Therefore, tan3A= [3tanA - tan^3A]/[1-3tan^2A]

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Can you help me find length bc please!!!!
masya89 [10]

Remark

You are dealing with a right angle triangle. You can use one of Sin(x) Cos(x) or Tan(x). Since the Hypotenuse is involved and since you are given the adjacent side, Cos(71) is what you will need.

Solution

H = BC

Cos(x) = adjactent / hypotenuse

Cos(71) = 6.3 / H            Multiply both sides by H

H * Cos(71) = 6.3            Divide by Cos(71)

H = 6.3 / cos(71)             Cos(71) = 0.32557

H = 6.3 / 0.32557          Divide

H = BC = 19.351                       Answer

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