Question

TgA=3/5 sinA-?срочно помогитеспасибо заранее ответ должен быть 3/5даю 25 баллов

Answer 1

1+сtg^2x=1/sin^2x

ctgx=1/tgx=5/3

1+25/9=1/sin^2x

34/9=1/sin^2x

34sin^2x=9

sin^2x=9/34

sinx=3/√34

Answer 2

To find the value of sinA, we need more information about the triangle. However, we can use the identity sin^2 A + cos^2 A = 1 to find the value of cosA and then use the given value of TgA to solve for sinA.

We know that TgA = sinA/cosA. Thus, cosA = sinA/TgA = (3/5)/TgA. Now, using the identity sin^2 A + cos^2 A = 1, we have:

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sin^2 A + [(3/5)/TgA]^2 = 1

Simplifying this equation, we get:

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sin^2 A = 1 - [(3/5)/TgA]^2

Taking the square root of both sides, we get:

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sin A = sqrt(1 - [(3/5)/TgA]^2)

Substituting the given value of TgA = 3/5, we get:

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sin A = sqrt(1 - [(3/5)/(3/5)]^2) = sqrt(1 - 1/1) = 0

Therefore, sinA = 0, and the answer to the question is -. It is not possible for sinA to be equal to 3/5 given the value of TgA = 3/5.