# Calculate tg a if a belongs to interval (pi, 3pi/2) and sin a = -4/5?

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sin a= -4/5

No let us assume that a is an angle in a right triangle.

sin a = opposite / hypotenuse = -4/5

Then the adjacent= sqrt[(5)^2 - (4)^3] = 3

Then cosa= adjacent/hypotenuse=-3/5 (because it is in the third quadrant)

Then tg a= sina/cosa = (-4/5) / (-3/5) = 4/3

sina = -4/5. a belongs to (pi, 3pi/2).

In (pi, 3pi/2) both sine and cosine ratios are -ve and tangent ratio is positive,

cos a = sqrt(1-sin^2a) = -sqrt{1- (-4/5)^2} = -sqrt (1-16/25) = -sqrt(9/25) = -3/5'

Therefore tga = sina/cosa =-4/5)/(-3/5) = 4/3

If a belongs to the third qudrant, that means that the values of the tangent function are positive.

The tangent function is a ratio:

tg a = sin a / cos a

We have the value of sin a but we don't know the value of cos a.

We can calculate the value of cos a, using the fundamental formula of trigonometry:

(sina)^2 + (cosa)^2 = 1

Because a is in the third quadrant, cos a<0.

cos a = -sqrt[1 - (sina)^2]

cos a = -sqrt(1 - 16/25)

cos a = -sqrt [(25-16)/25]

cos a = -sqrt (9/25)

cos a = - (3/5)

tg a = sin a / cos a

tg a = (-4/5) / (-3/5)

**tg a = 4/3**