# Calculate tg (a/2) if sin a = 3/5 and a is in the interval (pi/2, pi)?

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sin a= 3/5

We know that sina = opposite/hypotenuse = 3/5

Then, the adjacent^2= 5^2 -3^2 = 25-9=16

Then the adjacent= 4

Then cos a= adj./hypotenuse= 4/5.

Now, we know that:

tg(a/2)= sina / 1-cosa

= 3/5 / 1-4/5 = 3/5 / 1/5 = 3

Also tg(a/2) = (1-cosa)/sina= 1/5 / 3/5 = 1/3

We could use the formula: sin a = 2tg (a/2)/[1+(tg (a/2))^2]

We'll note tg(a/2) = t and we'll substitute sin a by it's value:

3/5 = 2t/(1+t^2)

We'll cross multiplying and we'll get:

10t = 3 + 3t^2

We'll move all terms to on side:

3t^2 - 10t + 3 = 0

We'll apply the quadratic formula:

t1 = [10+sqrt(100-36)]/6

t1 = (10+8)/6

t1 = 3

t2 = 2/6

t2 = 1/3

So, tg(a/2) = 3 and tg(a/2)=1/3