Question

Sin(a-b)=√10/10, ctg a=1/2, tgb?​

Answer 1

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Answer 2

Let's solve for the tangent (tg) of angle b, given that sin(a - b) = √10/10 and cotangent (ctg) of angle a is 1/2.

First, we need to find the values of sin(a) and cos(a) since we have sin(a - b) and ctg(a):

1. ctg(a) = 1/2 This means that tan(a) = 2.

2. We know that tan(a) = sin(a) / cos(a).

3. We can use the Pythagorean identity for sine and cosine: sin²(a) + cos²(a) = 1.

Now, let's find sin(a) and cos(a) using the information we have:

From ctg(a) = 1/2, we know that tan(a) = 2, which means sin(a)/cos(a) = 2.

Now, we square both sides of the equation sin(a)/cos(a) = 2:

(sin(a)/cos(a))² = 2² (sin(a)/cos(a))² = 4

Now, we use the Pythagorean identity:

sin²(a) + cos²(a) = 1

Let's express sin(a) in terms of cos(a) using the equation sin(a)/cos(a) = 2:

sin(a) = 2cos(a)

Now we can substitute sin(a) in the Pythagorean identity:

(2cos(a))² + cos²(a) = 1

4cos²(a) + cos²(a) = 1

Now, combine like terms:

5cos²(a) = 1

Now, isolate cos²(a):

cos²(a) = 1/5

Taking the square root of both sides:

cos(a) = ±√(1/5)

Now, let's determine the sign of cos(a). Since ctg(a) is positive (1/2), angle a must be in the first quadrant of the unit circle, where both sine and cosine are positive. Therefore, cos(a) is positive:

cos(a) = √(1/5)

Now that we have found sin(a) and cos(a), we can find sin(b) using the formula sin(a - b) = √10/10:

sin(a - b) = √(1/5) - sin(b) = √10/10

Now, solve for sin(b):

sin(b) = √(1/5) - √10/10

sin(b) = (√2/√10) - (√10/√10)

sin(b) = (√2 - √10)/√10

Now, we can find tangent (tg) of angle b:

tg(b) = sin(b)/cos(b)

tg(b) = [(√2 - √10)/√10] / (√(1/5))

tg(b) = [(√2 - √10)/√10] * (√5)

tg(b) = [(√2 - √10)*√5] / 10

So, tg(b) = [(√10√5 - √50)/10]