Given:17 sin ɓ -0 =0 where 90°<ɓ<360° ..without the use of a calculator and by means of a diagram, determine the value of 15/8 tan²ɓ?

First, let's solve for ɓ in the given equation:

17 sin ɓ = 0
sin ɓ = 0

Since sin ɓ = 0, this means that ɓ is either 180° or 360°.

Next, let's find the value of tan²ɓ:

tan ɓ = sin ɓ / cos ɓ
tan ɓ = 0 / cos ɓ
tan ɓ = 0

Therefore, tan ɓ is 0, and tan²ɓ is 0.

Finally, let's find the value of 15/8 tan²ɓ:

15/8 * 0 = 0

Therefore, the value of 15/8 tan²ɓ is 0.

Diagram:

[Image of a unit circle with an angle ɓ measured from the positive x-axis]

Solution:

1. Since ɓ lies in the second quadrant (90°<ɓ<360°), sin ɓ is positive.

2. From the equation 17 sin ɓ -0 =0, we have sin ɓ = 0. This means that ɓ is either 90° or 270°.

3. Since ɓ is greater than 90°, it must be 270°.

4. Therefore, cos ɓ = 0 and tan ɓ = ±∞.

5. Substituting tan ɓ = ∞ into 15/8 tan²ɓ, we get:
```
15/8 tan²ɓ = 15/8(∞)² = 15/8(∞) = ∞
```

Therefore, the value of 15/8 tan²ɓ is ∞.