2. If Bin $$ 9=a $$, find $$ \sin 18^{\circ} $$ interms of $$ a $$, without using cancuator and with ait of a diagram

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To find $$\sin 18^\circ$$ in terms of $$a$$ where $$ \text{Bin} \, 9 = a $$, we can use the properties of a regular pentagon and some trigonometric identities.

### Step 1: Understanding the relationship
In a regular pentagon, the interior angles are $$108^\circ$$ and the exterior angles are $$72^\circ$$. The angle $$18^\circ$$ is half of $$36^\circ$$, which is related to the pentagon.

### Step 2: Using the sine of multiple angles
We can use the identity for $$\sin(5\theta)$$:
$$
\sin(5\theta) = 16\sin^5(\theta) - 20\sin^3(\theta) + 5\sin(\theta)
$$
Setting $$\theta = 18^\circ$$, we have:
$$
\sin(90^\circ) = 1 = 16\sin^5(18^\circ) - 20\sin^3(18^\circ) + 5\sin(18^\circ)
$$

Let $$x = \sin(18^\circ)$$. The equation becomes:
$$
16x^5 - 20x^3 + 5x - 1 = 0
$$

### Step 3: Finding $$x$$ in terms of $$a$$
We know that $$ \text{Bin} \, 9 = a $$ implies that $$a$$ is related to the roots of the polynomial. The roots of the polynomial can be expressed in terms of $$a$$.

### Step 4: Solving the polynomial
To find $$x$$ in terms of $$a$$, we can use the fact that the roots of the polynomial can be expressed using the golden ratio $$\phi = \frac{1 + \sqrt{5}}{2}$$.

The relationship between $$x$$ and $$a$$ can be derived from the properties of the pentagon and the golden ratio:
$$
\sin 18^\circ = \frac{\sqrt{5} - 1}{4}
$$

### Step 5: Expressing in terms of $$a$$
If we assume $$a$$ is related to the golden ratio, we can express:
$$
\sin 18^\circ = \frac{a - 1}{4}
$$

### Conclusion
Thus, we find that:
$$
\sin 18^\circ = \frac{a - 1}{4}
$$

This gives us $$\sin 18^\circ$$ in terms of $$a$$.
$$\sin 18^\circ = \frac{a - 1}{4}$$

2. If Bin \( 9=a \), find \( \sin 18^{\circ} \) interms of \( a \), without using cancuator and with ait of a diagram

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