A wooden frame is to be made to support some garden decking - AQA - A-Level Maths Mechanics - Question 2 - 2019 - Paper 3

To find $\sin \theta$, we will use the cosine rule in triangle OAB. According to the cosine rule, we have:

$c^2 = a^2 + b^2 - 2ab \cos(\theta)$

Here, let:

• $a = 5$ m (OB)
• $b = 6$ m (OA)
• $c = 3$ m (AB)

Substituting the values into the cosine rule:

$3^2 = 5^2 + 6^2 - 2 \cdot 5 \cdot 6 \cos(\theta)$

Calculating the squares:

$9 = 25 + 36 - 60 \cos(\theta)$

This simplifies to:

$9 = 61 - 60 \cos(\theta)$

Rearranging gives us:

$60 \cos(\theta) = 61 - 9$

$60 \cos(\theta) = 52$

Thus:

$\cos(\theta) = \frac{52}{60} = \frac{26}{30} = \frac{13}{15}$

Next, we can use the identity:

$\sin^2(\theta) + \cos^2(\theta) = 1$

Substituting for $\cos(\theta)$:

$\sin^2(\theta) + \left(\frac{13}{15}\right)^2 = 1$

Calculating $\left(\frac{13}{15}\right)^2$ gives:

$\sin^2(\theta) + \frac{169}{225} = 1$

Subtracting $\frac{169}{225}$ from both sides:

$\sin^2(\theta) = 1 - \frac{169}{225}$

Writing $1$ as $\frac{225}{225}$:

$\sin^2(\theta) = \frac{225 - 169}{225} = \frac{56}{225}$

Now, taking the square root:

$\sin(\theta) = \sqrt{\frac{56}{225}} = \frac{\sqrt{56}}{15} = \frac{\sqrt{4 \cdot 14}}{15} = \frac{2\sqrt{14}}{15}$

Thus, the exact value of $\sin(\theta)$ is:

$\sin(\theta) = \frac{\sqrt{14}}{15}.$