To calculate the size of angle $\angle BCD$, we can use the cosine rule, as we are given three sides of triangle $BCD$:

• $BC = 7.4$ cm
• $BD = 6.5$ cm
• $CD = 4.7$ cm

The cosine rule is given by: $\cos(C) = \frac{b^2 + c^2 - a^2}{2bc}$ where:

• $C$ is the angle we are trying to find ($\angle BCD$),
• $a$ is the side opposite the angle ($BD$),
• $b$ and $c$ are the other two sides ($BC$ and $CD$).

Substitute the known values into the formula: $\cos(C) = \frac{6.5^2 + 4.7^2 - 7.4^2}{2 \times 6.5 \times 4.7}$

First, calculate each term:

• $6.5^2 = 42.25$
• $4.7^2 = 22.09$
• $7.4^2 = 54.76$

Now substitute: $\cos(C) = \frac{42.25 + 22.09 - 54.76}{2 \times 6.5 \times 4.7}$ $\cos(C) = \frac{9.58}{61.1}$ $\cos(C) \approx 0.1568$

To find the angle $C$: $C = \cos^{-1}(0.1568) \approx 80.98^\circ$

Rounding to 1 decimal place, the size of angle $BCD$ is approximately: $\boxed{81.0^\circ}$

Let me know if you would like further details or clarifications!

Here are 5 related questions:

1. How would you calculate the angle $\angle BDA$ using the cosine rule?
2. Can the sine rule be applied to this triangle? If so, how would you use it?
3. How would the answer change if side $BC$ were shorter?
4. What are the other angles in the quadrilateral $ABCD$?
5. How can you use the Pythagorean theorem to verify the distances?

Tip: Always double-check your units and rounding when working with angles and lengths in trigonometry.