If you are giving a compass and a ruler only, is not possible to construct 150°.

Here are the solutions to the questions:

54. If you are giving a compass and a ruler only, is not possible to construct $150^\circ$.
To construct $150^\circ$, one can first construct $180^\circ$ (a straight line) and then construct $30^\circ$ from one end. The remaining angle will be $150^\circ$. Since $30^\circ$ can be constructed by bisecting a $60^\circ$ angle (which is constructible), $150^\circ$ is constructible using only a compass and a ruler.
The statement claims it is not possible, which is incorrect.
The correct option is B. False.

55. In the figure below, $\angle BOC$ represents $30^\circ$.
The construction marks show:

1. An arc centered at O, passing through C.
2. An arc centered at C, with the same radius, intersecting the first arc at A. This implies $\triangle OAC$ is equilateral, so $\angle COA = 60^\circ$.
3. An arc centered at A, with the same radius, intersecting the first arc at B. This implies $\triangle OAB$ is equilateral, so $\angle AOB = 60^\circ$. Therefore, $\angle BOC = \angle COA + \angle AOB = 60^\circ + 60^\circ = 120^\circ$. The statement claims $\angle BOC$ represents $30^\circ$, which is incorrect. The correct option is B. False.

SECTION C

1. a) A($28^\circ$ N, $25^\circ$ W) and B($32^\circ$ S, $25^\circ$ W) are two places on the earth's surface. Calculate the distance of AB measured along the meridian. (Take R = 6400km, and $\pi = 3.142$).

Step 1: Determine the angular difference between points A and B.
Since both points are on the same meridian ($25^\circ$ W) but on opposite sides of the equator (North and South), the angular difference is the sum of their latitudes.

$$\theta = 28^\circ + 32^\circ = 60^\circ$$

Step 2: Calculate the distance along the meridian using the formula $L = \frac{\theta}{360^\circ} \times 2\pi R$.

$$L = \frac{60^\circ}{360^\circ} \times 2 \times 3.142 \times 6400 \text{ km}$$ $$L = \frac{1}{6} \times 2 \times 3.142 \times 6400 \text{ km}$$ $$L = \frac{1}{3} \times 3.142 \times 6400 \text{ km}$$ $$L = \frac{20108.8}{3} \text{ km}$$ $$L \approx 6702.93 \text{ km}$$

The distance of AB measured along the meridian is $\boxed{\text{6702.93 km}}$.

1. b) If $\sin \theta = \frac{5}{13}$, $0^\circ < \theta < 90^\circ$, find: i) $\cos \theta$. ii) $\tan \theta$.

Given $\sin \theta = \frac{5}{13}$ and $\theta$ is in the first quadrant. We can use the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$ or a right-angled triangle.
Using a right-angled triangle:
Opposite side = 5, Hypotenuse = 13.
Adjacent side = $\sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12$.