Solve the following trigonometry problems: 1. In triangle ABC, ABC = 80, CAB = 40, AC = 50. Find BC. 2. In triangle PQR, RPQ = 60, QRP = 50, RP = 8. Find PQ.

Here are the solutions to your trigonometry problems:

1. In the triangle ABC, ABC = 80, CAB = 40 and AC = 50. Find BC.

Step 1: Find the third angle of the triangle.
The sum of angles in a triangle is $180^\circ$.

$$\angle ACB = 180^\circ - \angle ABC - \angle CAB$$ $$\angle ACB = 180^\circ - 80^\circ - 40^\circ$$ $$\angle ACB = 60^\circ$$

Step 2: Apply the Sine Rule.
The Sine Rule states $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$. We want to find BC (side $a$).

$$\frac{BC}{\sin(\angle CAB)} = \frac{AC}{\sin(\angle ABC)}$$ $$\frac{BC}{\sin(40^\circ)} = \frac{50}{\sin(80^\circ)}$$

Step 3: Solve for BC.

$$BC = \frac{50 \times \sin(40^\circ)}{\sin(80^\circ)}$$ $$BC = \frac{50 \times 0.6428}{0.9848}$$ $$BC = \frac{32.14}{0.9848}$$ $$BC \approx 32.636$$

The length of BC is approximately $\boxed{\text{32.64 units}}$.

2. In the triangle PQR, RPQ = 60, QRP = 50 and RP = 8. Find PQ.

Step 1: Find the third angle of the triangle.

$$\angle PQR = 180^\circ - \angle RPQ - \angle QRP$$ $$\angle PQR = 180^\circ - 60^\circ - 50^\circ$$ $$\angle PQR = 70^\circ$$

Step 2: Apply the Sine Rule.
We want to find PQ (side $r$).

$$\frac{PQ}{\sin(\angle QRP)} = \frac{RP}{\sin(\angle PQR)}$$ $$\frac{PQ}{\sin(50^\circ)} = \frac{8}{\sin(70^\circ)}$$

Step 3: Solve for PQ.

$$PQ = \frac{8 \times \sin(50^\circ)}{\sin(70^\circ)}$$ $$PQ = \frac{8 \times 0.7660}{0.9397}$$ $$PQ = \frac{6.128}{0.9397}$$ $$PQ \approx 6.521$$

The length of PQ is approximately $\boxed{\text{6.52 units}}$.

3. Find the height of a flagpole which casts a shadow of 6.35m when the sun makes an angle of 65 to the horizontal.

Step 1: Identify the trigonometric relationship.
Let $h$ be the height of the flagpole and $s$ be the length of the shadow. The angle of elevation is $\theta = 65^\circ$. This forms a right-angled triangle where $h$ is the opposite side and $s$ is the adjacent side.

$$\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{s}$$

Step 2: Substitute the given values.

$$\tan(65^\circ) = \frac{h}{6.35 \text{ m}}$$

Step 3: Solve for $h$.

$$h = 6.35 \text{ m} \times \tan(65^\circ)$$ $$h = 6.35 \text{ m} \times 2.1445$$ $$h \approx 13.618 \text{ m}$$

The height of the flagpole is approximately $\boxed{\text{13.62 m}}$.

4. A rectangle is 9.2 m by 3.8 m. What angle does its diagonal make with its longer side?

Step 1: Identify the sides and the angle.
The longer side (adjacent) is $L = 9.2$ m.
The shorter side (opposite) is $W = 3.8$ m.
Let $\theta$ be the angle the diagonal makes with the longer side.

$$\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{W}{L}$$

Step 2: Substitute the values.

$$\tan(\theta) = \frac{3.8 \text{ m}}{9.2 \text{ m}}$$ $$\tan(\theta) \approx 0.41304$$

Step 3: Solve for $\theta$.

$$\theta = \arctan(0.41304)$$ $$\theta \approx 22.43^\circ$$

The angle is approximately \boxed{\text{22.43^\circ}}.

5. The diagonal and the longer side of a rectangle make an angle of 43.2. If the longer side is 12.6 cm, find the length of the shorter side.

Step 1: Identify the sides and the angle.
The longer side (adjacent) is $L = 12.6$ cm.
Let $W$ be the shorter side (opposite).
The angle $\theta = 43.2^\circ$.

$$\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{W}{L}$$

Step 2: Substitute the given values.

$$\tan(43.2^\circ) = \frac{W}{12.6 \text{ cm}}$$

Step 3: Solve for $W$.

$$W = 12.6 \text{ cm} \times \tan(43.2^\circ)$$ $$W = 12.6 \text{ cm} \times 0.9391$$ $$W \approx 11.833 \text{ cm}$$

The length of the shorter side is approximately $\boxed{\text{11.83 cm}}$.

6. A rhombus has diagonals of length 12 cm and 7 cm respectively. Find the larger angle of the rhombus.

Step 1: Understand the properties of a rhombus.
The diagonals of a rhombus bisect each other at right angles. This forms four congruent right-angled triangles.
The half-diagonals are the legs of these right-angled triangles.
Half of diagonal 1: $d_1/2 = 12 \text{ cm} / 2 = 6$ cm.
Half of diagonal 2: $d_2/2 = 7 \text{ cm} / 2 = 3.5$ cm.

Step 2: Find the angles within one of the right-angled triangles.
Let $\alpha$ be one of the angles at a vertex of the rhombus, bisected by a diagonal.
Consider the right-angled triangle formed by the half-diagonals.
Let $\theta$ be the angle opposite the 3.5 cm side and adjacent to the 6 cm side.

$$\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{3.5}{6}$$ $$$$