Solve the following trigonometry problems: 1. In triangle ABC, angle ABC = 80, angle CAB = 45, AC = 50. Find BC. 2. In triangle PQR, angle RPQ = 60, angle QRP = 50, RP = 6.5. Find PQ.

The sum of angles in a triangle is $180^\circ$.

The Sine Rule states $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$.
We want to find side $BC$ (let's call it $a$), which is opposite $\angle CAB$. We are given side $AC$ (let's call it $b$), which is opposite $\angle ABC$.

: In the triangle PQR, $\angle RPQ = 60^\circ$, $\angle QRP = 50^\circ$ and $RP = 6.5$. Find $PQ$.

We want to find side $PQ$ (opposite $\angle QRP$). We are given side $RP$ (opposite $\angle PQR$).

: Find the height of a flagpole which casts a shadow of $6.35 \text{ m}$ when the sun makes an angle of $65^\circ$ to the horizontal.

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 to the angle $\theta$.

Given $s = 6.35 \text{ m}$ and $\theta = 65^\circ$.

: A rectangle is $9.2 \text{ m}$ by $3.8 \text{ m}$. What angle does its diagonal make with its longer side?

The longer side is $9.2 \text{ m}$ and the shorter side is $3.8 \text{ m}$. The diagonal forms a right-angled triangle with these two sides. Let $\theta$ be the angle the diagonal makes with the longer side. In this triangle, the shorter side is opposite to $\theta$, and the longer side is adjacent to $\theta$.

: The diagonal and the longer side of a rectangle make an angle of $43.2^\circ$. If the longer side is $12.6 \text{ cm}$, find the length of the shorter side.

Let the angle be $\theta = 43.2^\circ$. The longer side is adjacent to this angle, and the shorter side is opposite to this angle in the right-angled triangle formed by the sides and the diagonal.

Given longer side $= 12.6 \text{ cm}$ and $\theta = 43.2^\circ$. Let the shorter side be $x$.

: A rhombus has diagonals of length $12 \text{ cm}$ and $7 \text{ cm}$ respectively. Find the larger angle of the rhombus.

The diagonals of a rhombus bisect each other at right angles. This creates four congruent right-angled triangles. The half-diagonals are the legs of these triangles.
Half of the longer diagonal $= \frac{12 \text{ cm}}{2} = 6 \text{ cm}$.
Half of the shorter diagonal $= \frac{7 \text{ cm}}{2} = 3.5 \text{ cm}$.

Let $\alpha$ be the angle opposite the $3.5 \text{ cm}$ side and adjacent to the $6 \text{ cm}$ side.

Let $\beta$ be the angle opposite the $6 \text{ cm}$ side and adjacent to the $3.5 \text{ cm}$ side.

Note that $\alpha + \beta = 30.25^\circ + 59.75^\circ = 90^\circ$, which is correct for a right-angled triangle.

The angles of the rhombus are formed by two of these angles from the right-angled triangles.
The smaller angle of the rhombus is $2\alpha$.

The larger angle of the rhombus is $2\beta$.

The sum of adjacent angles in a rhombus is $180^\circ$, so $60.5^\circ + 119.5^\circ = 180^\circ$.
The larger angle of the rhombus is $\boxed{119.5^\circ}$.

: Find the angle of elevation to the top of a $62 \text{ m}$ high building from point A, which is at ground level $155 \text{ m}$ from its base. What is the angle of depression from the top of the building to A?

Let $h$ be the height of the building ($62 \text{ m}$) and $d$ be the distance from the base ($155 \text{ m}$). Let $\theta$ be the angle of elevation.

The angle of depression from the top of the building to point A is equal to the angle of elevation from point A to the top of the building, due to alternate interior angles being equal when parallel lines (ground and horizontal line from top of building) are intersected by a transversal (line of sight).
Therefore, the angle of depression is also $\boxed{21.80^\circ}$.

: The angle of depression from the top of a $180 \text{ m}$ high vertical cliff to a boat B is $26^\circ$. Find how far the boat is from the base of the cliff.

Let $h$ be the height of the cliff ($180 \text{ m}$). Let $x$ be the distance of the boat from the base of the cliff. The angle of depression from the top of the cliff to the boat is $26^\circ$. This means the angle of elevation from the boat to the top of the cliff is also $26^\circ$.
In the right-angled triangle formed, $h$ is the opposite side and $x$ is the adjacent side to the angle of elevation.

: Sarah measures the angle of elevation to the top of a tree as $25^\circ$ from where she is standing at a point $150 \text{ m}$ from its base. Her eye level, where the angle measurement was taken, is $2.5 \text{ m}$ above the ground. Assuming the ground is horizontal, find the height of the tree.

Let $h'$ be the height of the tree above Sarah's eye level. The distance from the base of the tree to Sarah is $150 \text{ m}$. The angle of elevation is $25^\circ$.

The total height of the tree ($H$) is the height above eye level plus Sarah's eye level height.