Triangle KLM is similar to triangle LRS where angle K corresponds to angle Q and where angles L and R are right angles. If sin K = 105/233 and sin M = 203/233, what is the value of tan S?

Since $\triangle KLM \sim \triangle QRS$, their corresponding angles are equal. This means $\angle K = \angle Q$, $\angle L = \angle R$, and $\angle M = \angle S$. We are given that $\angle L = 90^\circ$ and $\angle R = 90^\circ$. In a right-angled triangle, the sum of the two acute angles is $90^\circ$, so $\angle K + \angle M = 90^\circ$.

Let the sides opposite to angles K, L, M be $k, l, m$ respectively.
In $\triangle KLM$, $\angle L = 90^\circ$.
We are given $\sin M = \frac{208}{233}$. In a right triangle, $\sin M = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{m}{l}$.
So, we can assume $m = 208$ and $l = 233$ (or a multiple thereof).
We are also given $\sin K = \frac{105}{233}$. In a right triangle, $\sin K = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{k}{l}$.
This implies $k = 105$ and $l = 233$.
These values are consistent: $k=105$, $m=208$, $l=233$.

Check if $k^2 + m^2 = l^2$:
$105^2 + 208^2 = 11025 + 43264 = 54289$.
$233^2 = 54289$.
The side lengths are consistent with a right triangle.

The tangent of angle M is $\tan M = \frac{\text{opposite}}{\text{adjacent}} = \frac{m}{k}$.

Since $\angle M = \angle S$, we have $\tan S = \tan M$.

The value of $\tan S$ is $\boxed{\text{}\frac{208}{105}\text{}}$.

9. In the diagram, $\triangle ABC$ is similar to $\triangle EDC$, with $\angle BAC$ corresponding to $\angle CED$ and $\angle ABC$ corresponding to $\angle CPE$. Which of the following must be true?
The options are: A) $AE \parallel BD$, B) $AE \perp BD$, C) $AB \parallel DE$, D) $AB \perp DE$.

Given $\triangle ABC \sim \triangle EDC$. This means their corresponding angles are equal:

• $\angle BAC = \angle EDC$
• $\angle ABC = \angle DEC$
• $\angle ACB = \angle ECD$ (These are vertically opposite angles, which are always equal, confirming the similarity).

Consider lines AB and DE intersected by the transversal line BD.
The angles $\angle ABC$ and $\angle EDC$ are alternate interior angles.
From the similarity, we know that $\angle ABC = \angle EDC$.

If alternate interior angles formed by a transversal intersecting two lines are equal, then the two lines are parallel.
Since $\angle ABC = \angle EDC$, it implies that line AB is parallel to line DE.

The statement that must be true is $AB \parallel DE$.
The correct option is $\boxed{\text{C) } AB \parallel DE}$.

10. If arc FG is congruent to arc GH and the measure of $\angle G$ is $30^\circ$, what is the measure of $\angle H$?

If two arcs in a circle are congruent (arc FG $\cong$ arc GH), then the chords that subtend these arcs are equal in length. So, chord FG = chord GH.

Since chord FG = chord GH, $\triangle FGH$ is an isosceles triangle.

In an isosceles triangle, the angles opposite the equal sides are equal. The angle opposite side FG is $\angle H$, and the angle opposite side GH is $\angle F$. Therefore, $\angle F = \angle H$.

The sum of the interior angles of any triangle is $180^\circ$.

Substitute $\angle F = \angle H$ and the given $\angle G = 30^\circ$:

The measure of $\angle H$ is $\boxed{\text{}75^\circ\text{}}$.

11. Triangle ABC is a right triangle and $\sin B = \frac{5}{13}$. What is the length of side BC?
The diagram shows a right triangle ABC, with the right angle at C. The hypotenuse AB has a length of 26.

In a right triangle, $\sin(\text{angle}) = \frac{\text{opposite side}}{\text{hypotenuse}}$.
For angle B, the opposite side is AC and the hypotenuse is AB.
So, $\sin B = \frac{AC}{AB}$.

We are given $\sin B = \frac{5}{13}$ and $AB = 26$.

Multiply both sides by 26:

In a right triangle, $AC^2 + BC^2 = AB^2$.
Substitute the known values $AC = 10$ and $AB = 26$:

The length of side BC is $\boxed{\text{}24\text{}}$.

12. All the dimensions shown are for the interior of the triangle. What is the value of $\cos x$?
The diagram shows an isosceles triangle with two equal sides of $24 \text{ ft}$ and a base of $32 \text{ ft}$. The base angles are labeled $x$.

Draw an altitude from the apex (the vertex between the two equal sides) to the base. In an isosceles triangle, this altitude bisects the base and forms two congruent right-angled triangles.

The hypotenuse of each right triangle is one of the equal sides of the isosceles triangle, which is $24 \text{ ft}$.
The base of each right triangle is half of the isosceles triangle's base: $\frac{32 \text{ ft}}{2} = 16 \text{ ft}$.
The angle $x$ is one of the acute angles in this right triangle.

In a right triangle, $\cos(\text{angle}) = \frac{\text{adjacent side}}{\text{hypotenuse}}$.
For angle $x$, the adjacent side is $16 \text{ ft}$ and the hypotenuse is $24 \text{ ft}$.

The value of $\cos x$ is $\boxed{\text{}\frac{2}{3}\text{}}$.