To find the ratio AD/BE in triangle ABC, we can use trigonometry and the information given in the problem.

First, let’s draw a diagram of triangle ABC with altitudes AD and BE:

   A
   |\
   | \
   |  \
D |   \ E
   |    \
   |     \
   |      \
---B------C

We are given that ∠BAC = 45° and ∠ABC = θ. Since AD and BE are altitudes, they are perpendicular to their respective bases, BC and AC. This means that ∠ADC = 90° and ∠BEC = 90°.

Now, let’s consider triangle ADC and triangle BEC separately:

In triangle ADC:

• ∠DAC = 45° (complementary to ∠BAC).
• ∠ADC = 90° (given).

In triangle BEC:

• ∠ECB = θ (given).
• ∠BEC = 90° (given).

We can use the trigonometric ratios in these right triangles to find the lengths of AD and BE.

For triangle ADC:

• AD/AC = tan(∠DAC) = tan(45°) = 1.

For triangle BEC:

• BE/BC = tan(∠ECB) = tan(θ).

Now, we have the ratios AD/AC and BE/BC. We want to find AD/BE, so let’s divide these ratios:

AD/BE = (AD/AC) / (BE/BC) = 1 / tan(θ) = 1 / (sin(θ) / cos(θ)) = cos(θ) / sin(θ).

Therefore, the correct answer is (c) √2 sinθ.

To solve this problem, we’ll use the information that the area of triangle ADC is half the area of triangle ABD. Since triangle ABC is equilateral, all its angles are 60 degrees, and we can use trigonometry to find the length of AD.

Let’s denote the length of AD as x. Since triangle ADC is half the area of triangle ABD, we can write:

(1/2) * AD * DC = (1/2) * AB * BD

We know that AB = 3 cm, and BD = BC – CD. Since triangle ABC is equilateral, BC = 3 cm, so BD = 3 – CD. We can rewrite the equation as:

(1/2) * x * DC = (1/2) * 3 * (3 – CD)

Now, we need to find DC in terms of x. We can use trigonometry to find it. In triangle ADC, we have:

tan(60 degrees) = DC / x

tan(60 degrees) = √3

So, DC = x * √3.

Substituting this back into the equation:

(1/2) * x * x * √3 = (1/2) * 3 * (3 – x * √3)

Simplify the equation:

x^2 * √3 = 9 – 3x * √3

Now, isolate x:

x^2 * √3 + 3x * √3 – 9 = 0

Divide the entire equation by √3:

x^2 + 3x – 3√3 = 0

We can use the quadratic formula to solve for x:

x = [-b ± √(b^2 – 4ac)] / (2a)

In this case, a = 1, b = 3, and c = -3√3. Plugging these values into the formula:

x = [-3 ± √(3^2 – 4 * 1 * (-3√3))] / (2 * 1)

x = [-3 ± √(9 + 12√3)] / 2

Now, simplify further:

x = (-3 ± √(9 + 12√3)) / 2

x = (-3 ± √3√7) / 2

So, the length of AD is x = (-3 ± √3√7) / 2, and since we are dealing with a length, the negative value is not meaningful in this context. Therefore, the length of AD is:

x = (3 + √3√7) / 2

So, the correct answer is (a) √7.

Let’s analyze the situation step by step:

1. Initially, the two ships and the port formed an equilateral triangle with sides of length 24 km. This means that each side of the equilateral triangle was 24 km.
2. When the slower ship traveled 8 km, the triangle formed by the new positions of the two ships and the port became right-angled. This implies that the slower ship moved 8 km along one of the sides of the original equilateral triangle, creating a right triangle with the other ship and the port.
3. Since the original triangle was equilateral, each of its angles was 60 degrees. Therefore, the right triangle formed is a 30-60-90 triangle.

Now, let’s consider the properties of a 30-60-90 triangle. In a 30-60-90 triangle, the side opposite the 30-degree angle is half the length of the hypotenuse, and the side opposite the 60-degree angle is √3/2 times the length of the hypotenuse.

In this case, the hypotenuse is the side of the original equilateral triangle, which has a length of 24 km. So, the shorter leg (opposite the 30-degree angle) is 12 km (half of 24 km), and the longer leg (opposite the 60-degree angle) is 12√3 km (√3/2 times 24 km).

Now, the slower ship has traveled 8 km along one of the legs of this right triangle, so the distance between the slower ship and the port (the remaining part of the shorter leg) is 12 km – 8 km = 4 km.

When the faster ship reaches the port, the distance between the other ship (slower ship) and the port is the remaining part of the longer leg, which is 12√3 km.

So, the correct answer is (c) 12.

To find the area of triangle EOD, we need to use the fact that the medians of a triangle divide each other into segments of a 2:1 ratio.

Given that medians BD and CE intersect at point O, O divides each median into two segments, with the ratio of 2:1.

Let M be the midpoint of side BC. Since O divides BD into a 2:1 ratio, we can say that BO = 2(OD). Similarly, since O divides CE into a 2:1 ratio, we can say that CO = 2(OE).

Now, let’s consider triangle ABC and triangle BOC (the smaller triangle formed by the medians). The area of triangle BOC is 1/4 the area of triangle ABC because it has 1/4 of the base (BC) and the same height (from O to side BC).

So, the area of triangle BOC = (1/4) * 108 sq. cm = 27 sq. cm.

Now, triangle BOC is similar to triangle EOD because they share the same height (the distance from O to side BC) and have parallel sides. Since the sides of triangle EOD are 1/3 of the sides of triangle BOC, the area of triangle EOD will be (1/3)^2 = 1/9 of the area of triangle BOC.

Area of triangle EOD = (1/9) * 27 sq. cm = 3 sq. cm.

So, the area of triangle EOD is 3 sq. cm.

To find the area of triangle ABC, we can use the ratios of the areas of triangles ADE, ABD, and ACE, along with the given information.

First, let’s express the ratios of areas:

1. Area of triangle ADE = 8 sq cm (given).
2. AD : BD = 2 : 1, which means the heights of triangles ADE and ABD are in the same ratio. Therefore, the area of triangle ABD is 2/3 of the area of triangle ADE. So, the area of triangle ABD is (2/3) * 8 sq cm = 16/3 sq cm.
3. AE : CE = 2 : 3, which means the heights of triangles ADE and ACE are in the same ratio. Therefore, the area of triangle ACE is 2/5 of the area of triangle ADE. So, the area of triangle ACE is (2/5) * 8 sq cm = 16/5 sq cm.

Now, we have found the areas of triangles ADE, ABD, and ACE. To find the area of triangle ABC, we need to add these areas together, as follows:

Area of triangle ABC = Area of triangle ABD + Area of triangle ACE + Area of triangle ADE

Area of triangle ABC = (16/3) sq cm + (16/5) sq cm + 8 sq cm

Now, let’s find a common denominator for the fractions:

Area of triangle ABC = (80/15) sq cm + (48/15) sq cm + (120/15) sq cm

Area of triangle ABC = (80 + 48 + 120) / 15 sq cm

Area of triangle ABC = 248 / 15 sq cm

So, the area of triangle ABC is 248/15 square centimeters.