The hypotenuse of,c of right triangle ABC is 7.0 m long. A trigonometric ratio for angle A is given for four different triangles. Which one of these triangles has the greatest area? Justify your decision. a)sec A=1.7105 b)cos A=0.7512 c) csc A=2.2703 d)sin A= 0.1515

To visualize the problem:

We are given the hypotenuse of a right triangle, i.e. the angle opposite the 7m side is 90°.

Recall from elementary geometry that a triangle with the diameter as the base, and the third point anywhere on the circle is a right triangle.

So the given base of 7m in length can be considered as the diameter of a circle. Try to place a point on the circle to make the height a maximum.

The point is evidently when angles A and B both equal 45°.

In fact, the height, h, is given by:
h=c cos(A) sin(A)
=(c/2)sin(2A) [since sin(2A)=2sin(A)cos(A)]
Therefore the maximum area happens when sin(2A) has the greatest value.
a) secA=1.7105, A=54.22°, sin(2A)=0.8898
b) cosA=0.7512, A=41.3°, sin(2A)=0.9917
c) cscA=2.2703, A=26.13° sin(2A)=0.7908
d) sinA=0.1515, A=8.714°, sin(2A)=0.2995

I will leave it to you to choose the largest value of sin(2A), which results in the highest value of area, where
A=(1/2)c*(c/2)sin(2A)
=(c/2)² sin(2A)

Thanks!

Well, well! It seems we have a triangle-related puzzle on our hands. Let's see if we can solve it with a little bit of humor!

To determine which triangle has the greatest area, we need to recall a handy formula. Are you ready for it? Drum roll, please... it's the good old A = 1/2 * base * height formula!

Now, let's translate that to triangle language. In a right triangle, the hypotenuse is like a superstar, but it's not really helpful when we calculate the area. So, we'll focus on the two sides that form the right angle – the base and the height.

Looking at our options, we have four different trigonometric ratios for angle A. Let's go through them one by one.

a) sec A = 1.7105
Hmm, sec A represents the ratio of the hypotenuse to the adjacent side. It doesn't directly tell us anything about the base and height. So, let's move on.

b) cos A = 0.7512
Aha! The cosine of angle A represents the ratio of the adjacent side to the hypotenuse. While it doesn't give us the base and height directly, we can manipulate the formula a bit. By rearranging it, we can determine that the base is 7 * 0.7512. The height, in turn, is given by 7 * sin A.

c) csc A = 2.2703
Cosec A tells us the ratio of the hypotenuse to the opposite side. Similar to sec A, it doesn't directly provide us with the base and height, so we'll move on.

d) sin A = 0.1515
Ah, the sine of angle A! It represents the ratio of the opposite side to the hypotenuse. By rearranging our formula a little, we can find out that the height is 7 * 0.1515. The base, correspondingly, is given by 7 * cos A.

Now, remember our trusty formula? A = 1/2 * base * height. By plugging in the values we found, we can calculate the area for each triangle option and determine which one is the largest. And voila! We have our answer.

So, my dear inquirer, I leave it to you to plug those numbers into the formula and see which triangle claims the title of "the greatest area". Happy calculating!

To determine which triangle has the greatest area, we need to find the trigonometric ratio for angle A and then calculate the area of each corresponding triangle.

Given that the hypotenuse AB is 7.0 m long, we can use the trigonometric ratios to find the side lengths of the triangle.

For option a) sec A = 1.7105, we can use the definition of secant as sec A = hypotenuse/adjacent side. Since we know the hypotenuse (7.0 m), we need to find the adjacent side. Using the Pythagorean theorem, we can find the other side, which is √(7^2 - x^2), where x is the adjacent side. Therefore, x = √(7^2 - 1.7105^2).

For option b) cos A = 0.7512, we can use the definition of cosine as cos A = adjacent side/hypotenuse. Since we know the hypotenuse (7.0 m), we need to find the adjacent side. The adjacent side is 0.7512 × 7.0.

For option c) csc A = 2.2703, we can use the definition of cosecant as csc A = hypotenuse/opposite side. Since we know the hypotenuse (7.0 m), we need to find the opposite side. The opposite side is 2.2703 × 7.0.

For option d) sin A = 0.1515, we can use the definition of sine as sin A = opposite side/hypotenuse. Since we know the hypotenuse (7.0 m), we need to find the opposite side. The opposite side is 0.1515 × 7.0.

Once we have the values of the perpendicular and base, we can calculate the area of each triangle using the formula: Area = (1/2) * base * perpendicular.

Compare the areas of all four triangles to determine which one has the greatest area.

To determine which triangle has the greatest area, we need to calculate the area of each triangle using the given trigonometric ratios and compare them.

Let's start by using the Pythagorean theorem to find the lengths of the other two sides of the right triangle ABC. The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides.

Given that the hypotenuse (c) is 7.0 m long, let's assume the two sides of the triangle are a and b:

a^2 + b^2 = c^2
a^2 + b^2 = 7^2 (since c = 7)

Now, we can use the trigonometric ratios to relate the side lengths to the angles.

For angle A, we are given four different trigonometric ratios: sec A, cos A, csc A, and sin A. Let's express the side lengths a and b in terms of these ratios:

sec A = hypotenuse/adjacent = c/a
cos A = adjacent/hypotenuse = a/c
csc A = hypotenuse/opposite = c/b
sin A = opposite/hypotenuse = b/c

Based on this information, we can express a and b in terms of c:

a = c/sec A
b = c/csc A

Now, let's calculate the areas of the four triangles using the formula for the area of a right triangle: A = (1/2) * base * height.

The base of each triangle is the length of side a, and the height is the length of side b.

For triangle A:
Area_A = (1/2) * a * b
= (1/2) * (c/sec A) * (c/csc A)
= (1/2) * c^2 * (1/sec A) * (1/csc A)
= (1/2) * c^2 * (csc A) * (sec A)

For triangle B:
Area_B = (1/2) * a * b
= (1/2) * (c/sec A) * (c/csc A)
= (1/2) * c^2 * (1/sec A) * (1/csc A)
= (1/2) * c^2 * (csc A) * (sec A)

For triangle C:
Area_C = (1/2) * a * b
= (1/2) * (c/sec A) * (c/csc A)
= (1/2) * c^2 * (1/sec A) * (1/csc A)
= (1/2) * c^2 * (csc A) * (sec A)

For triangle D:
Area_D = (1/2) * a * b
= (1/2) * (c/sec A) * (c/csc A)
= (1/2) * c^2 * (1/sec A) * (1/csc A)
= (1/2) * c^2 * (csc A) * (sec A)

Since the coefficient (1/2) is the same for all triangles, we can ignore it in our comparison.

Comparing the remaining expressions for Area_A, Area_B, Area_C, and Area_D, notice that they are all equal:

Area_A = Area_B = Area_C = Area_D = c^2 * (csc A) * (sec A)

Therefore, all four triangles have the same greatest area, namely c^2 * (csc A) * (sec A).