solve 41 43 and 44---**Question 41:**
In the given figure, the hypotenuse is 64 what is the cos x ?
[Chart/Diagram Description]
Type: Geometric figure (Right-angled triangle)
Main Elements:
- A right-angled triangle is shown.
- One acute angle is labeled 60 degrees.
- The other acute angle is labeled x degrees.
- The angle opposite one leg is marked as a right angle (90 degrees).
- The hypotenuse is the side opposite the right angle. The problem statement gives its length as 64, but it is not labeled on the figure.
- The note states the figure is not drawn to scale.
Note: Figure not drawn to scale.
Options:
A) sqrt(3)/2
B) 1/2
C) 2*sqrt(3)/3
D) 2
**Question 43:**
A quadratic function models the height, in feet, of an object above the ground in terms of time, in seconds, after the object is launched off an elevated surface. The model indicates that at a time of 9 seconds, the object is 294 feet above the ground. At a time of 13 seconds, the object is 310 feet above the ground. If the object was at a height of 24 feet when it was launched, what is the height, in feet, of the object above ground 18 seconds after being launched?
Options:
A) 96
B) 192
C) 222
D) 240
**Question 44:**
A rectangular pyramid has a square base with an area of 324 square meters. What is the surface area, in square meters, of one of the triangular faces if the rectangular pyramid has a volume of 4,320 cubic meters?
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Let's solve Question 41. We have a right triangle with angles 90 degrees, 60 degrees, and x. Since angles in a triangle sum to 180 degrees, we get x equals 30 degrees. We need to find cosine of x, which is cosine of 30 degrees. The value of cosine 30 degrees is square root of 3 over 2. So the answer is A.
Now let's solve Question 43. We have a quadratic height model h of t equals a t squared plus b t plus c. The initial height is 24 feet, so c equals 24. We're given two points: at 9 seconds the height is 294 feet, and at 13 seconds it's 310 feet. Solving this system of equations gives us a equals negative 2 and b equals 48. So our function is h of t equals negative 2 t squared plus 48 t plus 24. At t equals 18 seconds, h of 18 equals negative 648 plus 864 plus 24, which equals 240 feet. The answer is D.
Finally, let's solve Question 44. We have a rectangular pyramid with a square base area of 324 square meters. The side length is square root of 324, which equals 18 meters. Given the volume is 4320 cubic meters, we can find the height using the volume formula. The height equals 40 meters. To find the triangular face area, we need the slant height. Using the Pythagorean theorem, the slant height is 41 meters. The triangular face area is one half times 18 times 41, which equals 369 square meters.
Let's summarize our solutions. For Question 41, we used the angle sum property to find that x equals 30 degrees, then found cosine of 30 degrees equals square root of 3 over 2. For Question 43, we set up a quadratic height model and solved a system of equations to find the coefficients. For Question 44, we applied pyramid volume formulas and used the Pythagorean theorem to find the slant height. All three problems required systematic algebraic approaches to reach the correct answers.