Solve the following;
1). Peter is planning to build a house with an area of x²-12x +20. What will be the length of his house if the width is x - 2? 2)Merry is 15 kilometers away to her school. If the distance from the topmost of the school to Merry is 50 kilometers, what is the height of the school?
3) Luffy has a boat and the length of his triangular sail is 30√5 with an angle of 30°. What is the area of the triangular sail? 62. Zoro is planning to cut down a 20-foot-tall tree. What will be the safest distance if the tree fall with an angle of depression of 35"?
4. In the town of Gen. McArthur, there are two islets destinations namely: Minalungon and Minagkong. Assuming that the distance from pantalan to Minalungon is 10 km; Minalungon to Minagkong is 6 km; and Minagkong to pantalan is 20 km. Find each angle if we form a triangle from pantalan-Minalungon- Minagkong.
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1). The area of the house is given by A = x² - 12x + 20. The length of the house is the dimension perpendicular to the width, which is x - 2. To find the length, we can divide the area by the width:
Length = A / Width
Length = (x² - 12x + 20) / (x - 2)
2). We can use the Pythagorean theorem to solve for the height of the school. Let h be the height of the school, and d be the distance Merry is away from the school. Then:
h² + d² = 50² (by the Pythagorean theorem)
d = 15
Substituting d = 15 into the equation above, we get:
h² + 15² = 50²
h² = 50² - 15²
h = √(50² - 15²)
h ≈ 48.46 kilometers
3). The area of a triangle can be calculated using the formula A = (1/2)bh, where b is the base of the triangle and h is its height. The length of the sail is not relevant for finding the area. Instead, we need to use the angle and the length of the base of the triangle. Since the sail is a right triangle, we know that one angle is 90 degrees, and the other angle is 30 degrees. The base of the triangle is opposite the 30 degree angle, and its length is (1/2) the length of the sail:
b = (1/2)(30√5) = 15√5
Now we can find the area:
A = (1/2)bh = (1/2)(15√5)(30 sin 30) ≈ 112.5 square meters
4) We can use the Law of Cosines to find the angles of the triangle. Let A, B, and C be the vertices of the triangle corresponding to pantalan, Minalungon, and Minagkong, respectively. Let a, b, and c be the sides opposite to A, B, and C, respectively. Then:
a² = b² + c² - 2bc cos A
b² = a² + c² - 2ac cos B
c² = a² + b² - 2ab cos C
Substituting the given distances, we get:
a² = 6² + 20² - 2(6)(20) cos A
b² = 10² + 20² - 2(10)(20) cos B
c² = 6² + 10² - 2(6)(10) cos C
Simplifying each equation, we get:
a² = 316 - 240 cos A
b² = 500 - 400 cos B
c² = 136 - 120 cos C
To solve for the angles, we need to find the cosines first. We can use the Law of Cosines again to solve for them. For example, to find cos A, we can use the equation for a:
cos A = (6² + 20² - a²) / (2(6)(20))
cos A ≈ 0.9659
We can use similar calculations to find cos B and cos C. Then, we can use the inverse cosine function to find the angles:
A = cos⁻¹(0.9659) ≈ 16.26 degrees
B = cos⁻¹(0.8) ≈ 36.87 degrees
C = 180 - A