4. Solve the triangle(s), if possible. Round to the nearest tenth.
Solve the triangle. Side a = 8, side b = 10, side c = 12
6. Two boats leave the same dock at the same time. The first sails in a straight line
N40°W at 36 miles per hour and the second goes in a straight line S12°W at 62 miles per hour.
After 1 hour, how far apart are the boats, to the nearest tenth of a mile?
use the law of cosines for two of the angles, then the 3rd angle is trivial. For angle A,
10^2 + 12^2 - 2*10*12cosA = 8^2
similarly for B or C
For the boats, draw a diagram. Using the law of cosines, the distance d is found by
d^2 = 36^2 + 62^2 - 2*36*62 cos128°
4. To determine if a triangle is possible, we can use the triangle inequality theorem which states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.
In this case, let's check if the triangle is possible by applying the triangle inequality theorem:
a + b > c
8 + 10 > 12
18 > 12
b + c > a
10 + 12 > 8
22 > 8
c + a > b
12 + 8 > 10
20 > 10
Since all three inequalities are satisfied, the triangle is possible.
Now, let's use the law of cosines to find one of the angles:
c² = a² + b² - 2ab * cos(C)
c² = 8² + 10² - 2(8)(10) * cos(C)
c² = 64 + 100 - 2(80) * cos(C)
c² = 164 - 160 * cos(C)
cos(C) = (c² - a² - b²) / (-2ab)
cos(C) = (164 - 64 - 100) / (-2(8)(10))
cos(C) = 0 / (-160)
cos(C) = 0
C = arccos(0)
C = 90°
So, one of the angles of the triangle is 90 degrees.
Since we know the lengths of all three sides, we can use the law of cosines to find the remaining angle:
cos(A) = (b² + c² - a²) / (2bc)
cos(A) = (10² + 12² - 8²) / (2(10)(12))
cos(A) = 100 + 144 - 64) / (240)
cos(A) = 180 / 240
cos(A) = 0.75
A = arccos(0.75)
A ≈ 41.4°
To find the remaining angle, we can use the fact that the sum of the angles in a triangle is 180 degrees:
180° - 90° - 41.4° = 180° - 131.4° = 48.6°
So, the three angles of the triangle are approximately 90°, 41.4°, and 48.6°.
Now, to find the area of the triangle, we can use Heron's formula:
s = (a + b + c) / 2
s = (8 + 10 + 12) / 2
s = 30 / 2
s = 15
Area = √(s(s-a)(s-b)(s-c))
Area = √(15(15-8)(15-10)(15-12))
Area = √(15(7)(5)(3))
Area = √(1575)
Area ≈ 39.7
Therefore, the triangle has angles of approximately 90°, 41.4°, and 48.6°, and an area of approximately 39.7 square units.
6. To find the distance between the two boats after 1 hour, we can use the formula:
Distance = Speed × Time
For the first boat:
Distance1 = 36 miles/hour × 1 hour
Distance1 = 36 miles
For the second boat:
Distance2 = 62 miles/hour × 1 hour
Distance2 = 62 miles
To find the distance between the two boats, we can use the Law of Cosines with the angles given:
c² = a² + b² - 2ab × cos(C)
c² = Distance1² + Distance2² - 2 × Distance1 × Distance2 × cos(180° - (40° + 12°))
c² = 36² + 62² - 2 × 36 × 62 × cos(128°)
c² = 1296 + 3844 - 4464 × cos(128°)
c² = 5140 - 4464 × (-0.4910)
c² = 5140 - (-2192.344)
c ≈ √7332.344
c ≈ 85.65
Therefore, after 1 hour, the two boats are approximately 85.65 miles apart.
4. To solve the triangle with side lengths a = 8, b = 10, and c = 12, we can use the Law of Cosines. The Law of Cosines states that for any triangle with sides a, b, and c, and angle C opposite side c:
c² = a² + b² - 2ab * cos(C)
In this case, we have side lengths a = 8, b = 10, and c = 12. We are interested in finding the angle C. By rearranging the formula, we can find cos(C):
cos(C) = (a² + b² - c²) / (2ab)
Plugging in the values, we get:
cos(C) = (8² + 10² - 12²) / (2 * 8 * 10)
cos(C) = (64 + 100 - 144) / 160
cos(C) = 20 / 160
cos(C) = 1 / 8
C ≈ acos(1 / 8) ≈ 82.8°
Now, we can use the Law of Sines to find the other two angles:
sin(A) / a = sin(C) / c
sin(A) = (sin(C) / c) * a
sin(A) = (sin(82.8°) / 12) * 8
sin(A) ≈ 0.9289
A ≈ asin(0.9289) ≈ 68.3°
Similarly,
sin(B) / b = sin(C) / c
sin(B) ≈ 0.8485
B ≈ asin(0.8485) ≈ 57.2°
Therefore, the triangle has angles A ≈ 68.3°, B ≈ 57.2°, and C ≈ 82.8°.
6. To find how far apart the boats are after 1 hour, we can use the concept of relative velocity. The distance between the boats can be determined by calculating the resultant vector of their velocities.
First, we need to convert the given directions into their corresponding components:
The first boat is moving N40°W. This means it is moving in a direction 40° west of north, or 50° north of west. So, the x-component of its velocity is 36 * cos(50°) and the y-component of its velocity is 36 * sin(50°).
The second boat is moving S12°W. This means it is moving in a direction 12° west of south, or 78° south of west. So, the x-component of its velocity is 62 * cos(78°) and the y-component of its velocity is -62 * sin(78°) (negative due to south direction).
Now, we can calculate the respective displacements of the boats after 1 hour.
The displacement of the first boat is (36 * cos(50°)) * 1 = 36 * cos(50°) miles in the west direction and (36 * sin(50°)) * 1 = 36 * sin(50°) miles in the north direction.
The displacement of the second boat is (62 * cos(78°)) * 1 = 62 * cos(78°) miles in the west direction and (-62 * sin(78°)) * 1 = -62 * sin(78°) miles in the south direction.
To find the distance between the boats, we can use the distance formula:
Distance = √((x1 - x2)² + (y1 - y2)²)
Where (x1, y1) and (x2, y2) are the coordinates of the displacements of the first and second boats, respectively.
Distance = √((36 * cos(50°) - 62 * cos(78°))² + (36 * sin(50°) - (-62 * sin(78°)))²)
Calculating this expression will give you the distance between the two boats after 1 hour, rounded to the nearest tenth.