need help here please :)
5.02

1. Solve the triangle.
A = 50°, b = 13, c = 6 (2 points)

a) a ≈ 10.2, C ≈ 26.3, B ≈ 103.7
b) a ≈ 14, C ≈ 26.3, B ≈ 103.7
c) No triangles possible
d) a ≈ 14, C ≈ 30.3, B ≈ 99.7

2. Solve the triangle.
a = 12, b = 22, C = 95° (2 points)

a) c ≈ 26, A ≈ 27.6°, B ≈ 57.4°
b) c ≈ 26, A ≈ 54.4°, B ≈ 30.6°
c) c ≈ 25.1, A ≈ 27.6°, B ≈ 57.4°
d) c ≈ 25.1, A ≈ 31.6°, B ≈ 53.4°

3. Find the area of the triangle with the given measurements. Round the solution to the nearest hundredth if necessary.
A = 50°, b = 30 ft, c = 14 ft (2 points)

a) 160.87 ft2
b) 420 ft2
c) 134.99 ft2
d) 321.74 ft2

4. Determine whether a triangle can be formed with the given side lengths. If so, use Heron's formula to find the area of the triangle. (2 points)

240
121
263

a) 17,578.06
b) No triangle is formed.
c) 17,592.16
d) 14,499.68

5. An airplane leaves an airport and flies due west 150 miles and then 170 miles in the direction S 49.17°W. How far is the plane from the airport (round to the nearest mile)?
Image of a triangle with legs of 150 and 170 miles and a hypotenuse with unknown length. The points on the ends of the hypotenuse are entitled airport and plane. The angle opposite the hypotenuse shows a 90 degree angle and an additional south west bearing of 49.17 degrees

Question

14-02-2018
Mathematics

contestada

need help here please :)
5.02

1. Solve the triangle.
A = 50°, b = 13, c = 6 (2 points)

a) a ≈ 10.2, C ≈ 26.3, B ≈ 103.7
b) a ≈ 14, C ≈ 26.3, B ≈ 103.7
c) No triangles possible
d) a ≈ 14, C ≈ 30.3, B ≈ 99.7

2. Solve the triangle.
a = 12, b = 22, C = 95° (2 points)

a) c ≈ 26, A ≈ 27.6°, B ≈ 57.4°
b) c ≈ 26, A ≈ 54.4°, B ≈ 30.6°
c) c ≈ 25.1, A ≈ 27.6°, B ≈ 57.4°
d) c ≈ 25.1, A ≈ 31.6°, B ≈ 53.4°

3. Find the area of the triangle with the given measurements. Round the solution to the nearest hundredth if necessary.
A = 50°, b = 30 ft, c = 14 ft (2 points)

a) 160.87 ft2
b) 420 ft2
c) 134.99 ft2
d) 321.74 ft2

4. Determine whether a triangle can be formed with the given side lengths. If so, use Heron's formula to find the area of the triangle. (2 points)

240
121
263

a) 17,578.06
b) No triangle is formed.
c) 17,592.16
d) 14,499.68

5. An airplane leaves an airport and flies due west 150 miles and then 170 miles in the direction S 49.17°W. How far is the plane from the airport (round to the nearest mile)?
Image of a triangle with legs of 150 and 170 miles and a hypotenuse with unknown length. The points on the ends of the hypotenuse are entitled airport and plane. The angle opposite the hypotenuse shows a 90 degree angle and an additional south west bearing of 49.17 degrees

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zhaotianyu369 zhaotianyu369 · Answer 1 · 2018-02-14T15:22:44+01:00

1. A.Use law of cosines. cosA=(b^2+c^2-a^2)/(2bc) because A is the included angle between b and c. Plug in A=50 degrees, b=13, c=6. cos50=(13^2+6^2-a^2)/(2*13*6), a^2=104.7,a=10.2 approximately, so choose A.

2. A.Use law of cosines again. cosC=(a^2+b^2-c^2)/(2ab). C=95 degrees, a=12, b=22. Plug in, cos95=(12^2+22^2-c^2)/(2*12*22), solve the equation, c^2=674, c=26 approximately. The use law of sines to solve for angle A (also works for B), a/sinA=c/sinC, 12/sinA=26/sin95, sinA=0.46, A=arcsin(0.46)=27.6. Choose A.

3. Answer is A. Area=1/2bc*sinA, since A is the included angle between b and c. Plug in b=30, c=14, A=50 degrees, area=1/2*14*30*sin50=160. 87, so the answer is A.

4. D. As long as the sum of any two sides of the triangle is bigger than the third, the triangle exists. 240+121>263, 240+263>121, 263+121>240, so it exists. To use Heron's formula, first find the semiperimeter, (240+263+121)/2=312. A=\sqrt(312*(312-240)*(312-263)*(312-121))=14499.7 approximately, so choose D.

5. 300. The included angle between the two paths is C=49.17+90=139.17 degrees. The lengths of the two paths are a=150, b=170. c is the distance we want. Use the law of cosines, cosC=(a^2+b^2-c^2)/(2ab). Plug in, c^2=89989, c=300 approximately.