Distance between plane and airport is 134.4 miles.
Given : An airplane leaves an airport and flies due west 150 miles and then 170 miles in the direction S 49.17°W.
To find : How far is the plane from the airport.
Solution : Distance from airport to west is 150 miles and then 170 miles in the direction south and angle form is S 49.17° W
Refer the attached picture for clearance.
Applying law of cosines
where a= 150 miles
C= 49.17° angle in degree
c = distance between plane from the airport
Put values in the formula,
Therefore, Distance between plane and airport is 134.4 miles.
I used the simplest geometry I could (wasn't sure what level of mathematics you were).
If the picture doesn't come through I'm sorry.
Final answer is 291.10936206249 miles or approximately 291 miles.
If it's right, can I be the brainliest answer??
d = 251 nm
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.
c ≈ 24.9, A ≈ 28.7°, B ≈ 57.3°
in the picture
c = 26 , ∠A = 27.38° and ∠B = 57.62°
a = 12 , b = 22 and ∠C = 95°
To find: c , ∠A and ∠B
We use Law of cosine and Law of sine to solve the triangle.
Law of cosine states that c² = a² + b² - 2ab cos C
Law of sines states that
Now, using Law of cosine,
c² = 12² + 22² - 2(12)(22) cos 95
c² = 144 + 484 - 528(-0.087)
c² = 628 + 45.9
c² = 673.93
c = 25.96 ≈ 26
Now using Law of sines
Therefore, c = 26 , ∠A = 27.38° and ∠B = 57.62°
c ≈ 26, A ≈ 27.6°, B ≈ 57.4°
c ≈ 26, A ≈ 54.4°, B ≈ 30.6°
c ≈ 25.1, A ≈ 27.6°, B ≈ 57.4°
c ≈ 25.1, A ≈ 31.6°, B ≈ 53.4°