distance extremes are at the ends of the semi-axes. You know that a=93, and c/a = 1/60. And, b^c+c^2=a^2.
Hard to figure just what you are after with the ships, but apparently you want the angle ACB.
Angle CAB is arctan(120/80)+10°
To find the distance CB, use the law of cosines.
CB^2 = 150^2 + (40√13)^2 - 2(150)(40√13)cos(CAB)
Then using the law of sines,
sin(CAB)/150 = sin(CAB)/BC
1. The orbit of the earth is an ellipse with the sun at a focus, the semi-major axes 93 million miles and the eccentricity is 1/60. Find the greatest and least distance of the earth from the sun.
2. A battle shipping is sailing at a straight line towards east at 40km/hour from start and another smaller ship sailed north at 60km/hour. If both ship started at the same time, 2 hours later, the battleship is now at point C, the battleship wants to launch a missile towards B. If the smaller ship headed north (now at point A 2 hours later) found that point But is 10 degrees north of east and it is 150 km away. Solve the angle that the battleship should direct its missile with respect to point A in order to hit point B.
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