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Sagot :
distance between the location after one hour of Ship X and O that traveled to different direction 43 degrees from the pier (origin).
The sketch is a triangle, where the angle 43 degrees is between two sides (distances) 115 km and 98 km.
Notice that the angle is between two sides (if you sketch it, it's an angle depression)
Use Law of Cosines to find the third side (distance)
Let a and b are the sides, and cosine theta = 43 degrees.
a = 115 b = 98 theta = 43 degrees
c² = a² + b² - 2(a)(b)(cosC)
c² = (115)² + (98)² - 2 (115)(98)(cos 43degrees)
c² = 13 225 + 9 604 - 22 540 (0.707)
c² = 22 829 - 15 935.78
c² = 6 893.22
[tex] \sqrt{ c^{2} } = \sqrt{6 893.22} [/tex]
c = 83.025
c = 83 kms.
The distance between the two ships at 45 degrees after one hour is 83 km.
(I hope this helps)
The sketch is a triangle, where the angle 43 degrees is between two sides (distances) 115 km and 98 km.
Notice that the angle is between two sides (if you sketch it, it's an angle depression)
Use Law of Cosines to find the third side (distance)
Let a and b are the sides, and cosine theta = 43 degrees.
a = 115 b = 98 theta = 43 degrees
c² = a² + b² - 2(a)(b)(cosC)
c² = (115)² + (98)² - 2 (115)(98)(cos 43degrees)
c² = 13 225 + 9 604 - 22 540 (0.707)
c² = 22 829 - 15 935.78
c² = 6 893.22
[tex] \sqrt{ c^{2} } = \sqrt{6 893.22} [/tex]
c = 83.025
c = 83 kms.
The distance between the two ships at 45 degrees after one hour is 83 km.
(I hope this helps)
a²=b²+c²-2(b)(c) Cos A
a²=98²+115²-2(98)(115) Cos 47°
a=√7456.76
a=86.35 miles
115/sinC = 86.35/Sin47°
Sin C = Sin47°(115)/86.35 C= 76°.91
∠B= 180°-∠A+∠C
= 180°-47°+76°.91
= 56°.09
a²=98²+115²-2(98)(115) Cos 47°
a=√7456.76
a=86.35 miles
115/sinC = 86.35/Sin47°
Sin C = Sin47°(115)/86.35 C= 76°.91
∠B= 180°-∠A+∠C
= 180°-47°+76°.91
= 56°.09
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