Forces of 42 and 53 pounds act on a body so that the angle between the resultant and the larger force is 47 degrees. Find to the nearest degree two possible measures of the angle between the resultant and the smaller force.
Law of Cosine makes resultant = about 45 (can't be smaller that 53?)
Law of Sines makes angle opposite 53 = about 68 degrees (can't be acute?)
How do you do it????
Law of Cosine makes resultant = about 45 (can't be smaller that 53?)
Law of Sines makes angle opposite 53 = about 68 degrees (can't be acute?)
How do you do it????
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Sketch a triangle ∆ABC, with AB=53, BC=42, and ∠A=47°. The component force vectors are AB and BC, and the resultant is AC. That's the setup, since AB is the larger force.
The law of sines for A and C is:
(sin A)/BC = (sin C)/AB
C = arcsin [(sin A) * AB/BC] = arcsin( sqrt(2)/2 * 53/42 )
...which gives TWO solutions for ∠C, one acute and one obtuse. If your calculation of 68° is accurate, then thats 90° - 22°, so the other angle is 90° + 22° =112°.
The law of cosines says that
x^2 + 53^2 = 42^2 + 2*53*x*(cos 45°)
...where x is the length of AC (the resultant). That's a quadratic equation in x, which gives two possible values for AC: about 18.5 and about 56.4. This doesn't match your calculation. It also doesn't find the angle C, which is what was asked.
The law of sines for A and C is:
(sin A)/BC = (sin C)/AB
C = arcsin [(sin A) * AB/BC] = arcsin( sqrt(2)/2 * 53/42 )
...which gives TWO solutions for ∠C, one acute and one obtuse. If your calculation of 68° is accurate, then thats 90° - 22°, so the other angle is 90° + 22° =112°.
The law of cosines says that
x^2 + 53^2 = 42^2 + 2*53*x*(cos 45°)
...where x is the length of AC (the resultant). That's a quadratic equation in x, which gives two possible values for AC: about 18.5 and about 56.4. This doesn't match your calculation. It also doesn't find the angle C, which is what was asked.