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Object A can move at 50km/h, wants to intercept object B (currently $15^{\circ}$, east of north from A) moving at 26km/h, $40^{\circ}$ east of north. What angle should A take to intercept B? AB is 20km apart

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The provided answer looks like:


Choose x axis along 20km distance.

$26t \sin{(40-15)} = 50t \sin{\theta}$

$\theta = \sin^{-1}{\frac{11}{50}} = 12.7$

$15 + 12.7 = 27.7$


I took a different approach and used $\cos$ and got a different answer ... why is that?

$26t \cos{(40-15)} = 50t \cos{\theta}$

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  • $\begingroup$ Not a bad question, but could you show your approach as well for comparison? $\endgroup$ – David Z Sep 27 '12 at 17:11
  • $\begingroup$ @DavidZaslavsky, I simply changed sin to cos. Updated question too. I think my error might be I didnt take into account they start from different x positions. Solving by y is easier as they start on the same y position (0)? $\endgroup$ – Jiew Meng Sep 27 '12 at 23:49
  • $\begingroup$ Jiew Meng, you need to add to the right hand side of your equation a +20, to account for the distance AB. $\endgroup$ – Jaime Sep 28 '12 at 5:43
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Your equation is almost correct.

Sketch

Updated

Using x-axis along AB

Taking the x-axis along AB yields

$$ 50 t \sin (\theta-15^\circ) = 26 t \sin(40^\circ) $$ $$ \sin (\theta-15^\circ) = 0.52 \sin(40^\circ) $$

$$ \theta = 34.527^\circ $$

$$ \cos (\theta-15^\circ) = \sqrt{1-\sin^2 (\theta-15^\circ) } $$

and the y-axis perpendicular to AB

$$ 50 t \cos(\theta-15^\circ) = 20 + 26 t \cos(40^\circ) $$

$$ t = 0.735 $$

Using x-axis along AC (interception pt)

taking y-axis perpendicular to AC

$$ 50 t = 20 \cos(\theta-15^\circ)+26 t \cos(55^\circ-\theta) $$ $$ t = \frac{20 \cos(\theta-15^\circ)}{50 - 26 \cos(55^\circ-\theta)} $$

taking x-axis along AC $$ 20 \sin(\theta-15^\circ)=26 t \sin(55^\circ-\theta) $$

which when expanded you need to solve an equation of the form $$A\cos \theta + B \sin \theta = C$$ for $\theta$ with the same results as above.

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