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

enter image description here

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?

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

  • $\begingroup$ 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

Your equation is almost correct.



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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.