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# A coast guard cutter detects an unidentified ship at 20.0km in the direction 15.0 degrees east of north. The ship is traveling at 26.0km/h o

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A coast guard cutter detects an unidentified ship at 20.0km in the direction 15.0 degrees east of north. The ship is traveling at 26.0km/h on a course at 40.0 degrees east of north. The Coast Guard wishes to send a speed boat to intercept and inveatigate the vessel. If the speedboat travels at 50.0km/h , in what direction should it head?

Aug 19, 2015

#3
+95179
+10

Let the point (x,y ) be the point where the speed boat intercepts the vessel

Let t be the time taken to intercept.

Let alpha be the angle between East and the direction the speed boat travels (as shown in the diagram)

So the bearing of the speed boat will be  N(90-alpha)E

$$\\x=20cos75+26tcos50\qquad and \qquad x=50tcos\alpha\\ so\\ 20cos75+26tcos50=50tcos\alpha\\\\ 20cos75=50tcos\alpha-26tcos50\\\\ 20cos75=t(50cos\alpha-26cos50)\\\\ \frac{20cos75}{ (50cos\alpha-26cos50)}=t\\\\ AND\\ y=20sin75+26tsin50\qquad and \qquad y=50tsin\alpha\\\\ ...\\ \frac{20sin75}{ (50sin\alpha-26sin50)}=t\\ so\\ \frac{(20sin75)}{ (50sin\alpha-26sin50)}=\frac{(20cos75)}{ (50cos\alpha-26cos50)}\\\\$$

$$\\(20sin75)(50cos\alpha-26cos50)=(20cos75)(50sin\alpha-26sin50)\\\\ (20sin75)(50cos\alpha)-(20sin75)(26cos50)=(20cos75)(50sin\alpha)-(20cos75)(26sin50)\\\\ (1000sin75)(cos\alpha)-(520sin75cos50)=(1000cos75sin\alpha)-(520cos75sin50)$$

$$\\(1000sin75cos\alpha)-(1000cos75sin\alpha)=(520sin75cos50)-(520cos75sin50)\\\\ 1000(sin75cos\alpha-cos75sin\alpha)=520(sin75cos50-cos75sin50)\\\\ 1000sin(75-\alpha)=520sin(75-50)\\\\ sin(75-\alpha)=0.52sin(25)\\\\ 75-\alpha=asin(0.52sin(25))\\\\ -\alpha=asin(0.52sin(25))-75\\\\ \alpha=75-asin(0.52sin(25))\\\\$$

$${\mathtt{75}}{\mathtt{\,-\,}}\underset{\,\,\,\,^{{360^\circ}}}{{sin}}^{\!\!\mathtt{-1}}{\left({\mathtt{0.52}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{{360^\circ}}}{{sin}}{\left({\mathtt{25}}^\circ\right)}\right)} = {\mathtt{62.304\: \!975\: \!095\: \!575}}$$

$${\mathtt{90}}{\mathtt{\,-\,}}{\mathtt{62.304\: \!975\: \!095\: \!575}} = {\mathtt{27.695\: \!024\: \!904\: \!425}}$$   =     27 degrees 41minutes and 42 seconds

The speed boat needs to set a course for

$$N\;27^041'42"E$$

What a great question and Heureka and I even agree   :)))

Aug 20, 2015

#1
+20831
+10

A coast guard cutter detects an unidentified ship at 20.0km in the direction 15.0 degrees east of north. The ship is traveling at 26.0km/h on a course at 40.0 degrees east of north. The Coast Guard wishes to send a speed boat to intercept and inveatigate the vessel. If the speedboat travels at 50.0km/h , in what direction should it head?

We set t = time

$$\small{ \begin{array}{lcl} \text{Ship position: } \vec{p}= 20\cdot\binom { \sin{(15\ensurement{^{\circ}})} } { \cos{(15\ensurement{^{\circ}})} } \\\\ \text{Ship traveling: } \vec{s}= 26\cdot t \cdot\binom { \sin{(40\ensurement{^{\circ}})} } { \cos{(40\ensurement{^{\circ}})} } \\\\ \text{Guard cutter traveling: } \vec{c}= 50\cdot t\cdot\binom { \sin{(x)} } { \cos{(x)} } \\\\ \vec{p} + \vec{s} & = &\vec{c} \\\\ 20\cdot\binom { \sin{(15\ensurement{^{\circ}})} } { \cos{(15\ensurement{^{\circ}})} } + 26\cdot t \cdot\binom { \sin{(40\ensurement{^{\circ}})} } { \cos{(40\ensurement{^{\circ}})} } &=& 50\cdot t\cdot\binom { \sin{(x) } } { \cos{(x)} } \\\\ \hline 20\cdot \sin{(15\ensurement{^{\circ}})} + 26\cdot \sin{(40\ensurement{^{\circ}})}\cdot t &=&50\cdot t\cdot \sin{(x)} \\ 20\cdot \cos{(15\ensurement{^{\circ}})} + 26\cdot \cos{(40\ensurement{^{\circ}})}\cdot t &=&50\cdot t\cdot \cos{(x})} \\ \hline \text{Square both sides }\\ \left[ 20\cdot \sin{(15\ensurement{^{\circ}})} + 26\cdot \sin{(40\ensurement{^{\circ}})}\cdot t \right]^2 &=&50^2\cdot t^2 \sin^2{(x)} \\ \left[ 20\cdot \cos{(15\ensurement{^{\circ}})} + 26\cdot \cos{(40\ensurement{^{\circ}})}\cdot t \right]^2 &=&50^2\cdot t^2 \cos^2{(x})} \\ \hline \text{Add }\\ \left[ 20\cdot \sin{(15\ensurement{^{\circ}})} + 26\cdot \sin{(40\ensurement{^{\circ}})}\cdot t \right]^2 \\ + \left[ 20\cdot \cos{(15\ensurement{^{\circ}})} + 26\cdot \cos{(40\ensurement{^{\circ}})}\cdot t \right]^2 &=& 50^2\cdot t^2 [ \sin^2{(x)} + \cos^2{(x})} ] \\ \hline \sin^2{(x)} + \cos^2{(x})} = 1\\ \left[ 20\cdot \sin{(15\ensurement{^{\circ}})} + 26\cdot \sin{(40\ensurement{^{\circ}})}\cdot t \right]^2 \\ + \left[ 20\cdot \cos{(15\ensurement{^{\circ}})} + 26\cdot \cos{(40\ensurement{^{\circ}})}\cdot t \right]^2 &=& 50^2\cdot t^2\\ \hline 20^2\cdot \sin^2{(15\ensurement{^{\circ}})} + 2\cdot 20\cdot 26 \cdot \sin{(15\ensurement{^{\circ}})} \cdot \sin{(40\ensurement{^{\circ}})} \cdot t + 26^2\cdot t^2 \sin^2{(40\ensurement{^{\circ}})} \\ +20^2\cdot \cos^2{(15\ensurement{^{\circ}})} + 2\cdot 20\cdot 26 \cdot \cos{(15\ensurement{^{\circ}})} \cdot \cos{(40\ensurement{^{\circ}})} \cdot t +26^2 \cdot t^2 \cos^2{(40\ensurement{^{\circ}})} &=& 50^2\cdot t^2\\ \hline \sin^2{ (15\ensurement{^{\circ}}) } + \cos^2{ (15\ensurement{^{\circ}}) } = 1\\ \sin^2{ (40\ensurement{^{\circ}}) } + \cos^2{ (40\ensurement{^{\circ}}) } = 1\\ 20^2 + 2\cdot 20\cdot 26 \cdot [ \underbrace{ \cos{ (15\ensurement{^{\circ}}) } \cdot \cos{ (40\ensurement{^{\circ}}) } + \sin{ (15\ensurement{^{\circ}}) } \cdot \sin{ (40\ensurement{^{\circ}}) } }_{=\cos{(40\ensurement{^{\circ}}-15\ensurement{^{\circ}})}=\cos{25\ensurement{^{\circ}}}} ] \cdot t +26^2 t^2 &=& 50^2\cdot t^2\\ \hline \end{array} }$$

$$\small{ \begin{array}{lcl} 50^2\cdot t^2 - 26^2 \cdot t^2 -2\cdot 20\cdot 26 \cdot \cos{ (25\ensurement{^{\circ}}) }\cdot t -20^2 &=& 0\\\\ 1824\cdot t^2 - 1040 \cdot \cos{ (25\ensurement{^{\circ}}) }\cdot t - 400 &=& 0 \\\\ t_{1,2} &=& \frac { 1040\cdot \cos{ (25\ensurement{^{\circ}}) } \pm \sqrt{ 1040^2\cdot \cos^2{ (25\ensurement{^{\circ}}) } - 4\cdot 1824 \cdot (-400) } } { 2\cdot 1824 } \\\\ t_{1,2} &=& \frac { 1040\cdot \cos{ (25\ensurement{^{\circ}}) } \pm \sqrt{ 1040^2\cdot \cos^2{ (25\ensurement{^{\circ}}) } + 2918400 } } { 2\cdot 1824 } \\\\ t_{1,2} &=& \frac { 1040\cdot \cos{ (25\ensurement{^{\circ}}) } \pm \sqrt{3806819.53932 } } { 3648 } \\\\ t_{1,2} &=& \frac { 1040\cdot \cos{ (25\ensurement{^{\circ}}) } \pm 1951.10725982 } { 3648 } \\\\ t_{1,2} &=& \frac{ 942.560098518 \pm 1951.10725982 } { 3648 } \\\\ t &=& 0.79322021884~ \text{hours}\\ t &=& 47.5932131305~ \text{minutes}\\ \hline \end{array} }$$

$$\small{ \begin{array}{llcl} \text{Azimuth angle: } & \tan{(\alpha_{\text{azimuth}})} &=& \frac{ \sin{ (x) } }{ \cos{ (x) } } \\\\ & \tan{(\alpha_{\text{azimuth}})} &=& \dfrac{ 20\cdot \sin{(15\ensurement{^{\circ}})} + 26\cdot \sin{(40\ensurement{^{\circ}})}\cdot t } { 20\cdot \cos{(15\ensurement{^{\circ}})} + 26\cdot \cos{(40\ensurement{^{\circ}})}\cdot t } \\\\ & \tan{(\alpha_{\text{azimuth}})} &=& \dfrac{ 18.4330562411 } { 35.1172069869 } \\\\ & \tan{(\alpha_{\text{azimuth}})} &=& 0.52490097655 \\\\ &\alpha_{\text{azimuth}} &=& \arctan{ ( 0.52490097655 ) } \\ &\mathbf{\alpha_{\text{azimuth}} } & \mathbf{=} & \mathbf{27.6950249044\ensurement{^{\circ}} }\\ \end{array} }$$

The speed boat needs to set a course for

$$N\;27^0~41'~42"~E$$

Thank you Melody, i don't need to divide by 16, my mistake!

Aug 20, 2015
#2
+95179
0

That is a really interesting question.   And a great answer

I wish I could give you more points Heureka

Aug 20, 2015
#3
+95179
+10

Let the point (x,y ) be the point where the speed boat intercepts the vessel

Let t be the time taken to intercept.

Let alpha be the angle between East and the direction the speed boat travels (as shown in the diagram)

So the bearing of the speed boat will be  N(90-alpha)E

$$\\x=20cos75+26tcos50\qquad and \qquad x=50tcos\alpha\\ so\\ 20cos75+26tcos50=50tcos\alpha\\\\ 20cos75=50tcos\alpha-26tcos50\\\\ 20cos75=t(50cos\alpha-26cos50)\\\\ \frac{20cos75}{ (50cos\alpha-26cos50)}=t\\\\ AND\\ y=20sin75+26tsin50\qquad and \qquad y=50tsin\alpha\\\\ ...\\ \frac{20sin75}{ (50sin\alpha-26sin50)}=t\\ so\\ \frac{(20sin75)}{ (50sin\alpha-26sin50)}=\frac{(20cos75)}{ (50cos\alpha-26cos50)}\\\\$$

$$\\(20sin75)(50cos\alpha-26cos50)=(20cos75)(50sin\alpha-26sin50)\\\\ (20sin75)(50cos\alpha)-(20sin75)(26cos50)=(20cos75)(50sin\alpha)-(20cos75)(26sin50)\\\\ (1000sin75)(cos\alpha)-(520sin75cos50)=(1000cos75sin\alpha)-(520cos75sin50)$$

$$\\(1000sin75cos\alpha)-(1000cos75sin\alpha)=(520sin75cos50)-(520cos75sin50)\\\\ 1000(sin75cos\alpha-cos75sin\alpha)=520(sin75cos50-cos75sin50)\\\\ 1000sin(75-\alpha)=520sin(75-50)\\\\ sin(75-\alpha)=0.52sin(25)\\\\ 75-\alpha=asin(0.52sin(25))\\\\ -\alpha=asin(0.52sin(25))-75\\\\ \alpha=75-asin(0.52sin(25))\\\\$$

$${\mathtt{75}}{\mathtt{\,-\,}}\underset{\,\,\,\,^{{360^\circ}}}{{sin}}^{\!\!\mathtt{-1}}{\left({\mathtt{0.52}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{{360^\circ}}}{{sin}}{\left({\mathtt{25}}^\circ\right)}\right)} = {\mathtt{62.304\: \!975\: \!095\: \!575}}$$

$${\mathtt{90}}{\mathtt{\,-\,}}{\mathtt{62.304\: \!975\: \!095\: \!575}} = {\mathtt{27.695\: \!024\: \!904\: \!425}}$$   =     27 degrees 41minutes and 42 seconds

The speed boat needs to set a course for

$$N\;27^041'42"E$$

What a great question and Heureka and I even agree   :)))

Melody Aug 20, 2015
#4
+95179
+5

I added this into the "great questions to learn from" sticky thread :)

Aug 20, 2015
#5
+20831
0

Thank you Melody, i don't need to divide by 16, my mistake!

Aug 20, 2015
#6
+95179
+5

I thought your answer was the same as mine. I thought you gave the complementary angle. :)

Aug 20, 2015
#7
+94526
0

Very nice problem and good answers from heureka and Melody.....!!!

Aug 23, 2015
#8
+95179
0

Thanks Chris :))

Aug 24, 2015