Questions   
Sort: 
 #108
avatar+118723 
+5

@@ End of Day Wrap    Thurs 20/8/15     Sydney, Australia Time   12:22pm       ♪ ♫   

Yes it is Friday already  

Hi all,

Our fabulous answers were delivered by Radix, Heureka, Geno3141, Alan, Asinus and MathsGod1. Thank you 

 

If you would like to comment on other site issues please do so on the Lantern Thread.  Thank you.    

 

Interest Posts:

 

1) Puzzle time                         FTJ+                                                      Thanks MathsGod1

2) Using algebra to simplify a numeric expression.    Really Cool           Thanks Heureka

3) Why is -3^2 = -9  NOT +9 ?                                                              Thanks Alan and Melody

4) Great trig, bearings and speed, question and answer                           Thanks Heureka and Melody

5) Line question - co-ordinate geometry (unusual)                                  Melody

 

                      ♫♪  ♪ ♫                                ♬ ♬ MELODY ♬ ♬                       ♫♪  ♪ ♫ 

Aug 20, 2015
 #3
avatar+118723 
+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

 

 

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

 

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

 

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

 

$${\mathtt{75}}{\mathtt{\,-\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}^{\!\!\mathtt{-1}}{\left({\mathtt{0.52}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{25}}^\circ\right)}\right)} = {\mathtt{62.304\: \!975\: \!095\: \!575}}$$

 

$${\mathtt{90}}{\mathtt{\,-\,}}{\mathtt{62.304\: \!975\: \!095\: \!575}} = {\mathtt{27.695\: \!024\: \!904\: \!425}}$$   =     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
avatar+118723 
+5
Aug 20, 2015

0 Online Users