CPhill

Two ships leave a Port O. One ship travels on a bearing of 3400 to a point P which is 50 km from O. The other ship travels on a bearing of 0600 to a point Q, 85 km from O. (a) Draw a diagram to represent the position of the port and the two ships. On your diagram, carefully label North, the given angles and the distances travelled. (b) Calculate the distance PQ in km (c) Determine the bearing of P from Q.

I'm assuming that 3400 = 340.0    and that 0600 = 60.0

Then the angle between them is just 80°

We can use the Law of Cosines to find PQ =

PQ^2  = 85^2 + 50^2 - 2(85)(50)cos(80)

PQ^2 = 8248.9904898305

PQ = about 90.824 km

And the bearing from P to Q is given by [90 + sin-1(4.485/90.824)]° = [90 + 2.83]° = 92.83°

Here's a diagram.....("North" is at the "top' )

2x=1.5(x+6)   distribute the 1.5 across the terms in the parentheses

2x = 1.5x + 9  subtract 1.5x from  both sides

.5x  = 9       note that  .5= 5/10  = 1/2

(1/2) x = 9      multiply both sides by 2

x = 18

And that's it !!!

No prob....

Ignore the signs....take the "bigger' thing  and subtract the "smaller" thing....so we have

27.4 - 15.8  =  11.6

Now..ask yourself the question....what was the sign on the "bigger" thing.....a "negative," right??

Then...that's the sign we use for our answer....!!!  ( -11.6 )

This method always works when we have numbers with different signs......

Let x be the age of the daughter.....then 4x is the age of the mother

And in 12 years, we will have

(4x + 12) = 2(x + 12)  simplify

4x + 12 = 2x + 24    rearrange

2x = 12   divide by 2

x = 6

So...her mother is 4x = 4(6) = 24

8b + 2x = 6x - 4a    subtract 2x from both sides.....add 4a to both sides

8b + 4a = 4x          divide everything by 4

2b + a  = x

We need to see if the sum of the squares of the two shorter sides  =  the square of the longest side

20^2 + 21^2 = 29^2  ???

400 + 441 = 841

And that's true.....so....we have a right triangle....

I've done it with Calculus, before......but...since "Mathematician" hasn't had trig......I doubt that Calculus would do him much good, either......LOL!!!

Just ignore this junk, jillybean.....it's NEVER going away....believe me !!!

Note in the pic below that FH = (1/2)s    where s is the side length

And also that DA =  AF = r = 4.5

So we have, by the Pythagorean Theorem

AF^2 =  (1/2)s^2 + AH^2

4.5^2 = 20.25 = (1/2)s^2 + AH^2

AH^2 = 20.25 - (s/2)^2

AH = √(20.25 - (s/2)^2 )

Therefore, again using the Pythagorean Theorem, we have

s^2 = (s/2)^2 + [DA + AH]^2

s^2 = (s/2)^2 + ((4.5 + √(20.25 - (s/2)^2 ))^2

(3/4)s^2 = 20.25 + 9 √(20.25 - (s/2)^2 ) + 20.25 - (s/4)^2   (add (s/4)^2 to each side)

s^2 = 40.5 + 9 √(20.25 - (s/2)^2 )  subtract 40.5 from each side

s^2 - 40.5  = 9 √(20.25 - (s/2)^2 )   square both sides

(s^2 - 40.5)^2 = 81 (20.25 - (s^2)/4)

s^4 - 81s^2 + 1640.25 = 1640.25 - (81/4)s^2

s^4 - 81s^2 + (81/4)s^2 = 0

s^4 - (243/4)s^2 = 0

Factor

(s^2) (s^2 - (243/4)) =0   one solution is s = 0 .... reject this

So we have

(s^2 - 243/4) = 0

s = (1/2)√243 = (1/2)(9)√3 = 4.5√3 = 7.7942286340599478

Here's a pic....(the angles are not exact..it's just to give you a feel for the geometry of the problem)