Question 1208283: A ship sailing from a port P travels 50Km in a direction 070° to reach a port T. The ship then sails from T in a direction bearing 130° and a distance of 80km to a point Q. How far will the ship sail if it travels directly from P to Q?
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52803)   (Show Source):
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
Exact Answer = 10*sqrt(129) kilometers
Approximate Answer = 113.5782 kilometers
Ask your teacher how s/he wants you to round the approximate value.
----------------------------------------------------------------------------------------------------------
Explanation
The compass bearing has 000° pointing directly north.
As you turn eastward, i.e. rotate clockwise, the angle increases.
045° points to the northeast
090° points east
135° points southeast
And so on.
With that in mind, here is what the diagram looks like
Angle APT = 70° and Angle BTQ = 130° are given
Points A and B are directly north of P and T respectively.
Angle BTP = 110° is found by solving the equation angleAPT+angleBTP = 180. Note how vertical segments AP and BT are parallel, which means the consecutive interior angles are supplementary.
Angle PTQ = 120° is determined by noting that the three angles around point T must add to 360 (you'll solve this equation: anglePTB+angleBTQ+anglePTQ = 360)
Focus on triangle PTQ.
To find x, the length of segment PQ, we can use the Law of Cosines.
c^2 = a^2 + b^2 - 2*a*b*cos(C)
x^2 = 50^2 + 80^2 - 2*50*80*cos(120)
x^2 = 50^2 + 80^2 - 2*50*80*(-1/2)
x^2 = 12900
x = sqrt(12900)
x = sqrt(100*129)
x = sqrt(100)*sqrt(129)
x = 10*sqrt(129) which is the exact distance
x = 113.578166916005
x = 113.5782 kilometers which is the approximate distance
Make sure that your calculator is set to degrees mode.
Ask your teacher how s/he wants you to round this approximate value.
--------------------------------------------------------------------------
Another Approach.
Place P at the origin (0,0)
This is how we'll locate point T.
T = P + 50*( sin(70), cos(70) )
T = (0,0) + 50*( 0.9396926, 0.3420201 )
T = ( 50*0.9396926, 50*0.3420201 )
T = ( 46.98463, 17.101005 )
The decimal values are approximate.
Make sure that your calculator is set to degrees mode.
Normally cosine is associated with the x coordinate; however, the compass bearing angles have 000° pointing north (rather than east), so we have a 90° rotation. This 90° rotation swaps the roles of sine and cosine. Cosine is a 90° phase-shifted version of sine.
We'll follow a similar method to find where point Q is located.
Q = T + 80*( sin(130), cos(130) )
Q = ( 46.98463, 17.101005 ) + 80*( 0.7660444, -0.6427876 )
Q = ( 46.98463, 17.101005 ) + ( 80*0.7660444, 80*(-0.6427876) )
Q = ( 46.98463, 17.101005 ) + ( 61.283552, -51.423008 )
Q = ( 46.98463+61.283552, 17.101005+(-51.423008) )
Q = ( 108.268182, -34.322003 )
The decimal values are approximate.
Here is the calculation template for point Q's coordinates in one single line
Q = ( 50*sin(70)+80*sin(130), 50*cos(70)+80*cos(130) )
That line is based off of this
Q = P + 50*( sin(70), cos(70) ) + 80*( sin(130), cos(130) )
The key takeaways are these locations
P = (0,0)
Q = ( 108.268182, -34.322003 ) which is approximate
Use the distance formula to find that
PQ = sqrt( (108.268182)^2 + (-34.322003)^2 ) = 113.5782 kilometers approximately
The answer will vary depending on the rounding precision.
More practice with similar questions
https://www.algebra.com/algebra/homework/Trigonometry-basics/Trigonometry-basics.faq.question.1200155.html
and
https://www.algebra.com/algebra/homework/Triangles/Triangles.faq.question.1182009.html
|
|
|
