The locations, given in polar coordinates, for two ships are (8 mi, 63º) and (8 mi, 123º). Find the distance between the two shi
2 answers:
We know that
The arrangement forms an isosceles triangle with equal legs of 8 miles.
The angle between the legs is equal to
°
Therefore, the other two angles are
Angles = (180-60)/2 = 120/2 = 60°
It can, therefore, be noted that all angles are equal and thus the resulting triangle is actually an equilateral triangle and thus all the sides are equal.
Hence
the answer is
the distance between the two ships is 8 miles apart
alternative Method
Applying the law of cosines
<span>c²=a²+b²-2*a*b*cos C
</span>where
a=8 miles
b=8 miles
C is the angle between the legs-------> 123-63------> 60 degrees
c is the distance between the two ships
so
c²=8²+8²-2*8*8*cos 60------> c²=64-------> c=√64------> c=8 miles
The arrangement forms an isosceles triangle with equal legs of 8 miles.
The angle between the legs is equal to
Therefore, the other two angles are
Angles = (180-60)/2 = 120/2 = 60°
It can, therefore, be noted that all angles are equal and thus the resulting triangle is actually an equilateral triangle and thus all the sides are equal.
Hence
the answer is
the distance between the two ships is 8 miles apart
alternative Method
Applying the law of cosines
<span>c²=a²+b²-2*a*b*cos C
</span>where
a=8 miles
b=8 miles
C is the angle between the legs-------> 123-63------> 60 degrees
c is the distance between the two ships
so
c²=8²+8²-2*8*8*cos 60------> c²=64-------> c=√64------> c=8 miles
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so the perimeter of the polygon is the sum of all its sides, namely, AB + BC + CD + DA.
now, let's check how long each side is,
![\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\ \quad \\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &A&(~{{ -6}} &,&{{ -4}}~) % (c,d) &B&(~{{ -3}} &,&{{ 6}}~) \end{array} \\\\\\ d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}\\\\ -------------------------------\\\\ AB=\sqrt{[-3-(-6)]^2+[6-(-4)]^2} \\\\\\ AB=\sqrt{(-3+6)^2+(6+4)^2} \\\\\\ AB=\sqrt{3^2+10^2}\implies \boxed{AB=\sqrt{109}}\\\\ -------------------------------](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%5C%5C%20%5Cquad%20%5C%5C%0A%5Cbegin%7Barray%7D%7Bccccccccc%7D%0A%26%26x_1%26%26y_1%26%26x_2%26%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0A%26A%26%28~%7B%7B%20-6%7D%7D%20%26%2C%26%7B%7B%20-4%7D%7D~%29%20%0A%25%20%20%28c%2Cd%29%0A%26B%26%28~%7B%7B%20-3%7D%7D%20%26%2C%26%7B%7B%206%7D%7D~%29%0A%5Cend%7Barray%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%20%3D%20%5Csqrt%7B%28%7B%7B%20x_2%7D%7D-%7B%7B%20x_1%7D%7D%29%5E2%20%2B%20%28%7B%7B%20y_2%7D%7D-%7B%7B%20y_1%7D%7D%29%5E2%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0AAB%3D%5Csqrt%7B%5B-3-%28-6%29%5D%5E2%2B%5B6-%28-4%29%5D%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0AAB%3D%5Csqrt%7B%28-3%2B6%29%5E2%2B%286%2B4%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0AAB%3D%5Csqrt%7B3%5E2%2B10%5E2%7D%5Cimplies%20%5Cboxed%7BAB%3D%5Csqrt%7B109%7D%7D%5C%5C%5C%5C%0A-------------------------------)
![\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\ \quad \\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &B&(~{{ -3}} &,&{{6}}~) % (c,d) &C&(~{{ 4}} &,&{{ 0}}~) \end{array} \\\\ -------------------------------\\\\ BC=\sqrt{[4-(-3)]^2+[0-6]^2}\implies BC=\sqrt{(4+3)^2+(0-6)^2} \\\\\\ BC=\sqrt{7^2+(-6)^2}\implies \boxed{BC=\sqrt{85}}\\\\ -------------------------------](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%5C%5C%20%5Cquad%20%5C%5C%0A%5Cbegin%7Barray%7D%7Bccccccccc%7D%0A%26%26x_1%26%26y_1%26%26x_2%26%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0A%26B%26%28~%7B%7B%20-3%7D%7D%20%26%2C%26%7B%7B6%7D%7D~%29%20%0A%25%20%20%28c%2Cd%29%0A%26C%26%28~%7B%7B%204%7D%7D%20%26%2C%26%7B%7B%200%7D%7D~%29%0A%5Cend%7Barray%7D%0A%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0ABC%3D%5Csqrt%7B%5B4-%28-3%29%5D%5E2%2B%5B0-6%5D%5E2%7D%5Cimplies%20BC%3D%5Csqrt%7B%284%2B3%29%5E2%2B%280-6%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0ABC%3D%5Csqrt%7B7%5E2%2B%28-6%29%5E2%7D%5Cimplies%20%5Cboxed%7BBC%3D%5Csqrt%7B85%7D%7D%5C%5C%5C%5C%0A-------------------------------)
![\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\ \quad \\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &D(~{{ 2}} &,&{{-1}}~) % (c,d) &A&(~{{ -6}} &,&{{ -4}}~) \end{array}\\\\ -------------------------------\\\\ DA=\sqrt{[-6-2]^2+[-4-(-1)]^2}\\\\\\ DA=\sqrt{(-6-2)^2+(-4+1)^2} \\\\\\ DA=\sqrt{(-8)^2+(-3)^2}\implies \boxed{DA=\sqrt{73}}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%5C%5C%20%5Cquad%20%5C%5C%0A%5Cbegin%7Barray%7D%7Bccccccccc%7D%0A%26%26x_1%26%26y_1%26%26x_2%26%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0A%26D%28~%7B%7B%202%7D%7D%20%26%2C%26%7B%7B-1%7D%7D~%29%20%0A%25%20%20%28c%2Cd%29%0A%26A%26%28~%7B%7B%20-6%7D%7D%20%26%2C%26%7B%7B%20-4%7D%7D~%29%0A%5Cend%7Barray%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0ADA%3D%5Csqrt%7B%5B-6-2%5D%5E2%2B%5B-4-%28-1%29%5D%5E2%7D%5C%5C%5C%5C%5C%5C%20DA%3D%5Csqrt%7B%28-6-2%29%5E2%2B%28-4%2B1%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0ADA%3D%5Csqrt%7B%28-8%29%5E2%2B%28-3%29%5E2%7D%5Cimplies%20%5Cboxed%7BDA%3D%5Csqrt%7B73%7D%7D)
sum those sides up, and that's the perimeter of the polygon.
so the perimeter of the polygon is the sum of all its sides, namely, AB + BC + CD + DA.
now, let's check how long each side is,
sum those sides up, and that's the perimeter of the polygon.
Answer:
Step-by-step explanation:
This is a right triangle trig problem. The base of the right triangle is the distance that Donna if from the flagpole; the flagpole is the side opposite the reference angle which was given as 26, and we are looking for the height of the flagpole, h. The trig ratio that uses the side opposite over the side adjacent is the tangent ratio, specifically:
and
123tan(26) = h so
h = 60.0 rounded to the nearest tenth. But that is only the height from her line of vision and up, not the whole height. In order to find the whole height, we have to add in her height up to her line of vision which is 5.3 feet. Therefore, the height of the flagpole is
60.0 + 5.3 = 65.3 feet.