Question 1121308
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Since the ship bears N49°E of A, angle A of the triangle must be 41°.  Likewise, based on the bearing of the ship from B, angle B of the triangle must be 29°.  Then, because the sum of the angles in any triangle is 180°, the third angle of the triangle must be 110°.


Use the Law of Sines to calculate the distance from A to the Ship, hereafter *[tex \Large x].


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\sin\,110^\circ}{76}\ =\ \frac{\sin\,29^\circ}{x}]


Construct the NS line through the Ship to create a right triangle where *[tex \Large x] is the hypotenuse and angle A is one of the acute angles.


Then the distance from the ship to the EW line that we will call *[tex \Large h] is the height of triangle ASB.  Calculate *[tex \Large h] by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h\ =\ x\sin(41^\circ)]


Once you have calculated *[tex \Large h], calculate the desired area by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A\ =\ \frac{76h}{2}\text{ km^2}]


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John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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