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2. An aeroplane flying horizontally 1 km above the ground is observed at an elevation of 60 degrees. After 10 seconds , its elevation is observed to be 30 degees. Find the speed of the aeroplane in km/hr.

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- #1

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2. An aeroplane flying horizontally 1 km above the ground is observed at an elevation of 60 degrees. After 10 seconds , its elevation is observed to be 30 degees. Find the speed of the aeroplane in km/hr.

explain also

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You are triangulating distances, so you need a function that returns a distance.

Building B is the location of fire p, a distance p'Q and p'P in a straight line (the road the trucks need to use). You need to triangulate the three points on the surface first with something like p'Q = 60(pQ) for a radial measure of the location of firestation Q, and similarly for P.

Then p and p' are measuring functions, and you use the overhead projection onto BPQ on the ground (Cartesian) plane after doing the same with that. You don't "need" pi but it's a handy symbol in circular functions. 60deg is pi/3

IOW 20k is the distance function value for PQ which is the hypotenuse of right triangle BPQ

Building B is the location of fire p, a distance p'Q and p'P in a straight line (the road the trucks need to use). You need to triangulate the three points on the surface first with something like p'Q = 60(pQ) for a radial measure of the location of firestation Q, and similarly for P.

Then p and p' are measuring functions, and you use the overhead projection onto BPQ on the ground (Cartesian) plane after doing the same with that. You don't "need" pi but it's a handy symbol in circular functions. 60deg is pi/3

IOW 20k is the distance function value for PQ which is the hypotenuse of right triangle BPQ

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Tell me how to solve both of them and the answers.

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Use the http://en.wikipedia.org/wiki/Law_of_sines" [Broken]. The law of sines for a triangle is just the relation that the ratio of the sine of an angle by the length of the opposite side of the triangle is a constant for each triangle. With this law you can calculate the size of both sides of the triangle, or quicker you can just compare the two sides in question to see which one is bigger.

Let PB be the distance from fire station P to building B. Let QB be the distance from Q to B, then

[tex]\frac{PB}{sin(60\,degrees)}=\frac{QB}{sin(45\,degrees)}[/tex]

[tex]\frac{2PB}{\sqrt{3}}=\frac{2QB}{\sqrt{2}}\Rightarrow PB=QB\frac{\sqrt{3}}{\sqrt{2}}>QB[/tex]

which is what is to be expected from the angles.

The second problem is done in a similar manner. Calculate the distance it has moved, and just divide by the time to get the average speed.

Let PB be the distance from fire station P to building B. Let QB be the distance from Q to B, then

[tex]\frac{PB}{sin(60\,degrees)}=\frac{QB}{sin(45\,degrees)}[/tex]

[tex]\frac{2PB}{\sqrt{3}}=\frac{2QB}{\sqrt{2}}\Rightarrow PB=QB\frac{\sqrt{3}}{\sqrt{2}}>QB[/tex]

which is what is to be expected from the angles.

The second problem is done in a similar manner. Calculate the distance it has moved, and just divide by the time to get the average speed.

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