PLASE HLP In a large harbour there are many docks, labeled alphabetically, which are built side by side in a straight line. Dock A and Dock K are located 2.3 km apart along this line. There is a Dock L on an island outside the harbour such that is 45° and is 64°. How far is the island dock away from each of the other two docks?

distance island dock to Dock A = 4.99 km

distance island dock to Dock K = 6.35 km

Step-by-step explanation:

Always make a scetch to visualize the situation.

You need to construct two triangle both with a streight angle, so you can use Pythagoras to calculate the unknown distances between the island dock L, and each of the other two docks A an K.

I chose to introduce an extra letter, the letter C. In total you have the letters A K L and the letter C.

The letter C has a streight angle of 90° between ACL and it has the same streight angle of 90° with KCL. It is crucial that you see that the distance of LC is exactly the same in triangle LAC and that LC has exactly the same distance in the other triangleLKC.

The distance between AK = 2.3 km.

I define the distance between K and point C as 2.3 + x, because the distance x is unknown.

KC = 2.3 + x

Further more, when you make a picture, you can see that the distance between A and point C = x.

From such a picture, it would show clearly, that K is further away in respect to L then point A. From the picture it would be clear that the angle of LKC is smaller then the angle of LAC, so LKC = 45° and LAC = 64°.

Because angle LKC = 45° and we choose C to have an angle of 90°, the TRIANGLE LKC must be a special triangle... In any triangle, the sum of the three angles together, must add up to 180° .

If that is true, then we have 45 + 90 + 45 (because that adds up to 180). Now that means triangle LKC must have two equal sides (because of the same angels of 45° ).

So we know the distance KC = LC and we already defined KC = 2.3 + x.

Now we know enough to solve the problem.

AK = 2.3 km

angle of LKC = 45°

angle of LAC = 64°

AC = x

KC = 2.3 + x

LC = KC

LC = 2.3 + x

Try to calculate the distance x by using tan. After that you can use Pythagoras to find the other distances.

tan(LKC) = ( LC ) / ( KC )

tan(LKC) = ( x+2.3 ) / ( x+2.3 )

That is not helpful. Let's try the other triangle...

tan(LAC) = LC / AC

tan(LAC) = ( x+2.3 ) / x

tan(64) = ( x+2.3 ) / x

Solve the equation which means you try to find the value for x.

x * tan(64) = ( x+2.3 )

tan(64) * x -x = 2.3

tan(64) * x - 1* x = 2.3

Try to get x outside of the braquets...

x* ( tan(64) - 1 ) = 2.3

x* (2.0503038415793 - 1 ) = 2.3

1.0503038415793 * x = 2.3

x = 2.3 / 1.0503038415793

x = 2.19

Now use Pythagoras a² + b² = c² in triangle LAC to find distance LA.

LA² = AC² + LC²

AC = x = 2.19

LC = 2.3 + x = 4.39

LA² = 2.19² + 4.39²

LA = SQRT( 4.79 + 20.16 )

LA = SQRT( 24.95 )

LA = 4.99 km

Now use Pythagoras a² + b² = c² in triangle LKC to find distance LK.

LK² = KC² + LC²

KC = 2.3 + x = 4.39

LC = 2.3 + x = 4.39

LK² = 4.39² + 4.39²

LK = SQRT( 20.16 + 20.16 )

LK = SQRT( 40.32 )

LK = 6.35 km

step-by-step explanation:

a rhombus is a parallelogram enjoying his symmetry.