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julian

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I'm trying to relate different ways of getting a value for the "width" of a non-parallelogram. The non-parallelogram is given in figure 1: note that the left-hand side edge is vertical, the right-hand side is tilted 4 degrees away from vertical, and the bottom edge is 24 degrees below the horizontal.

(i) One way of obtaining a value for the "width" is measuring the distance between the two sides midway up while having the ruler parallel to the bottom edge. This is represented by the solid line with arrow in figure 2 from point ##P## to point ##P'##. We label the value this distance by ##w_A##.

(ii) Another way would be to measure the distance but having the ruler horizontal (that is, have the ruler at right-angles to the left-hand side edge). This is represented by the dashed line with arrow in figure 2 from point ##P## to point ##P''##. We label this distance by ##w_B##.

(iii) Another way would be to find the least distance from the point P to the right-hand side edge. We label this distance by ##w_C##.

Note that for a perfect rectangle we would have ##w_A = w_B = w_C##!

Figure 3 is a close up of the triangle ##PP'P''## from figure 2. Obviously this is a non-right-angled triangle but can be split into two right angled triangles. From this I can relate ##w_B## to ##w_A##:

##

w_B = w_A \cos 24^0 + w_A \sin 24^0 \; \tan 4^0

##

(see figure 3 for calculation). Is that correct?

In order to address part (iii) I want to find a formula for the distance from the point ##P## to the point ##Q## as a function of ##\phi## as indicated in figure 4. Wondering if I could use the Law of Sines on the triangle ##PQP''## in figure 4 to do this and how to go about it? I could then use simple calculus to find minimum or/and plot the function on wolfram given ##w_A = 5.1##.

(i) One way of obtaining a value for the "width" is measuring the distance between the two sides midway up while having the ruler parallel to the bottom edge. This is represented by the solid line with arrow in figure 2 from point ##P## to point ##P'##. We label the value this distance by ##w_A##.

(ii) Another way would be to measure the distance but having the ruler horizontal (that is, have the ruler at right-angles to the left-hand side edge). This is represented by the dashed line with arrow in figure 2 from point ##P## to point ##P''##. We label this distance by ##w_B##.

(iii) Another way would be to find the least distance from the point P to the right-hand side edge. We label this distance by ##w_C##.

Note that for a perfect rectangle we would have ##w_A = w_B = w_C##!

Figure 3 is a close up of the triangle ##PP'P''## from figure 2. Obviously this is a non-right-angled triangle but can be split into two right angled triangles. From this I can relate ##w_B## to ##w_A##:

##

w_B = w_A \cos 24^0 + w_A \sin 24^0 \; \tan 4^0

##

(see figure 3 for calculation). Is that correct?

In order to address part (iii) I want to find a formula for the distance from the point ##P## to the point ##Q## as a function of ##\phi## as indicated in figure 4. Wondering if I could use the Law of Sines on the triangle ##PQP''## in figure 4 to do this and how to go about it? I could then use simple calculus to find minimum or/and plot the function on wolfram given ##w_A = 5.1##.

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