Measuring Diagonals (part 2)

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Check out part 1 here: Measuring diaognals without Sine Law - this is an equivalent construct.

Suppose we construct a triangle with lengths $5,7,\sqrt{32}$ on the cartesian plane. Using Heron's Area formula, we can verify A = $(0,4)$ (ie: height = 4).

Now we rotate triangle around $(4,0)$ so that one corner touches the Y-axis (ie: second figure).

Question: what is the new position of A$(x,y)$? The top figure is easily solved, but the 2nd figure requires quadratic sums with two possible answers. (see Part 1 link above).

note: law of Cosines is the generalization of Pythagoras Theorem. However, we aren't using this rule yet.

Triangle with rotation

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Well, I don't even see why trig has to be involved in this question. If you want to find the coordinates, use the distance formula:

between $(0,\sqrt{33})$ and $(x,y)$:

$\sqrt{x^2 + (y-\sqrt{33})^2} = 25$

between $(x,y)$ and $(4,0)$

$\sqrt{(x-4)^2 + y^2} = 32$

solve for x and y.