Zoagliaj

Answered

2022-07-20

I want to find out the angle for the expression ${a}^{3}+{b}^{3}={c}^{3}$.

Answer & Explanation

kamphundg4

Expert

2022-07-21Added 20 answers

Step 1

There is no single angle corresponding to the relationship ${a}^{3}+{b}^{3}={c}^{3}$.

Suppose that a triangle has sides of lengths a,b, and c such that ${a}^{3}+{b}^{3}={c}^{3}$. We know from the law of cosines that if θ is the angle opposite the side of length c, then ${c}^{2}={a}^{2}+{b}^{2}-2ab\mathrm{cos}\theta $, so $\mathrm{cos}\theta =\frac{{a}^{2}+{b}^{2}-{c}^{2}}{2ab}\phantom{\rule{thickmathspace}{0ex}}.$.

Now let’s look at just a few examples. If $a=b=1$, then $c=\sqrt[3]{2}$, and $\mathrm{cos}\theta =\frac{2-{2}^{2/3}}{2}\approx 0.20630\phantom{\rule{thickmathspace}{0ex}}.$.

Step 2

If $a=1$ and $b=2$, then $c=\sqrt[3]{9}$, and $\mathrm{cos}\theta =\frac{5-{9}^{2/3}}{4}\approx 0.16831\phantom{\rule{thickmathspace}{0ex}}.$.

If $a=1$ and $b=3$, then $c=\sqrt[3]{28}$, and $\mathrm{cos}\theta =\frac{10-{28}^{2/3}}{6}\approx 0.12985\phantom{\rule{thickmathspace}{0ex}}.$.

As you can see, these values of $\mathrm{cos}\theta $ are all different, so the angles themselves are also different.

There is no single angle corresponding to the relationship ${a}^{3}+{b}^{3}={c}^{3}$.

Suppose that a triangle has sides of lengths a,b, and c such that ${a}^{3}+{b}^{3}={c}^{3}$. We know from the law of cosines that if θ is the angle opposite the side of length c, then ${c}^{2}={a}^{2}+{b}^{2}-2ab\mathrm{cos}\theta $, so $\mathrm{cos}\theta =\frac{{a}^{2}+{b}^{2}-{c}^{2}}{2ab}\phantom{\rule{thickmathspace}{0ex}}.$.

Now let’s look at just a few examples. If $a=b=1$, then $c=\sqrt[3]{2}$, and $\mathrm{cos}\theta =\frac{2-{2}^{2/3}}{2}\approx 0.20630\phantom{\rule{thickmathspace}{0ex}}.$.

Step 2

If $a=1$ and $b=2$, then $c=\sqrt[3]{9}$, and $\mathrm{cos}\theta =\frac{5-{9}^{2/3}}{4}\approx 0.16831\phantom{\rule{thickmathspace}{0ex}}.$.

If $a=1$ and $b=3$, then $c=\sqrt[3]{28}$, and $\mathrm{cos}\theta =\frac{10-{28}^{2/3}}{6}\approx 0.12985\phantom{\rule{thickmathspace}{0ex}}.$.

As you can see, these values of $\mathrm{cos}\theta $ are all different, so the angles themselves are also different.

suchonosdy

Expert

2022-07-22Added 3 answers

Step 1

You do get slightly different answers for the Pythagorean Theorem on the surface of a sphere of radius 1, or the hyperbolic plane of curvature -1. On the sphere, a right triangle with geodesic lengths of legs a,b and hypotenuse c obeys $\mathrm{cos}c=\mathrm{cos}a\phantom{\rule{thickmathspace}{0ex}}\mathrm{cos}b,$, while in the hyperbolic plane with curvature -1 it becomes $\mathrm{cosh}c=\mathrm{cosh}a\phantom{\rule{thickmathspace}{0ex}}\mathrm{cosh}b.$.

Step 2

In both cases, if you write out the functions as power series in a, b, c, you see that the limit as a, b, c all shrink to nearly 0 is the traditional Pythagorean Theorem.

You do get slightly different answers for the Pythagorean Theorem on the surface of a sphere of radius 1, or the hyperbolic plane of curvature -1. On the sphere, a right triangle with geodesic lengths of legs a,b and hypotenuse c obeys $\mathrm{cos}c=\mathrm{cos}a\phantom{\rule{thickmathspace}{0ex}}\mathrm{cos}b,$, while in the hyperbolic plane with curvature -1 it becomes $\mathrm{cosh}c=\mathrm{cosh}a\phantom{\rule{thickmathspace}{0ex}}\mathrm{cosh}b.$.

Step 2

In both cases, if you write out the functions as power series in a, b, c, you see that the limit as a, b, c all shrink to nearly 0 is the traditional Pythagorean Theorem.

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