the bottom or base edge b,

the top left edge a,

the top right edge c.

Let a=c+2 and b=10.

the top left edge a,

the top right edge c.

Let a=c+2 and b=10.

Jameson Lucero
2022-07-12
Answered

the bottom or base edge b,

the top left edge a,

the top right edge c.

Let a=c+2 and b=10.

the top left edge a,

the top right edge c.

Let a=c+2 and b=10.

You can still ask an expert for help

Isla Klein

Answered 2022-07-13
Author has **12** answers

Taking the origin as the intersection of a and b, with the equation in terms of the angle $\theta $ at that vertex:

Through the cosine law we have:

$(a-2{)}^{2}={a}^{2}+100-20a\mathrm{cos}\theta $

$-4a+4=100-20a\mathrm{cos}\theta $

$a=\frac{96}{20\mathrm{cos}\theta -4}$

With our particular setup, a is the radius. So this is a polar equation, where

$r(\theta )=\frac{96}{20\mathrm{cos}\theta -4}=\frac{24}{5\mathrm{cos}\theta -1}$

we can convert to cartesian coordinates if you want:

$5r\mathrm{cos}\theta -r=24$

$(5r\mathrm{cos}\theta -24{)}^{2}={r}^{2}$

$(5x-24{)}^{2}={x}^{2}+{y}^{2}$

${y}^{2}=24{x}^{2}-240x+576$

${y}^{2}=24(x-5{)}^{2}-24$

$(x-5{)}^{2}-\frac{{y}^{2}}{24}=1$

which is a hyperbola. You want the rightmost branch (so $a>0$), which is valid when $\mathrm{cos}\theta <\frac{1}{5}$

Through the cosine law we have:

$(a-2{)}^{2}={a}^{2}+100-20a\mathrm{cos}\theta $

$-4a+4=100-20a\mathrm{cos}\theta $

$a=\frac{96}{20\mathrm{cos}\theta -4}$

With our particular setup, a is the radius. So this is a polar equation, where

$r(\theta )=\frac{96}{20\mathrm{cos}\theta -4}=\frac{24}{5\mathrm{cos}\theta -1}$

we can convert to cartesian coordinates if you want:

$5r\mathrm{cos}\theta -r=24$

$(5r\mathrm{cos}\theta -24{)}^{2}={r}^{2}$

$(5x-24{)}^{2}={x}^{2}+{y}^{2}$

${y}^{2}=24{x}^{2}-240x+576$

${y}^{2}=24(x-5{)}^{2}-24$

$(x-5{)}^{2}-\frac{{y}^{2}}{24}=1$

which is a hyperbola. You want the rightmost branch (so $a>0$), which is valid when $\mathrm{cos}\theta <\frac{1}{5}$

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