Question

98 POINTS MATH!

. Find the range of possible values for the variable.

Please show work

Answer 1

Let
`a` be the length of the leg with one tick mark and
`b` the length of the leg with two tick marks.

In the upper triangle, the law of cosines says

`6^2=a^2+b^2-2ab\cos43^\circ`

In the lower triangle, it says

`4^2=a^2+b^2-2ab\cos(4y-5)^\circ`

Subtract the second equation from the first to eliminate
`a^2+b^2`:

`6^2-4^2=-2ab(\cos43^\circ-\cos(4y-5)^\circ)`

`10=ab(\cos(4y-5)^\circ-\cos43^\circ)`

`a` and
`b` are lengths so they must both be positive. 10 is also positive, so in order to preserve the sign on both sides of this equation, we must have

`\cos(4y-5)^\circ-\cos43^\circ>0`

`\cos(4y-5)^\circ>\cos43^\circ`

Now we have to be a bit careful. If
`x` is an acute angle, then as
`x` gets larger, the value of
`\cos x` gets smaller. So if we have two angles
`\theta` and
`\varphi`, with
`0^\circ<\theta<\varphi<90^\circ`, then we would have
`\cos\theta>\cos\varphi`.

This means in our inequality, taking the inverse cosine of both sides would reverse the inequality:

`(4y-5)^\circ<43^\circ\implies y<12^\circ`

We know that
`(4y-5)^\circ` is an angle in a triangle, so it must be some positive measure:

`(4y-5)^\circ>0^\circ\implies y>\frac54^\circ`

So we must have

`\frac54^\circ<y<12^\circ`