Question

98 POINTS MATH!

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 $0^\circ$, then we would have $\cos\theta>\cos\varphi$.

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

$(4y-5)^\circ$

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

$(4y-5)^\circ>0^\circ\implies y>\dfrac54^\circ$

So we must have

$\dfrac54^\circ$