Answer:
See explanation
Step-by-step explanation:
Consider the first triangle:
1. From the right triangle with acute angle [tex]\theta_1:[/tex]
[tex]\dfrac{c}{h}=\tan\theta_1\Rightarrow c=h\tan\theta_1[/tex]
2. From the right triangle with acute angle [tex]\alpha:[/tex]
[tex]\dfrac{c+a}{h}=\tan\alpha\Rightarrow c+a=h\tan\alpha[/tex]
Thus,
[tex]h\tan\theta_1+a=h\tan\alpha\Rightarrow a=h\tan\alpha-h\tan\theta_1[/tex]
Consider the second triangle:
1. From the right triangle with acute angle [tex]\theta_2:[/tex]
[tex]\dfrac{d}{h}=\tan\theta_2\Rightarrow d=h\tan\theta_2[/tex]
2. From the right triangle with acute angle [tex]\beta:[/tex]
[tex]\dfrac{d+b}{h}=\tan\beta\Rightarrow b+d=h\tan\beta[/tex]
Thus,
[tex]h\tan\theta_2+b=h\tan\beta\Rightarrow b=h\tan\beta-h\tan\theta_2[/tex]
Now, since [tex]\alpha>\beta,[/tex] we have [tex]\tan\alpha>\tan\beta[/tex] and [tex]\sin\alpha>\sin\beta.[/tex]
If
[tex]\dfrac{\sin\alpha}{\sin\theta_1}=\dfrac{\sin\beta}{\sin\theta_2}[/tex]
and
[tex]\sin\alpha>\sin\beta,[/tex]
then
[tex]\sin\theta_1>\sin\theta_2.[/tex]
This means [tex]\theta_1>\theta_2[/tex] and [tex]\tan\theta_1>\tan \theta_2.[/tex]
Hence,
[tex]h\tan\theta_1>h\tan \theta_2\\ \\-h\tan\theta_1<-h\tan \theta_2[/tex]
Now,
[tex]h\tan\beta<h\tan\alpha[/tex]
and
[tex]-h\tan\theta_1<-h\tan \theta_2[/tex]
so
[tex]h\tan\beta-h\tan\theta_1<h\tan\alpha-h\tan \theta_2\\\\b<a[/tex]
Note that this solution is true only for acute angles [tex]\alpha,\ \beta,\ \theta_1,\ \theta_2[/tex]
