Question

Could you make proof of "a > b" please ?

Answer 1

Answer:

See explanation

Step-by-step explanation:

Consider the first triangle:

1. From the right triangle with acute angle $\theta_1:$

$\dfrac{c}{h}=\tan\theta_1\Rightarrow c=h\tan\theta_1$

2. From the right triangle with acute angle $\alpha:$

$\dfrac{c+a}{h}=\tan\alpha\Rightarrow c+a=h\tan\alpha$

Thus,

$h\tan\theta_1+a=h\tan\alpha\Rightarrow a=h\tan\alpha-h\tan\theta_1$

Consider the second triangle:

1. From the right triangle with acute angle $\theta_2:$

$\dfrac{d}{h}=\tan\theta_2\Rightarrow d=h\tan\theta_2$

2. From the right triangle with acute angle $\beta:$

$\dfrac{d+b}{h}=\tan\beta\Rightarrow b+d=h\tan\beta$

Thus,

$h\tan\theta_2+b=h\tan\beta\Rightarrow b=h\tan\beta-h\tan\theta_2$

Now, since $\alpha>\beta,$ we have $\tan\alpha>\tan\beta$ and $\sin\alpha>\sin\beta.$

If

$\dfrac{\sin\alpha}{\sin\theta_1}=\dfrac{\sin\beta}{\sin\theta_2}$

and

$\sin\alpha>\sin\beta,$

then

$\sin\theta_1>\sin\theta_2.$

This means $\theta_1>\theta_2$ and  $\tan\theta_1>\tan \theta_2.$

Hence,

$h\tan\theta_1>h\tan \theta_2\\ \\-h\tan\theta_1$

Now,

$h\tan\beta$

and

$-h\tan\theta_1$

so

$h\tan\beta-h\tan\theta_1$

Note that this solution is true only for acute angles $\alpha,\ \beta,\ \theta_1,\ \theta_2$