Question

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

Answer 1

See explanation

Explanation:

Consider the first triangle:

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

`(c)/(h)=\tan\theta_1\Rightarrow c=h\tan\theta_1`

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

`(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:`

`(d)/(h)=\tan\theta_2\Rightarrow d=h\tan\theta_2`

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

`(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

`(\sin\alpha)/(\sin\theta_1)=(\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<-h\tan \theta_2`

Now,

`h\tan\beta<h\tan\alpha`

and

`-h\tan\theta_1<-h\tan \theta_2`

so

`h\tan\beta-h\tan\theta_1<h\tan\alpha-h\tan \theta_2\\\\b<a`

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