Answer:
See below.
Step-by-step explanation:
To use the given diagram to show that h = (d tan α tan β)/(tan β - tan α), begin by using the tangent trigonometric ratio to express YW in terms of h and tan β.
[tex]\boxed{\begin{array}{l}\underline{\textsf{Tangent trigonometric ratio}}\\\\\sf \tan(\theta)=\dfrac{O}{A}\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$\theta$ is the angle.}\\\phantom{ww}\bullet\;\textsf{$O$ is the side opposite the angle.}\\\phantom{ww}\bullet\;\textsf{$A$ is the side adjacent the angle.}\end{array}}[/tex]
In right triangle TYW, the side opposite angle β is TW and the side adjacent angle β is YW. Therefore:
[tex]\tan \beta = \dfrac{TW}{YW}\\\\\\\tan \beta =\dfrac{h}{YW}\\\\\\YW=\dfrac{h}{\tan \beta}[/tex]
Now use the tangent trigonometric ratio again to express h in terms of tan α, d and YW.
In right triangle TWX, the side opposite angle α is TW and the side adjacent angle α is XW. Side XW can be expressed as d + YW. Therefore:
[tex]\tan \alpha = \dfrac{TW}{XW}\\\\\\\tan \alpha=\dfrac{h}{d+YW}\\\\\\h=\tan\alpha(d+YW)[/tex]
Now, substitute the expression for YW into the equation for h:
[tex]h=\tan\alpha \left(d+\dfrac{h}{\tan \beta}\right)[/tex]
Rearrange to isolate h:
[tex]h= d\tan\alpha+\dfrac{h\tan\alpha}{\tan \beta} \\\\\\h-\dfrac{h\tan\alpha}{\tan \beta}=d\tan\alpha \\\\\\ \dfrac{h\tan\beta}{\tan\beta}-\dfrac{h\tan\alpha}{\tan \beta}=d\tan\alpha \\\\\\ \dfrac{h\tan\beta-h\tan\alpha}{\tan \beta}=d\tan\alpha \\\\\\ h\tan\beta-h\tan\alpha=d\tan\alpha\tan \beta \\\\\\ h(\tan\beta-\tan\alpha)=d\tan\alpha\tan \beta \\\\\\ h=\dfrac{d\tan\alpha\tan \beta}{\tan\beta-\tan\alpha}[/tex]
Therefore, we have used the given diagram to show that:
[tex]h=\dfrac{d\tan\alpha\tan \beta}{\tan\beta-\tan\alpha}[/tex]
