Answer:
Following are the response to this questions:
Step-by-step explanation:
Please find the graph file in the attachment.
Given:
P=2
B=-5
H=?
[tex]H=\sqrt{P^2+B^2}[/tex]
[tex]=\sqrt{2^2+(-5)^2}\\\\=\sqrt{4+25}\\\\=\sqrt{29}\\\\[/tex]
Using formula:
[tex]\to \ cosec \theta \ or\ \ csco \theta =\frac{H}{P}\\\\\to \cos \theta=\frac{B}{H}\\\\\to \tan \theta=\frac{p}{B}\\\\[/tex]
So,
[tex]\to \ cosec \theta \ or\ \ csco \theta =\frac{\sqrt{29}}{2}\\\\\to \cos \theta=\frac{-5}{\sqrt{29}} =\frac{-5}{\sqrt{29}}\times \frac{\sqrt{29}}{\sqrt{29}}=-\frac{5\sqrt{29}}{29}\\\\\to \tan \theta=\frac{2}{-5}= -\frac{2}{5}\\\\[/tex]
