I. GIVEN TRIANGLE POR, DETERMINE THE FOLLOWING.
please help me out with this

This is a lengthy question ill go through all the formulas
This is a right triangle which means u can use pythagorean theorem to find QP
formula :
[tex]a^{2} +b^{2} =c^{2}[/tex]
Note that [tex]c^{2}[/tex] is always the hypotenuse(the hyp is always opposite to ninety degrees)
so lets substitute
=> [tex]a^{2} +17^{2} =28^{2}[/tex]
=> [tex]a^{2} +289 =784[/tex]
subtract 289 both sides
[tex]= > a^{2}=495[/tex]
Now square root both
[tex]\sqrt{ a}^{2}=\sqrt{495}[/tex]
==> [tex]3\sqrt{55}[/tex] or 22.2 (answer)
But for the other three u have to use another formula any of sin, tan,cos while using For example angle Q and then use any and do inverse it will be very easy after u know all the formulas of Sin,cos, tan
SOHCAHTOA
^
Thats an acronym for the formula
Sin= Opposite/ hyp
Cos= Adjustent/Hyp
Tan= Opposite/Adjustent