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can someone help please? this is trigonometry

Can Someone Help Please This Is Trigonometry class=

Sagot :

Answer:

Solution,

hypotenuse (h)=x

perpendicular(p)=9

base(b)=53°

Using pythagoras theorem,

[tex] {h}^{2} = {p}^{2} + {b}^{2} [/tex]

[tex] {x}^{2} = {9}^{2} + {53}^{2} [/tex]

[tex] {x}^{2} = 81 + 2809[/tex]

[tex] {x}^{2} = 2890[/tex]

[tex]x = \sqrt{2890} [/tex]

[tex]x = 53.75[/tex]

hope it helps you