Nico123o
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What is sec∅ of the following triangle?

See attachment. The most detailed and helpful response will receive Brainliest! :D

What is sec∅ of the following triangle? See attachment. The most detailed and helpful response will receive Brainliest! :D class=

Answer :

ibnuamam104

[tex]sec(\theta) = \frac{1}{ \cos(\theta) } \\ \cos(\theta) = \frac{adj}{hyp}, \: \: \sec(\theta) = \frac{hyp}{adj} [/tex]

Using the Pythagoras theorem,

[tex] {hyp}^{2} = {opp}^{2} + {adj}^{2} \\ {hyp}^{2} = {7}^{2} + {24}^{2} \\ {hyp}^{2} = 49 + 576 \\ {hyp}^{2} = 625 \\ hyp = \sqrt{625} \\ hyp = 25[/tex]

Plugging in our values,

[tex] \sec(\theta) = \frac{hyp}{adj} \\ \sec(\theta) = \frac{25}{7} [/tex]

So the last option is the correct one.

Nayefx

Answer:

[tex] \displaystyle \sec( \theta) = \frac{25}{7} [/tex]

Step-by-step explanation:

remember that,

[tex] \displaystyle \sec( \theta) = \frac{hypo}{adj} [/tex]

since we aren't given hypo we can consider Pythagoras theorem given by

[tex] \displaystyle {a}^{2} + {b}^{2} = {c}^{2} [/tex]

[tex] \displaystyle \implies c = \sqrt{ {a}^{2} + {b}^{2} }[/tex]

substitute the value of a and b:

[tex] \displaystyle c = \sqrt{ {7}^{2} + {24}^{2} }[/tex]

simplify squares:

[tex] \displaystyle c = \sqrt{49 + 576}[/tex]

simplify addition:

[tex] \displaystyle c = \sqrt{625}[/tex]

simplify square root:

[tex] \displaystyle c = 25[/tex]

with respect to [tex]\theta[/tex] adj is 7 and hypo is 25 thus

substitute:

[tex] \displaystyle \sec( \theta) = \frac{25}{7} [/tex]

and we are done!

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