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Find the value of cos J rounded to the nearest hundredth, if necessary.

Find The Value Of Cos J Rounded To The Nearest Hundredth If Necessary class=

Sagot :

DEVIL005

[tex] \green{\huge{\red{\boxed{\green{\mathfrak{QUESTION}}}}}} [/tex]

Find the value of cos J rounded to the nearest hundredth

[tex]\bold{ \red{\star{\blue{GIVEN }}}}[/tex]

HYPOTHESES= 13

PERPENDICULAR= 12

BASE = ?

[tex]\bold{\blue{\star{\red{TO \: \: FIND}}}}[/tex]

[tex] \cos(j) = ?[/tex]

[tex] \bold{ \green{ \star{ \orange{FORMULA \: USED}}}}[/tex]

  1. PYTHAGORAS THEROM
  2. [tex] \cos(j) = \frac{ BASE}{HYPOTENUSE}[/tex]

[tex] \huge\mathbb{\red A \pink{N}\purple{S} \blue{W} \orange{ER}}[/tex]

FOR BASE:-

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

[tex] {13}^{2} = {12}^{2} + {x}^{2} \\ 169 - 144 = {x}^{2} \\ 25 = {x}^{2} \\ \sqrt{25} = x \\ 5 = x = base[/tex]

NOW :-

[tex] \cos(j) = \frac{ BASE}{HYPOTENUSE}[/tex]

[tex] \cos(j) = \frac{5}{13} \\ \cos(j) = 0.34[/tex]

Answer:

cosJ ≈ 0.38

Step-by-step explanation:

This is a 5- 12- 13 right triangle

with HI = 12, HJ = 13, IJ = 5

cosJ = [tex]\frac{adjacent}{hypotenuse}[/tex] = [tex]\frac{IJ}{HJ}[/tex] = [tex]\frac{5}{13}[/tex] ≈ 0.38 ( to nearest hundredth )

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