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In the figure the triangle is right triangle at *Y* and *N* is the foot of the perpendicular from *Y* to XZ. Given that XY=6 cm and XZ=10 cm. What is the length of XN

In The Figure The Triangle Is Right Triangle At Y And N Is The Foot Of The Perpendicular From Y To XZ Given That XY6 Cm And XZ10 Cm What Is The Length Of XN class=

Sagot :

Caiying0043idn

Answer:

3.6cm

Step-by-step explanation:

YZ = 10² - 6² = 8

sinX = 8/10 = 53.13º

cos 53.13º = XN/6

XN = 3.6cm

TheAnimeGirlidn

Answer:

[tex] \underline{ \underline{ \red{ \large {\tt{✧ G \: I \: V \: E \: N}}} }}: [/tex]

  • XY = 6 cm , XZ = 10 cm

[tex] \underline{ \underline{ \purple{ \large{ \tt{✧ T \: O \: \: F\: I\: N\: D}}}}} : [/tex]

  • Length of XN

[tex] \underline{ \underline{ \pink{\large {\tt{✧ S \: O \: L \: U\: T\: I \: O\: N}}}} }: [/tex]

  • Let the length of XN be ' x ' & that of NZ be ' 10 - x '

In rt. XYZ :

  • Hypotenuse (h ) = 10 , Perpendicular ( p ) = YZ & Base ( b ) = 6

Using Pythagoras theorem:

[tex] \large{ \sf{p = \sqrt{ {h}^{2} - {b}^{2} } }}[/tex]

Plug the values & then simplify!

⇢ [tex] \large{ \sf{YZ = \sqrt{( {10)}^{2} - {(6)}^{2} } }}[/tex]

⇢ [tex] \large{ \sf{YZ = \sqrt{100 - 36}}}[/tex]

⇢ [tex] \large{ \sf{YZ = \sqrt{64}}} [/tex]

⇢ [tex] \large{ \sf{YZ = 8} \: cm} \: [/tex]

In rt. YNZ ,

  • Hypotenuse = 8 cm , Perpendicular = YN , base = 10 - x

Using Pythagoras theorem :

⇾ [tex] \large{ \sf{ {p}^{2} = {h}^{2} - {b}^{2} }}[/tex]

⇾ [tex] \large{ \sf{ {YN}^{2} = {8}^{2} - {(10 - x)}}}^{2} [/tex]

⇾ [tex] \large{ \sf{ {YN}^{2} = 64 - \{ {(10)}^{2} - 2 \times 10 \times x + {(x)}^{2} \} }}[/tex]

⇾ [tex] \large{ \sf{ {YN}^{2} = 64 - \{100 - 20x + {x}^{2} \}}} [/tex]

⇾ [tex] \large{ \sf{ {YN}^{2} = 64 - 100 + 20x - {x}^{2} }}[/tex]

⇾ [tex] \large{ \sf{ {YN}^{2} = 20x - {x}^{2} - 36}} \longrightarrow \: eq (i)[/tex]

In rt. XYN ,

  • Hypotenuse = 6 , Perpendicular = YN , base = x

Using Pythagoras theorem :

[tex] \large{ \sf{ {p}^{2} = {h}^{2} - {b}^{2} }}[/tex]

⇾ [tex] \large{ \sf{ {YN}^{2} = {(6)}^{2} - {(x)}}}^{2} [/tex]

⇾ [tex] \large{ \sf{ {YN}^{2} = 36 - {x}^{2} \longrightarrow \: eq(ii) }}[/tex]

Now , From Equation ( i ) & Equation ( ii )

⤑ [tex] \large{ \sf{20x - {x}^{2} - 36 = 36 - {x}^{2} }}[/tex]

⤑ [tex] \large{ \sf{20x - \cancel{ {x}^{2} } - 36 = 36 - \cancel{ {x}^{2} }}}[/tex]

⤑ [tex] \large{ \sf{20x = 36 + 36}}[/tex]

⤑ [tex] \large{ \sf{20x = 72}}[/tex]

⤑ [tex] \large{ \sf{x = 3.6}}[/tex] cm

Hence, The length of XN is [tex] \large{ \boxed{\red{ \bold{ \tt{3.6 \: cm}}}}}[/tex]

[ Correct me if I am wrong ]

♨ Hope I helped! ♡

♪ Have a wonderful day / night ! ☃

☥ [tex] \underbrace{ \overbrace{{ \mathfrak{Carry \: On \: Learning}}}}[/tex] ✎

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