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
Answers
Answer:
3.6cm
Step-by-step explanation:
YZ = 10² - 6² = 8
sinX = 8/10 = 53.13º
cos 53.13º = XN/6
XN = 3.6cm
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 ) = 6Using 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 - xUsing 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 = xUsing 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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