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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Step-by-step explanation:
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Answer:
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Step-by-step explanation:
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Step-by-step explanation:
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Answers
Answer:
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Answer:
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Step-by-step explanation:
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Answers
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Step-by-step explanation:
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Answers
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Step-by-step explanation:
In order to determine the outlier value, create a box plot.
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Answers
Answer:
А. y - 110 = 70 (x - 1)
Step-by-step explanation:
А. y - 110 = 70 (x - 1)
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Answers
Answer:
Yes, a 60,60,60 angle triangle equals 180
Step-by-step explanation:
Every triangle has to equal 180 degrees or it's not a triangle at all. This forms 1 triangle.
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Answers
Answer:
Step-by-step explanation:
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Answers
Answer:
47.5Step-by-step explanation:
Median - The middle number in a list of numbers starting from least to greatest.
Knowing this, we can start.
The first step is to rewrite all the numbers from least to greatest
15, 19, 28, 32, 43, 52, 66, 72, 72, 81
Now cross off a number from the least and greatest until there is 1 or 2 numbers left
19, 28, 32, 43, 52, 66, 72, 72
28, 32, 43, 52, 66, 72
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If you are left we 1 number, that is your answer. If you are left with 2 numbers, there is an extra step
In the case that you have 2 numbers, you need to add them together, and divide by 2
43 + 52 = 95
95 ÷ 2 = 47.5
Thus 47.5 is our answer
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Answers
A triangle has sides with lengths of 10 millimeters, 17 millimeters, and 18 millimeters. Is it a right triangle?
Answers
Answer:
no, it is not
if it were a right triangle, 10² + 17² = 18²
but this is not the case
Step-by-step explanation:
we would use Pythagoras theorem to determine if it is a right angle
The Pythagoras theorem : a² + b² = c²
where a = length
b = base
c = hypotenuse
10² + 17² =
100 + 289 = 389
square root of 389 is 19.72 and not 18. thus, it is not a right triangle