From given figure, in ∆MNK, if ∠MNK = 90°, ∠M = 45°, MK = 6, then for finding value of MN and KN, complete the following activity.Activity: In ∆MNK, ∠MNK = 90°, ∠M = 45° …...[Given] ∴ ∠K = □ .....[R - Geometry Mathematics 2

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Sum

From given figure, in ∆MNK, if ∠MNK = 90°, ∠M = 45°, MK = 6, then for finding value of MN and KN, complete the following activity.


Activity:

In ∆MNK, ∠MNK = 90°, ∠M = 45°  …...[Given]

∴ ∠K = `square`      .....[Remaining angle of ∆MNK]

By theorem of 45° – 45° – 90° triangle,

∴ `square = 1/sqrt(2)` MK and `square = 1/sqrt(2)` MK

∴ MN = `1/sqrt(2) xx square` and KN = `1/sqrt(2) xx 6`

∴ MN = `3sqrt(2)` and KN = `3sqrt(2)`

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Solution

In ∆MNK, ∠MNK = 90°, ∠M = 45°   …...[Given]

∴ ∠K = 45°      .....[Remaining angle of ∆MNK]

By theorem of 45° – 45° – 90° triangle,

MN = `1/sqrt(2)` MK and KN = `1/sqrt(2)` MK

∴ MN = `1/sqrt(2) xx 6` and KN = `1/sqrt(2) xx 6`

∴ MN = `(1 xx 6 xx sqrt(2))/(sqrt(2) xx sqrt(2))` and KN = `(1 xx 6 xx sqrt(2))/(sqrt(2) xx sqrt(2))`    .......[Multiply numerator and denominator by `sqrt(2)`]

∴ MN = `(6 xx sqrt(2))/2` and KN = `(6 xx sqrt(2))/2`

∴ MN = `3sqrt(2)` and KN = `3sqrt(2)`

  Is there an error in this question or solution?
Chapter 2: Pythagoras Theorem - Q.2 (A)

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From given figure, In ∆ABC, If ∠ABC = 90° ∠CAB=30°, AC = 14 then for finding value of AB and BC, complete the following activity.

Activity: In ∆ABC, If ∠ABC = 90°, ∠CAB=30°

∴ ∠BCA = `square`

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∴ `square = 1/2` AC and `square = sqrt(3)/2` AC

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Activity: In ∆ADC, if ∠ADC = 90°, ∠C = 45°    ......[Given]

∴ ∠DAC = `square`   .....[Remaining angle of ∆ADC]

By theorem of 45° – 45° – 90° triangle,

∴ `square = 1/sqrt(2)` AC and `square = 1/sqrt(2)` AC

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(ii) ML = ?

(iii) MN = ?


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