Question

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

Activity: In ∆MNK, If ∠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)`

Answer 1

In ∆MNK, If ∠MNK = 90°, ∠M = 45°    ...(Given)

∴ ∠K = \[\boxed{45°}\]   ...(Remaining angle of ∆MNK)

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

∴ \[\boxed{\text{MN}}\] = `1/sqrt(2)` MK and \[\boxed{\text{KN}}\] = `1/sqrt(2)` MK

∴ MN = \[\frac{1}{{2}}\times\boxed{6}\] 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)`