Question

From the given figure, In ∆ABC, If AD ⊥ BC, ∠C = 45°, AC = `8sqrt(2)`, BD = 5, then for finding value of AD and BC, complete the following activity.

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

∴ AD =`1/sqrt(2) xx square` and DC = `1/sqrt(2) xx 8sqrt(2)`

∴ AD = 8 and DC = 8

∴ BC = BD + DC

= 5 + 8

= 13

Answer 1

In ∆ADC, If ∠ADC = 90°, ∠C = 45°   ...(Given)

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

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

∴ \[\boxed{\text{AD}}\] = `1/sqrt(2)` AC and \[\boxed{\text{DC}}\] = `1/sqrt(2)` AC

∴ AD = \[\frac{1}{\sqrt{2}}\times\boxed{8\sqrt{2}}\] and DC = `1/sqrt(2) xx 8sqrt(2)`

∴ AD = 8 and DC = 8

∴ BC = BD + DC

= 5 + 8

= 13