1.

In the figure `AB=AC` angle `BAD=30^@ and AE=AD,` then the value of `x` is

Answer» In the given figure,
`AB = AC`
`:. /_ABC = ACB `
Let `/_ABC = ACB = theta`
Also, `AE = AD`
`:. /_ADE = /_AED`
Let, `/_ADE = /_AED = phi`
`:. /_CED = 180-phi`
Now, in `Delta CED`,
`x+theta+(180-phi) = 180`
`=> phi-theta =x ->(1)`
Now, in `Delta ABC`,
`/_ABC+/_ACB+/_BAC = 180`
`theta+theta+/_BAD+/_DAC = 180`
`2 theta+30 +/_DAC = 180`
`/_DAC = 150-2theta->(2)`
In `Delta ADE`
`2 phi+/_DAC = 180`
`/_DAC = 180-2phi->(3)`
From (2) and (3),
`150-2theta = 180-2phi`
`=>phi-theta = 15->(4)`
From (1) and (4),`x = 15^@`


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