Question

Please solve q 18?

Answer 1

`angleCAD=60^@`

Explanation:

Let `angleDCQ=x`,
given `DC=DQ, => angleDQC=angleDCQ=x`,
let `angleCDB=y and angleBDP=z, => angleCDP=y+z`
given `CP=CD, => angleCPD=angleCDP=y+z`
In `DeltaQDC, 2x+y=180^@ ----- Eq(1)`
As `ABCD` is a cyclic quadrilateral,
`=> angleBCA=angleBDA=z`
In `DeltaPDC, x+2y+3z=180^@ ----- Eq(2)`
`Eq(1)=Eq(2), => 2x+y=x+2y+3z`,
`=> color(red)(x=y+3z)`
Substituting `x=y+3z` into `Eq(2)`, we get:
`y+3z+2y+3z=180^@`
`=> 3y+6z=180, => 3(y+2z)=180`
`=> color(red)(y+2z=180/3=60^@)`
In `DeltaADC, angleCAD=180-(x+y+z)`
`=180-(y+3z+y+z)`
`=180-(2y+4z)`
`=180-2(y+2z)`
`=180-(2*60)=180-120=60^@`