In a parallelogram `A B C D ,` the diagonals bisect each other at `Odot` If `/_A B C=30^0,\ /_B D C=10^0a n d\ /_C A B=70^0,` Find: `/_D A B ,\ /_A D C ,\ /_B C D ,\ /_A O D ,\ /_D O C ,\ /_B O C ,\ /_A O B ,\ ` `/_A C D ,\ /_C A B ,\ /_A D B`
Given that
` ∠ABC = 30^2, ∠ABC = ∠ADC = 30^0`
[We know that measure of opposite angles are equal in a parallelogram]
`∠BDC = 10^0`
`∠CAB =70^0`
`∠BDA = ∠ADB = ∠ADC – ∠BDC `
` 30^2 – 10^2 = 20^2`
`∠DAB = 180^0 – 30^0 = 150^0`
` ∠ADB = ∠DBC = 20^0` (alternate angles)
` ∠BCD = ∠DAB = 150^0`
[we know, opposite angles are equal in a parallelogram]
` ∠DBA = ∠BDC = 10^0`
[we know, Alternate interior angles are equal]
In` ΔABC`
`∠CAB + ∠ABC + ∠BCA = 180^0`
[since, sum of all angles of a triangle is 180^0]
`70^0 + 30^0 + ∠BCA = 180^0`
`∠BCA = 180^0 – 100^0 = 80^0`
` ∠DAB = ∠DAC + ∠CAB = 70^0 + 80^0 = 150^0 `
`∠BCD = 150^0` (opposite angle of the parallelogram)
` ∠DCA = ∠CAB = 70^0 `
In `ΔDOC ∠BDC + ∠ACD + ∠DOC = 180^0`
[since, sum of all angles of a triangle is 180^0]
`10^0 + 70^0 + ∠DOC = 180^0 `
`∠DOC = 180^0- 80^0`
`∠ACB = 80^0`

Explanation

Given that
` ∠ABC = 30^2, ∠ABC = ∠ADC = 30^0`
[We know that measure of opposite angles are equal in a parallelogram]
`∠BDC = 10^0`
`∠CAB =70^0`
` `∠BDA = ∠ADB = ∠ADC – ∠BDC `
` 30^2 – 10^2 = 20^2`
`∠DAB = 180^0 – 30^0 = 150^0`
` ∠ADB = ∠DBC = 20^0` (alternate angles)
` ∠BCD = ∠DAB = 150^0`
[we know, opposite angles are equal in a parallelogram]
` ∠DBA = ∠BDC = 10^0`
[we know, Alternate interior angles are equal]
In` ΔABC`
`∠CAB + ∠ABC + ∠BCA = 180^0`
[since, sum of all angles of a triangle is 180^0]
`70^0 + 30^0 + ∠BCA = 180^0`
`∠BCA = 180^0 – 100^0 = 80^0`
` ∠DAB = ∠DAC + ∠CAB = 70^0 + 80^0 = 150^0 `
`∠BCD = 150^0` (opposite angle of the parallelogram)
` ∠DCA = ∠CAB = 70^0 `
In `ΔDOC ∠BDC + ∠ACD + ∠DOC = 180^0`
[since, sum of all angles of a triangle is 180^0]
`10^0 + 70^0 + ∠DOC = 180^0 `
`∠DOC = 180^0- 80^0`
`∠ACB = 80^0`