Question

Q6: In the figure, ABC is a
straight line


Given that ADB = BDC, find
(i) BDC
(ii) CBD

​

Answer 1

Given

• DCB= DCA= 20°
• DAC= DAB= 90°
• ABC is a Straight line.
• ADB = BDC

To Find

• BDC
• CBD

Solution

Now In ∆DAC we have,

• DAC=90°
• DCA=20°

$\: \: \mathfrak {As \: We \: know \: that}$

$\sf \: \angle \: DAC+ \angle \: DCA+ \angle \: ADC= 180° \: \: \: - - - angle \: sum \: property$

$\sf 90+20+ \angle \: ADC= 180$

$\sf \angle \: ADC= 180 - 110$

$\sf \angle \: ADC= 70 ^{0}$

$\\\sf \angle \: ADC= \angle \: ADB+\angle \: BDC\: \: \: \: --- db \: is \: the \: bisector \: of \: adc$

$\sf \: 70=2\angle \: BDC$

$\sf \: \angle \: BDC = 35 ^{0}$

Now In ∆ DBC We have,

$\\ \sf \: ∠BDC+∠BCD+∠CBD= 180 \: \: Angle \: Sum \: Property$

$\sf \: 35+20+∠CBD= 180$

$\sf \: ∠CBD= 180-55$

$\sf \: ∠CBD=125°$

$~~~~~~~~~~~~~~~\\\sf \: \angle \: BDC = 35 ^{0}\\~~~ \sf \angle \: CBD = 155 ^{0}$