Question

In triangle ABC, ∠ABC = 15 deg. D is a point on BC such that AD = BD. What is the measure of ∠ADC (in degrees)?

A) 15 B) 30 C) 45 D) 60

Answer 1

Answer:

$\sf\:The\:measure\:of\:\angle\:ADC\:is\:B\:)\:30^{\circ}$

Step-by-step-explanation:

NOTE: Refer to the attachment for the diagram. Diagram provided in the attachment is rough figure.

We have given that,

$\sf\:In\:\triangle\:ABC\:,\:\angle\:ABC\:=\:15^{\circ}\\\\\sf\:Also,\:point\:D\:is\:on\:BC\:such\:that\:AD\:=\:BD$

Now,

Join AD. [ Construction ]

Now,

$\sf\:In\:\triangle\:ABD,\:\\\\\sf\:AD\:=\:BD\:\:\:-\:-\:[\:Given\:]\\\\\therefore\sf\:\angle\:ABD\:\cong\:\angle\:BAD\\\\\implies\sf\:\measuredangle\:ABD\:=\:\measuredangle\:BAD\:=\:15^{\circ}\\\\\sf\:Now,\\\\\sf\:\angle\:ABD\:+\:\angle\:BAD\:+\:\angle\:ADB\:=\:180^{\circ}\:\:\:[\:Angle\:sum\:property\:of\:triangle\:]\\\\\therefore\sf\:15^{\circ}\:+\:15^{\circ}\:+\:\angle\:ADB\:=\:180\\\\\implies\sf\:30^{\circ}\:+\:\angle\:ADB\:=\:180^{\circ}\\\\\implies\sf\:\angle\:ADB\:=\:180^{\circ}\:-\:30^{\circ}\\\\\implies\boxed{\red{\sf\:\angle\:ADB\:=\:150^{\circ}}}$

Now,

$\sf\:In\:\triangle\:ABC,\\\\\sf\:\angle\:ADB\:+\:\angle\:ADC\:=\:180^{\circ}\:\:\:-\:-\:-\:[\:Angles\:in\:linear\:pair\:]\\\\\implies\sf\:150^{\circ}\:+\:\angle\:ADC\:=\:180^{\circ}\\\\\implies\sf\:\angle\:ADC\:=\:180^{\circ}\:-\:150^{\circ}\\\\\implies\boxed{\red{\sf\:\angle\:ADC\:=\:30^{\circ}}}$

$\therefore\sf\:The\:measure\:of\:\angle\:ADC\:is\:B\:)\:30^{\circ}$

Answer 2

Step-by-step explanation:

tnx for free points ill follow u