Question

Triangle ABC is right angled at C, and CD is perpendicular AB. Also <A = 65°. Find
(i) <ACD (ii) <BCD (iii) <CbD​

Answer 1

Step-by-step explanation:

ACD- 180-65= 105° .... Property of triangle

BCD- 105°

Answer 2

Answer:

Step-by-step explanation:

In ∆ADC ,

angle A + angle B + angle C = 180°

sum of measure of angle  of triangle

therefore ,

90+65+angleACD=180°

angle ACD = 180°-155°

angle ACD = 25°

In ∆ABC ,

Angle A + angle B + angle C = 180°

therefore, AngleCBD = 25°

In ∆CBD , by applying SMAT

we can find Angle BCD=65°

hope it helps......