Question

in figure two triangle ABC and and ACD are joined find angle 1+angle 2 + angle 3 + angle 4 + angle 5+ angle 6 and hence find the sum of Angle Bad and angle abc angle BCD and, angle ACD in figure no7is the question​

Answer 1

Answer:

(i) We know that the angles in the same segment of a circle are equal

From the figure we know that ∠ABD and ∠ACD are in the segment AD

∠ABD = ∠ACD = 54

(ii) We know that the angles in the same segment of a circle are equal

From the figure we know that ∠BAD and ∠BCD are in the segment BD

∠BAD = ∠BCD = 43

(iii) In △ ABD

Using the angle sum property

∠BAD + ∠ADB + ∠DBA = 180

By substituting the values

43 + ∠ADB + 54 = 180

On further calculation

∠ADB = 180 - 43 - 54

By subtraction

∠ADB = 180 - 97

So we get

∠ADB = 83

It can be written as

∠BDA = 83