In the given figure, AD divides ∠BAC in the ratio 1: 3 and AD = DB.

1.	In the given figure, AD divides ∠BAC in the ratio 1: 3 and AD = DB. Determine the value of x.
Answer» It is given that AD divides ∠BAC in the ratio 1: 3 So let us consider ∠BAD and ∠DAC as y and 3y According to the figure we know that BAE is a straight line From the figure we know that ∠BAC and ∠CAE form a linear pair of angles So we get ∠BAC + ∠CAE = 180^o We know that ∠BAC = ∠BAD + ∠DAC So it can be written as ∠BAD + ∠DAC + ∠CAE = 180^o By substituting the values we get y + 3y + 108^o = 180^o On further calculation 4y = 180^o – 108^o By subtraction 4y = 72^o By division y = 72/4 y = 18^o We know that the sum of all the angles in triangle ABC is 180^o. So we can write it as ∠ABC + ∠BCA + ∠BAC = 180^o It is given that AD = DB so we can write it as ∠ABC = ∠BAD From the figure we know that ∠BAC = y + 3y = 4y By substituting the values y + x + 4y = 180^o On further calculation 5y + x = 180^o By substituting the value of y 5 (18^o) + x = 180^o By multiplication 90^o + x = 180^o x = 180^o – 90^o By subtraction we get x = 90^o Therefore, the value of x is 90.

In the given figure, AD divides ∠BAC in the ratio 1: 3 and AD = DB. Determine the value of x.

Answer»

It is given that AD divides ∠BAC in the ratio 1: 3

So let us consider ∠BAD and ∠DAC as y and 3y

According to the figure we know that BAE is a straight line

From the figure we know that ∠BAC and ∠CAE form a linear pair of angles

So we get

∠BAC + ∠CAE = 180^o

We know that

∠BAC = ∠BAD + ∠DAC

So it can be written as

∠BAD + ∠DAC + ∠CAE = 180^o

By substituting the values we get

y + 3y + 108^o = 180^o

On further calculation

4y = 180^o – 108^o

By subtraction

4y = 72^o

By division

y = 72/4

y = 18^o

We know that the sum of all the angles in triangle ABC is 180^o.

So we can write it as

∠ABC + ∠BCA + ∠BAC = 180^o

It is given that AD = DB so we can write it as ∠ABC = ∠BAD

From the figure we know that ∠BAC = y + 3y = 4y

By substituting the values

y + x + 4y = 180^o

On further calculation

5y + x = 180^o

By substituting the value of y

5 (18^o) + x = 180^o

By multiplication

90^o + x = 180^o

x = 180^o – 90^o

By subtraction we get

x = 90^o

Therefore, the value of x is 90.