In △ PQR, if ∠P – ∠Q = 42° and ∠Q – ∠R = 21°, find �

1.	In △ PQR, if ∠P – ∠Q = 42° and ∠Q – ∠R = 21°, find ∠P, ∠Q and ∠R.
Answer» It is given that ∠P – ∠Q = 42^o It can be written as ∠P = 42^o+ ∠Q We know that the sum of all the angles in a triangle is 180^o. So we can write it as ∠P + ∠Q + ∠R = 180^o By substituting ∠P = 42^o + ∠Q in the above equation 42^o + ∠Q +∠Q + ∠R = 180^o On further calculation 42^o + 2 ∠Q + ∠R = 180^o 2 ∠Q + ∠R = 180^o – 42^o By subtraction we get 2 ∠Q + ∠R = 138^o …. (i) It is given that ∠Q – ∠R = 21^o It can be written as ∠R = ∠Q – 21^o By substituting the value of ∠R in equation (i) 2 ∠Q + ∠Q – 21^o = 138^o On further calculation 3 ∠Q – 21^o = 138^o 3 ∠Q = 138^o+ 21^o By addition 3 ∠Q = 159^o By division ∠Q = 159/3 ∠Q = 53^o By substituting ∠Q = 53^o in ∠P = 42^o+ ∠Q So we get ∠P = 42^o+ 53^o By addition ∠P = 95^o By substituting ∠Q in ∠Q – ∠R = 21^o 53^o – ∠R = 21^o On further calculation ∠R = 53^o – 21^o By subtraction ∠R = 32^o Therefore, ∠P = 95^o, ∠Q = 53^o and ∠R= 32^o

In △ PQR, if ∠P – ∠Q = 42° and ∠Q – ∠R = 21°, find ∠P, ∠Q and ∠R.

Answer»

It is given that ∠P – ∠Q = 42^o

It can be written as

∠P = 42^o+ ∠Q

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

So we can write it as

∠P + ∠Q + ∠R = 180^o

By substituting ∠P = 42^o + ∠Q in the above equation

42^o + ∠Q +∠Q + ∠R = 180^o

On further calculation

42^o + 2 ∠Q + ∠R = 180^o

2 ∠Q + ∠R = 180^o – 42^o

By subtraction we get

2 ∠Q + ∠R = 138^o …. (i)

It is given that ∠Q – ∠R = 21^o

It can be written as

∠R = ∠Q – 21^o

By substituting the value of ∠R in equation (i)

2 ∠Q + ∠Q – 21^o = 138^o

On further calculation

3 ∠Q – 21^o = 138^o

3 ∠Q = 138^o+ 21^o

By addition

3 ∠Q = 159^o

By division

∠Q = 159/3

∠Q = 53^o

By substituting ∠Q = 53^o in ∠P = 42^o+ ∠Q

So we get

∠P = 42^o+ 53^o

By addition

∠P = 95^o

By substituting ∠Q in ∠Q – ∠R = 21^o

53^o – ∠R = 21^o

On further calculation

∠R = 53^o – 21^o

By subtraction

∠R = 32^o

Therefore, ∠P = 95^o, ∠Q = 53^o and ∠R= 32^o