the given figure ABC and pqr to triangle find the value of angle A minus angle P + Angle B minus angle R + angle C minus angle q
Answers
Answer:
is given that ∠P−∠Q=42
∘
It can be written as
∠P=42
∘
+∠Q
We know that the sum of all the angles in a triangle is 180
∘
.
So we can write it as
∠P+∠Q+∠R=180
∘
By substituting ∠P=42
∘
+∠Q in the above equation
42
∘
+∠Q+∠Q+∠R=180
∘
On further calculation42
∘
+2∠Q+∠R=180
∘
2∠Q+∠R=180
∘
−42
∘
By subtraction we get
2∠Q+∠R=138
∘
.(i)
It is given that ∠Q−∠R=21
∘
It can be written as
∠R=∠Q−21
∘
By substituting the value of ∠R in equation (i)
2∠Q+∠Q−21
∘
=138
∘On further calculation
3∠Q−21
∘
=138
∘
3∠Q=138
∘
+21
∘
By addition
3∠Q=159
∘
By division
∠Q=159/3
∠Q=53
∘
By substituting ∠Q=53
∘
in ∠P=42
∘
+∠Q
So we get
∠P=42
∘
+53
∘
By addition
∠P=95
∘
By substituting ∠Q in ∠Q−∠R=21
∘
53
∘
−∠R=21
∘
On further calculation
∠R=53
∘
−21
∘
By subtraction
∠R=32
∘
Therefore, ∠P=95
∘
,∠Q=53
∘
and ∠R=32
∘
.