Question

A student when asked to measure two exterior angles of $$\Delta ABC$$ observed that the exterior angles at A and B are of $$103^o$$ and $$74^o$$ respectively. Is this possible? Why or why not?

Answer 1

Given: $$\angle ACD=103^{\circ}$$ and $$\angle CBE=74^{\circ}$$

Solution: Clearly, $$\angle ACD+\angle BAC=180^{\circ}\quad$$[Linear pair]

$$\Rightarrow 103^{\circ}+\angle BAC=180^{\circ}$$

$$\Rightarrow \angle BAC=180^{\circ}-103^{\circ}$$

$$\Rightarrow \angle BAC=77^{\circ}$$

Similarly, $$\angle CBE+\angle ABC=180^{\circ}\quad$$[Linear pair]

$$\Rightarrow 74^{\circ}+\angle ABC=180^{\circ}$$

$$\Rightarrow \angle ABC=180^{\circ}-74^{\circ}$$

$$\Rightarrow \angle ABC=106^{\circ}$$

Now, Sum of internal angles is given by

$$\angle ABC+\angle BAC=106^{\circ}+77^{\circ}=183^{\circ}$$

It means that the sum of internal angles at $$A$$ and $$B$$ is greater than $$180^{\circ}$$ which cannot be possible.