Question

In figure $\mathrm{\angle }ABC={69}^0,\mathrm{\angle }ACB={31}^0,find\mathrm{\angle }BDC.$

Answer 1

Hint: Here, to find the value of $\mathrm{\angle }BDC$ we first apply angle sum property of triangle to get third angle of triangle as its two angles are given and then using the concept of circle that angles subtended by chord on arc of a circle are equal.

Complete step-by-step answer:
Here, it is given that $\mathrm{\angle }ABC={69}^0,\mathrm{\angle }ACB={31}^0$
Now, in triangle ABC, by angle sum property of triangle we have
 $\mathrm{\angle }BAC+\mathrm{\angle }ABC+\mathrm{\angle }ACB={180}^0$
Substituting values in above equation. We have,
 $$\begin{gathered} \mathrm{\angle }BAC+{69}^0+{31}^0={180}^0 \\ \Rightarrow \mathrm{\angle }BAC+{100}^0={180}^0 \\ \Rightarrow \mathrm{\angle }BAC={180}^0-{100}^0 \\ \Rightarrow \mathrm{\angle }BAC={80}^0 \end{gathered}$$
Also, we know that in a circle angle subtended by a chord on the same arc of a circle are equal.
Hence, from the figure we see that $\mathrm{\angle }BACand\mathrm{\angle }BDC$ are angles subtended by chord BC on the arc of a circle.
 $\therefore \mathrm{\angle }BAC=\mathrm{\angle }BDC$
But, the value of $\mathrm{\angle }BACis{80}^0.$ Calculated above. Therefore using this in above we have,
 $\mathrm{\angle }BDC={80}^0$
Hence, from above we see that the required value of $\mathrm{\angle }BDCis{80}^0.$

Note: From the figure we see that chord BC subtends two angles on the arc of the circle. Let the value of $\mathrm{\angle }BDC=x$ , then as $\mathrm{\angle }BDC=\mathrm{\angle }BAC$ implies that value of $\mathrm{\angle }BAC=x$ . And, now in triangle ABC by using angle sum property of triangle and using it to form an equation by taking sum of all three angles equal to ${180}^0$ and then solving the equation so formed to find value of ‘x’ and hence required value of $\mathrm{\angle }BDC.$