Question

# In the given figure $$\angle{ABC}={69}^{\circ}$$, $$\angle{ACB}={31}^{\circ}$$.Find $$\angle{BDC}$$

Solution

## In $$\triangle ABC$$$$\angle {BAC}+\angle {ABC}+\angle {ACB}={180}^{o}$$ (Angle sum property)$$\angle {BAC}+{69}^{o}+{31}^{o}={180}^{o}$$$$\angle {BAC}={180}^{o}-{100}^{o}={80}^{o}$$For segment $$BADCB$$,$$V{BDC}$$ and $$\angle {BAC}$$ are angles in the same segment. So, they must be equal$$\therefore$$ $$\angle {BDC}=\angle {BAC}$$ (Angles in the same segment are equal)$$\angle {BDC}={80}^{o}$$Mathematics

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