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Question

In the given figure $$\angle{ABC}={69}^{\circ}$$, $$\angle{ACB}={31}^{\circ}$$.Find $$\angle{BDC}$$
1252303_e8a25ad62ffb41ff928871b557d00448.PNG


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)
$$\angle {BDC}={80}^{o}$$

1343195_1252303_ans_b2c1dcbc2b534b6984277612a065bcb1.PNG

Mathematics

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