Question

30. In the given figure, ZDBC = 58° and BD is a diameter of the circle.
Calculate : (i) ZBDC (ii) ZBEC (iii) ZBAC
(2014)
[Hint : (i) Since BD is a diameter, so ZBCD = 90°.
In ADBC, ZBDC = 180° - (90° + 58°) = 32°.
(ii) From cyclic quad. ABEC, we have
ZBEC = (180° - 32°) = 148º.
(iii) ZBAC = ZBDC = 32º. (Angles in the same segment)]
A580
E​

Answer 1

Answer:

In △ABC AB=AC

⇒∠B=∠C (Angles opposite to equal sides are equal)

Now using angle sum property

∠A+∠B+∠C=180

∘

⇒80

∘

+∠C+∠C=180

∘

⇒2∠C=180

∘

−80

∘

⇒∠C=

2

100

∘

=50

∘

now ∠C+∠x=180

∘

(Angles made on straight line (AC) are supplementary)

⇒50

∘

+∠x=180

∘

⇒∠x=180

∘

−50

∘

=130

∘

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