Solution:
(i) According to the given question, We know that
![\[\angle DCE\text{ }=\text{ }{{90}^{o}}~\angle CDE\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1de8e5dab5679f1fccd7f7f29a9c4788_l3.png "Rendered by QuickLaTeX.com")
![\[=\text{ }{{90}^{o}}-\text{ }{{40}^{o}}~=\text{ }{{50}^{o}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9d966b67f2fbe40747e2f393c14b1718_l3.png "Rendered by QuickLaTeX.com")
Hence,
![\[\angle DEC\text{ }=\angle OCB\text{ }=\text{ }{{50}^{o}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d3ce0210e385f9267b080d9d7a54dd26_l3.png "Rendered by QuickLaTeX.com")
(ii) In
![\[\vartriangle BOC\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-10338782b5ce598bc939b873127f2a0e_l3.png "Rendered by QuickLaTeX.com")
, we have
![\[\angle AOC\text{ }=\angle OCB\text{ }+\angle OBC\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9ff0ce56c67bd8fd69f88461adacf722_l3.png "Rendered by QuickLaTeX.com")
[Exterior angle property of a triangle]
![\[\angle OBC\text{ }=\text{ }{{80}^{o}}-\text{ }{{50}^{o}}~=\text{ }{{30}^{o}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-41720579926f965090269801c70e7d06_l3.png "Rendered by QuickLaTeX.com")
[Given ∠AOC = 80o]
Hence,
![\[\angle ABC\text{ }=\text{ }{{30}^{o}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e97ceab0b14d0b853560f34906eab252_l3.png "Rendered by QuickLaTeX.com")
In the figure given alongside, AB and CD are straight lines through the centre O of a circle. If ∠AOC = 80degree and ∠CDE = 40degree. Find the number of degrees in: (i) ∠DCE; (ii) ∠ABC.
and 
, internal resistance 
and 
, connected in parallel. The equivalent emf of the combination is- (A) 
(B) 
(C) 
(D) 
and 
are collinear, prove that 
and 
be three points such that area of a 
is 4 sq units, find the value of 
.
and 
is 35 sq units.
and 
are collinear.
and 
are collinear.
and 
are collinear.
and 