Solution:
(i) In
![\[\vartriangle ABD,\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0de07d895eaf877c759a7343c9a5d53d_l3.png "Rendered by QuickLaTeX.com")
![\[\angle DAB\text{ }+\angle ABD\text{ }+\angle ADB\text{ }=\text{ }{{180}^{o}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-fa0e7159e9f5aea73f26ec672c8ac2ed_l3.png "Rendered by QuickLaTeX.com")
![\[{{65}^{o}}~+\text{ }{{70}^{o}}~+\angle ADB\text{ }=\text{ }{{180}^{o}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3d9e9a83a20f29f70e70ab986e56d623_l3.png "Rendered by QuickLaTeX.com")
![\[{{135}^{o}}~+\angle ADB\text{ }=\text{ }{{180}^{o}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-8048278a0cf9be32804053a94c0d2f7d_l3.png "Rendered by QuickLaTeX.com")
![\[\angle ADB\text{ }=\text{ }{{180}^{o}}-\text{ }{{135}^{o}}~=\text{ }{{45}^{o}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e8ed40336bd10bc4a17a9631f6bf84ab_l3.png "Rendered by QuickLaTeX.com")
So now,
![\[\angle ADC\text{ }=\angle ADB\text{ }+\angle BDC\text{ }=\text{ }{{45}^{o}}~+\text{ }{{45}^{o}}~=\text{ }{{90}^{o}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ba10153fc34c5f457fd729ab44c21642_l3.png "Rendered by QuickLaTeX.com")
As
![\[\angle ADC\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a196e195ce49c508fd627176c1c2095a_l3.png "Rendered by QuickLaTeX.com")
is the angle of semi-circle for
![\[AC\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-bacd40acf7ad6dcf4035236de9692ef1_l3.png "Rendered by QuickLaTeX.com")
as the diameter of the circle.
(ii)
![\[\angle ACB\text{ }=\angle ADB\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c9e7f556872b093b0b67e9b4c92e7e6d_l3.png "Rendered by QuickLaTeX.com")
[Angles in the same segment of a circle]
Thus,
![\[\angle ACB\text{ }=\text{ }{{45}^{o}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6f0b931f7adbe6f9e30945108e4e3dfb_l3.png "Rendered by QuickLaTeX.com")
Solution:
(i) In
So now,
As
is the angle of semi-circle for
as the diameter of the circle.
(ii)
[Angles in the same segment of a circle]
Thus,
