Question

In the figure, AB is a diameter of a circle with the centre O and CD $\parallel$ BA. If $\angle BAC={20}^o$, find the values of
(i) $\angle BOC$
(ii) $\angle DOC$
(iii) $\angle DAC$
(iv) $\angle ADC$

• A. $\angle BOC={40}^o,\angle DOC={100}^o,\angle DAC={50}^o,\angle ACD={110}^o$
• B. $\angle BOC={40}^o,\angle DOC={120}^o,\angle DAC={50}^o,\angle ACD={110}^o$
• C. $\angle BOC={40}^o,\angle DOC={100}^o,\angle DAC={40}^o,\angle ACD={110}^o$
• D. $\angle BOC={40}^o,\angle DOC={100}^o,\angle DAC={50}^o,\angle ACD={120}^o$

Answer 1

The correct option is A $\angle BOC={40}^o,\angle DOC={100}^o,\angle DAC={50}^o,\angle ACD={110}^o$

Given,
AB is a diameter of a circle with the centre O and CD BA

$\angle BAC={20}^o$

Now,

(i)we know that the angle at the centre is twice the angle at the circumference subtended by the same arc.
Therefore,

$\angle BAC={20}^o$

$\angle BOC={20}^o\times 2$

$={40}^o$

(ii)Now,

$\angle DOA$ is the angle at the centre and $\angle DCA$ is the angle at the circumference.
Therefore,

$\angle DOA={40}^o$

Now,
$\angle DOC={180}^o-\angle DOA-\angle BOC={180}^o-{40}^o-{40}^o$
$={100}^o$

(iii)Now,

we know that the angle at the centre is twice the angle at the
circumference subtended by the same arc.
Therefore,

$\angle DAC=\frac{1}{2}\angle DOC$

$=(\frac{1}{2}\times {100}^o)$

$={50}^o$

(iv) CD, BA and AC is the transversal

$\therefore \angle ACD=\angle CAB={20}^o$

$△ACD$ we have,

$\angle ADC+\angle ACD+\angle DAC={180}^o$

$=>\angle ADC+{20}^o+{50}^o={180}^o$

$=>\angle ADC=({180}^o-{70}^o)$

$={110}^o$