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Question 12 Marks
In the given figure, if $\angle\text{BAC}=60^\circ$ and $\angle\text{BCA}=20^\circ,$ find $\angle\text{ADC}.$
Answer
Using angle sum property in $\triangle\text{ABC}, $$\angle\text{B}=180^\circ-(60^\circ+20^\circ)=100^\circ$
In cyclic quadrilateral ABCD, we have:$\angle\text{B}+\angle\text{C}=180^\circ$
$\angle\text{D}=180^\circ-100^\circ=80^\circ$
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Question 22 Marks
In the given figure, $\triangle\text{ABC}$ is an equilateral triangle. Find $\text{m}\angle\text{BEC}.$
Answer
Since, $\triangle\text{ABC}$ is an equilateral triangle Then, $\angle\text{BAC}=60^\circ$$\therefore\angle\text{BAC}+\angle\text{BEC}=180^\circ$ [Opposite angle of cyclic quad.]
$\Rightarrow60^\circ+\angle\text{BEC}=180^\circ$
$\Rightarrow\angle\text{BEC}=180^\circ-60^\circ=120^\circ$
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2 Mark Question - Maths STD 9 Questions - Vidyadip