MCQ
In the given figure, straight lines $AB$ and $CD$ intersect at $O.$ If $\angle\text{AOC}+\angle\text{BOD}=130^\circ$ then $\angle\text{AOD}=?$
- A$65^\circ $
- ✓$115^\circ$
- C$110^\circ$
- D$125^\circ$
✓
Answer
Correct option: B.
$115^\circ$
$\angle\text{AOC}+\angle\text{BOD}=1306^\circ$ (given)
But $\angle\text{AOC}=\angle\text{BOD}$ (Vartically Opposite angles)
$\Rightarrow2\angle\text{AOC}=130^\circ$
$\Rightarrow\angle\text{AOC}=65^\circ$
Since $COD$ is a straight line,
$\angle\text{AOC}+\angle\text{AOD}=180^\circ$
$\Rightarrow65^\circ+\angle\text{AOD}=180^\circ$
$\Rightarrow\angle\text{AOD}=115^\circ$
But $\angle\text{AOC}=\angle\text{BOD}$ (Vartically Opposite angles)
$\Rightarrow2\angle\text{AOC}=130^\circ$
$\Rightarrow\angle\text{AOC}=65^\circ$
Since $COD$ is a straight line,
$\angle\text{AOC}+\angle\text{AOD}=180^\circ$
$\Rightarrow65^\circ+\angle\text{AOD}=180^\circ$
$\Rightarrow\angle\text{AOD}=115^\circ$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
In the given figure, sides $CB$ and $BA$ of $\triangle\text{ABC}$ have been produced to D and E respectively such that $\angle\text{ABD}=110^\circ$ and $\text{CAE}=135^\circ.$ Then $\angle\text{ACB}=?$
→If two acute angles of a right triangle are equal, then each acute is equal to:
The number obtained on rationalising the denominator $\frac{1}{\sqrt{7}-2}$ of is:
→Mode of the data $15, 17, 15, 19, 14, 18, 15, 14, 16, 15, 14, 20, 19, 14, 15$ is:
→The distance of the point $(0, 3)$ from $x-$axis is:
→In the adjoining figure, $AB = AC$ and $\text{AD}\bot\text{BC}.$ The rule by which $\triangle\text{ABD}\cong\triangle\text{ACD}$ is:
→Which of the following needs a proof?
Three angles of a quadrilateral are $75^\circ , 90^\circ $ and $75^\circ $. The fourth angle is:
→The coefficient of $x^2$ in $\left(2-3 x^2\right)\left(x^2-5\right)$ is:
→If $AB$ is a chord of a circle, $P$ and $Q$ are the two points on the circle different from $A$ and $B,$ then:
→