MCQ
In the given figure, $AOB$ is a straight line. If $\angle\text{AOC}=(3\text{x}-10)^\circ,\angle\text{COD}=50^\circ$ and $\angle\text{BOD}=(\text{x}+20)^\circ$ then $\angle\text{AOC}=?$
- A$40^\circ$
- B$60^\circ$
- ✓$80^\circ$
- D$50^\circ$
✓
Answer
Correct option: C.
$80^\circ$
Since $AOB$ is a straight line,
$\angle\text{AOC}+\angle\text{COD}+\angle\text{BOD}=180^\circ$
$\Rightarrow(3\text{x}-10)+50+(\text{x}+20)=180^\circ$
$\Rightarrow4\text{x}+60=180$
$\Rightarrow4\text{x}=120$
$\Rightarrow\text{x}=30$
So, $\angle\text{AOC}=3\text{x}-10=3(30)-10=80^\circ$
$\angle\text{AOC}+\angle\text{COD}+\angle\text{BOD}=180^\circ$
$\Rightarrow(3\text{x}-10)+50+(\text{x}+20)=180^\circ$
$\Rightarrow4\text{x}+60=180$
$\Rightarrow4\text{x}=120$
$\Rightarrow\text{x}=30$
So, $\angle\text{AOC}=3\text{x}-10=3(30)-10=80^\circ$
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
The equation y = - 4 in two variables x and y, can be written as
In the given figure, straight lines $AB$ and $CD$ intersect at $O.$ If $\angle\text{AOC}+\angle\text{BOD}=130^\circ$ then $\angle\text{AOD}=?$
→In a histogram the class intervals or the group are taken along:
The graph given below shows the frequency distribution of the age of $22$ teachers in a school. The number of teachers whose age is less than $40$ years is:
→The zeroes of the polynomial $p(x) = x(x - 2) (x + 3)$ are:
→The value of $\sqrt[4]{(64)^{-2}}$ is:
→In the given figure, the measure of $\angle\text{a}$ is:
→If the perimeter and base of an isosceles triangle are $11\ cm$ and $5\ cm$ respectively, then its area is:
→The abscissa of a point is positive in the
Point $(3, 4)$ lies on the graph of the equation $3y = kx + 7$. The value of $k$ is:
→