Question
In figure, lines AB and CD intersect at O. If $\angle\text{AOC}+\angle\text{BOE}=70^\circ$ and $\angle\text{BOD}=40^\circ,$ find $\angle\text{BOE}$ and reflex $\angle\text{COE}.$
✓
Answer
In the figure, $\angle\text{AOC},\angle\text{BOE}$ and $\angle\text{COE} $ from a linear pair. Thus,
$\angle\text{AOC}+\angle\text{BOE}+\angle\text{COE}=180^\circ$
It is given that $\angle\text{AOC}+\angle\text{BOE}=70^\circ,$ on substituting this value, we get:$70^\circ+\angle\text{COE}=180^\circ$
$\angle\text{COE}=180^\circ-70^\circ$
$\angle\text{COE}=110^\circ$
Thus, reflex $\angle\text{COE}=360^\circ-110^\circ$ Therefore, reflex $\angle\text{COE}=250^\circ.$
$\angle\text{AOC}+\angle\text{BOE}+\angle\text{COE}=180^\circ$It is given that $\angle\text{AOC}+\angle\text{BOE}=70^\circ,$ on substituting this value, we get:$70^\circ+\angle\text{COE}=180^\circ$
$\angle\text{COE}=180^\circ-70^\circ$
$\angle\text{COE}=110^\circ$
Thus, reflex $\angle\text{COE}=360^\circ-110^\circ$ Therefore, reflex $\angle\text{COE}=250^\circ.$
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
Find the median of the data:
31, 38, 27, 28, 36, 25, 35, 40
31, 38, 27, 28, 36, 25, 35, 40
Prove that the sum of exterior angles of a triangle, obtained by extending its sides in the same direction is 360°.
If two straight lines intersect each other, then prove that the ray opposite the bisector of one of the angles so formed bisects the vertically-opposite angle.
Find the values of x in each of the following:$2^{5\text{x}}\div2^\text{x}\sqrt[5]{2^{20}}$
→Read the bar graph given in and answer the following questions:
- What information is given by the bar graph?
- What was the crop-production of rice in 1970–71?
- What is the difference between the maximum and minimum production of rice?
Find the median of the data:
41, 43, 127, 99, 71, 92, 71, 58, 57
41, 43, 127, 99, 71, 92, 71, 58, 57
Find the angle whose complement is one-third of its supplement.
Factorise:
Prove that $\frac{0.85\times0.85\times0.85+0.15\times0.15\times0.15}{0.85\times0.85-0.85\times0.15+0.15\times0.15}=1$
→Prove that $\frac{0.85\times0.85\times0.85+0.15\times0.15\times0.15}{0.85\times0.85-0.85\times0.15+0.15\times0.15}=1$
Simplify:$\frac{(25)^{\frac{3}{2}}\times(243)^{\frac{3}{5}}}{(16)^{\frac{5}{4}}\times(8)^{\frac{4}{3}}}$
→The exterior angles, obtained on producing the base of a triangle both ways are $104^{\circ}$ and $136^{\circ}$. Find all the angles of the triangle.
→