Question
In figure, $AOC$ is a line, find $x.$
✓
Answer
Since $\angle\text{AOB}$ and $\angle\text{BOC}$ are linear pairs,
$\angle\text{AOB}+\angle\text{BOC}=180^\circ$
$70 + 2x = 180$
$2x = 180 - 70$
$2x = 110$
$\text{x}=\frac{110}{2}$
$x = 55$
Hence, the value of $x$ is $55^\circ $
$\angle\text{AOB}+\angle\text{BOC}=180^\circ$
$70 + 2x = 180$
$2x = 180 - 70$
$2x = 110$
$\text{x}=\frac{110}{2}$
$x = 55$
Hence, the value of $x$ is $55^\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
Factorise: $(3 a-2 b)^3+(2 b-5 c)^3+(5 c-3 a)^3$
→Look at the Fig. Show that length AH > sum of lengths of AB + BC + CD.
Represent $\sqrt{9.3}$ on the number line.
→Simplify the following expressions:
$(x+y-2 z)^2-x^2-y^2-3 z^2+4 x y$
→$(x+y-2 z)^2-x^2-y^2-3 z^2+4 x y$
The height of a cone is $16 \ cm$ and its base radius is $12 \ cm$. Find the curved surface area and the total surface area of the cone (Use $\pi = 3.14).$
→Simplify: $\Big(\frac{32}{243}\Big)^{-\frac{4}{5}}$
→Define the following terms: Obtuse angle.
How many triangles can be drawn having its angles as $53^{\circ}, 64^{\circ}$ and $63^{\circ}$ ? Give reason for your answer.
→Multiply: $3\sqrt{5}$ by $2\sqrt{5}$
→Factorise:
$a^3+8 b^3+64 c^3-24 a b c$
→$a^3+8 b^3+64 c^3-24 a b c$