Question
✓
Answer
- x + y - 180
Solution:
In the given figure, $\angle\text{ABF}+\angle\text{ABC}=180^\circ$ (Linear pair of angles)
$∴\text{x}^\circ+\angle\text{ABC}=180^\circ$
$⇒\angle\text{ABC}=180^\circ−\text{x}°...\ (\text{i})$
Also, $\angle\text{ACG}+\angle\text{ACB}=180^\circ$ (Linear pair of angles)
$∴\text{y}′+\angle\text{ACB}=180^\circ$
$⇒\angle\text{ACB}=180^\circ−\text{y}′\ ...\ \text{(ii)}$
Also, $\angle\text{BAC}=\angle\text{DAE}=\text{z}° \ ....\ \text{(iii)}$ (Vertically opposite angles)
In $\triangle\text{ABC}$
$\angle\text{BAC}+\angle\text{ABC}+\angle\text{ACB}=180^\circ$ (Angle sum property)
$∴\text{z}^\circ+180−\text{x}^\circ+180^\circ−\text{y}^\circ=180^\circ$ [Using (1), (2) and (3)]
$⇒ \text{z} = \text{x} + \text{y} – 180$
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
If the ratio of the volumes of two spheres is 1 : 8 then the ratio of their surface area is:
- 1 : 2
- 1 : 4
- 1 : 8
- 1 : 16
$\Big(\frac{2}{3}\Big)^\text{x}\Big(\frac{3}{2}\Big)^{2\text{x}}=\frac{81}{16}$ then x =
→- 2
- 3
- 4
- 1
In Fig. if $l1 \ || \ l2,$ what is the value of $x$ ?
→In the given figure, AOB is a straight line. If $\angle\text{AOC}=(3\text{x} +10)^\circ$ and $\angle\text{BOC}=(4\text{x}-26)^\circ,$ then $\angle\text{BOC}=?$
→- 96º
- 86º
- 76º
- 106º
In a triangle, an exterior angle at a vertex is 95º and its one of the interior opposite angle is 55º, then the measure of the other interior angle is:
If $\sqrt2=1.4142,$ then $\sqrt{\frac{\sqrt2-1}{\sqrt2+1}}$ is equal to:
→-
0.1718
-
5.8282
-
0.4142
-
2.4142
If two acute angles of a right triangle are equal, then each acute is equal to:
In the following, write the correct answer.
In $\triangle\text{ABC}$ if BC = AB and $\angle\text{B}=80^{\circ}$ then is equal to:
→In $\triangle\text{ABC}$ if BC = AB and $\angle\text{B}=80^{\circ}$ then is equal to:
$\frac{1}{\sqrt9-\sqrt8}$ is equal to:
→- $3+2\sqrt2$
- $\frac{1}{3+2\sqrt2}$
- $3-2\sqrt2$
- $\frac{3}{2}-\sqrt2$