Question
In figure AB || CD and $\angle\text{1}$ and $\angle\text{2}$ are in the ratio 3 : 2. Determine all angles from 1 to 8.
✓
Answer
Let:$\angle\text{1}=3\text{x}$ and $\angle\text{2}=2\text{x}$
$\angle\text{1}$ and $\angle\text{2}$ are linear pair of angle. Now, $\angle\text{1}$ and $\angle\text{2}$ $\Rightarrow\ 3\text{x}+2\text{x}=180$
$\Rightarrow\ 5\text{x}=180$
$\Rightarrow\ \text{x}=\frac{180}{5}$
$\Rightarrow\ \text{x}=36$
$\angle\text{1}=3\text{x}=108^\circ,\angle\text{2}=2\text{x}=72^\circ$
We know, Vertically opposite angles are equal$\angle\text{1}=\angle\text{3}=108^\circ$
$\angle\text{2}=\angle\text{4}=72^\circ$
$\angle\text{6}=\angle\text{7}=108^\circ$
$\angle\text{5}=\angle\text{8}=72^\circ$
We also know, corresponding angles are equal$\angle\text{1}=\angle\text{5}=108^\circ$
$\angle\text{2}=\angle\text{6}=72^\circ$
$\angle\text{1}$ and $\angle\text{2}$ are linear pair of angle. Now, $\angle\text{1}$ and $\angle\text{2}$ $\Rightarrow\ 3\text{x}+2\text{x}=180$
$\Rightarrow\ 5\text{x}=180$
$\Rightarrow\ \text{x}=\frac{180}{5}$
$\Rightarrow\ \text{x}=36$
$\angle\text{1}=3\text{x}=108^\circ,\angle\text{2}=2\text{x}=72^\circ$
We know, Vertically opposite angles are equal$\angle\text{1}=\angle\text{3}=108^\circ$
$\angle\text{2}=\angle\text{4}=72^\circ$
$\angle\text{6}=\angle\text{7}=108^\circ$
$\angle\text{5}=\angle\text{8}=72^\circ$
We also know, corresponding angles are equal$\angle\text{1}=\angle\text{5}=108^\circ$
$\angle\text{2}=\angle\text{6}=72^\circ$
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
Calculate the value of x in the following figures.
Using rulers and compasses only, draw an angle of measure 135°.
The income and expenditure for 5 years of a family is given in the following data:
Represent the above data by a bar graph.
|
Years
|
1995–96
|
1996–97
|
1997–98
|
1998–99
|
1999–2000
|
|
Income (Rs. In thousands)
|
100
|
140
|
150
|
170
|
210
|
|
Expenditure (Rs. in thousands)
|
80
|
130
|
145
|
160
|
190
|
Without actual division, show that $\left(x^3-3 x^2-13 x+15\right)$ is exactly divisible by $\left(x^2+2 x-3\right)$
→A transversal $EF$ of line $AB$ and line $CD$ intersects the lines at points $P$ and $Q$ respectively. Ray $PR$ and ray $QS$ are parallel and bisectors of $\angle BPQ$ and $\angle PQC$ respectively. Prove that line $AB \|$ line $CD.$
Given: Ray $PR \|$ ray $QS$
Ray $PR$ and ray $QS$ are the bisectors of $\angle BPQ$ and $\angle PQC$ respectively.
To prove: line $AB \|$ line $CD$
→Given: Ray $PR \|$ ray $QS$
Ray $PR$ and ray $QS$ are the bisectors of $\angle BPQ$ and $\angle PQC$ respectively.
To prove: line $AB \|$ line $CD$
Solve the following equations:$8^{\text{x}+1}=16^{\text{y}+2}$ and $\Big(\frac{1}{2}\Big)^{3+\text{x}}=\Big(\frac{1}{4}\Big)^{3\text{y}}$
→Represent $\sqrt{10.5}$ on the number line.
→In figure, AB = AC and CP ∥ BA and AP is the bisector of exterior $\angle\text{CAD}$ of $\triangle\text{ABC}.$ Prove that:
→- $\angle\text{PAC}=\angle\text{BCA}.$
- ABCP is a parallelogram.
Find the cube of the following binomial expressions:$\frac{3}{\text{x}}-\frac{2}{\text{x}^2}$
→VWithout actual division, prove that $2 x^4-5 x^3+2 x^2-x+2$ is divisible by $x^2-3 x+2$.
→