Question
In a parallelogram $ABCD,$ if $\angle\text{A}=(3\text{x}-20)^\circ,\angle\text{B}=(\text{y}+15)^\circ,\angle\text{C}=(\text{x}+40)^\circ,$ then find the values of $x$ and $y.$
✓
Answer
In parallelogram $ABCD,$ $\angle\text{A}$ and $\angle\text{C}$ are opposite angles.
We know that in a parallelogram, the opposite angles are equal.
Therefore, $\angle\text{C}=\angle\text{A}$
We have $\angle\text{A}=(3\text{x}-20)^\circ$ and $\angle\text{C}=(\text{x}+40)^\circ$
Therefore, $x + 40^\circ = 3x - 20^\circ x - 3x = -40^\circ - 20^\circ -2x = -60^\circ x = 30^\circ$
Therefore, $\angle\text{A}=(3\text{x}-20)^\circ$
$\angle\text{A}=[3(30)-20]^\circ$
$\angle\text{A}=70^\circ$
Similarly, $\angle\text{C}=70^\circ$ Also, $\angle\text{B}=(\text{y}+15)^\circ$
Therefore, $\angle\text{D}=\angle\text{B}$
$\angle\text{D}=(\text{y}+15)^\circ$ By angle sum property of a quadrilateral, we have: $\angle\text{A}+\angle\text{B}+\angle\text{C}+\angle\text{D}=360^\circ$
$70^\circ+(\text{y}+15)^\circ+70^\circ+(\text{y}+15)^\circ=360^\circ$
$140^\circ+2(\text{y}+15)^\circ=360^\circ$
$2(\text{y}+15)^\circ=360^\circ-140^\circ$
$2(\text{y}+15)^\circ=220^\circ$
$(\text{y}+15)^\circ=110^\circ$
$\text{y}=95^\circ$
Hence the required values for $x$ and $y$ are $30^\circ$ and $95^\circ$ respectively.
We know that in a parallelogram, the opposite angles are equal.
Therefore, $\angle\text{C}=\angle\text{A}$
We have $\angle\text{A}=(3\text{x}-20)^\circ$ and $\angle\text{C}=(\text{x}+40)^\circ$
Therefore, $x + 40^\circ = 3x - 20^\circ x - 3x = -40^\circ - 20^\circ -2x = -60^\circ x = 30^\circ$
Therefore, $\angle\text{A}=(3\text{x}-20)^\circ$
$\angle\text{A}=[3(30)-20]^\circ$
$\angle\text{A}=70^\circ$
Similarly, $\angle\text{C}=70^\circ$ Also, $\angle\text{B}=(\text{y}+15)^\circ$
Therefore, $\angle\text{D}=\angle\text{B}$
$\angle\text{D}=(\text{y}+15)^\circ$ By angle sum property of a quadrilateral, we have: $\angle\text{A}+\angle\text{B}+\angle\text{C}+\angle\text{D}=360^\circ$
$70^\circ+(\text{y}+15)^\circ+70^\circ+(\text{y}+15)^\circ=360^\circ$
$140^\circ+2(\text{y}+15)^\circ=360^\circ$
$2(\text{y}+15)^\circ=360^\circ-140^\circ$
$2(\text{y}+15)^\circ=220^\circ$
$(\text{y}+15)^\circ=110^\circ$
$\text{y}=95^\circ$
Hence the required values for $x$ and $y$ are $30^\circ$ and $95^\circ$ respectively.
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 length and breadth of a hall are in the ratio $4 : 3$ and its height is $5.5$ meters. The cost of decorating its walls (including doors and windows) at ₹ $6.60$ per square meter is ₹ $5082$. Find the length and breadth of the room.
→The mean of the following distribution is $50$.
Find the value of a and hence the frequencies of $30$ and $70$.
→|
X
|
f | ||||||||||
|
|
A godown measures 40m × 25m × 15m. Find the maximum number of wooden crates each measuring 1.5m × 1.25m × 0.5m that can be stored in the godown.
Factorize the following expressions: $x^3 + 6x^2 + 12x + 16$
→Given below are the cumulative frequencies showing the weights of 685 students of a school. Prepare a frequency distribution table.
→|
Weight (in kg)
|
No. of students
|
|
Below $30$
|
$0$
|
|
Below $30$
|
$24$
|
|
Below $35$
|
$78$
|
|
Below $40$
|
$183$
|
|
Below $45$
|
$294$
|
|
Below $50$
|
$408$
|
|
Below $55$
|
$543$
|
|
Below $60$
|
$621$
|
|
Below $65$
|
$674$
|
|
Below $70$
|
$685$
|
In Fig. $X$ and $Y$ are the mid-points of $AC$ and $AB$ respectively, $QP || BC$ and $CYQ$ and $BXP$ are straight lines. Prove that $ar\ (ABP) = ar\ (ACQ).$
→If $\text{x}^2+\frac{1}{\text{x}^2}=79,$ find the value of $\text{x}+\frac{1}{\text{x}}.$
→If $a+b=10$ and $a b=16$, find the value of $a^2-a b+b^2$ and $a^2+a b+b^2$.
→In figure, $\angle\text{AOC}$ and $\angle\text{BOC}$ from a linear pair. If $a - 2b = 30°$, find $a$ and $b.$ 
→If $\text{x}=2+\sqrt3,$ find the value of $\text{x}^3+\frac{1}{\text{x}^3}.$
→