Let the required angle be $x,$ then its complement $= (90 - x)$
Given that $x = (90 - x) + 30$
$\Rightarrow 2x = 120$
$\Rightarrow x = 60^\circ $
Let the angles be $3x$ and $7x.$
We know that sum of two interior angles on the same side of transversal is $180.$
$3x + 7x =180$
$\Rightarrow 10x = 180$
$\Rightarrow x = 18$
Therefore, the greater angle is $7x = 7 \times 18 = 126$
Let the angle in degrees be $x$. Then, its supplement $= (180^\circ − x)$
Given, $\text{x}=\frac{1}{5}(180^\circ-\text{x})$
$\Rightarrow 5x = 180^\circ − x$
$\Rightarrow 6x = 180^\circ $
$\Rightarrow x = 30^\circ $
$\angle$ between the lines $x + y - 3 = 0\ \&\ x - y + 3 = 0$ is $90^\circ $
$\Rightarrow\alpha=90^\circ$
As, $\beta$ is acuteTherefore $\alpha>\beta$
Given, $PQ || SR$ and $PR$ is transversal.
$\therefore \angle \text{QPR}=\text{SRP}$ [Alternate interior angles]
$\Rightarrow \text{a}=20^\circ$
Also , $SP || RQ$ and $PR$ is transversal.
$\therefore \angle \text{SPR}=\text{QRP}$ [Alternate interior angles]
$\Rightarrow \text{b}=50^\circ$
Let the $\angle{\text{x}}$ be the required angle
Thus, according to the question
$180 - x = 3 (90 - x)$
$\Rightarrow 180 - x = 270 - 3x$
$\Rightarrow 2x = 90$
$\Rightarrow x = 45$
$\angle \text{AOD}+\angle \text{BOD}=180^\circ$ [Linear pair angles]
$\Rightarrow (7\text{x}-20)^\circ+3\text{x}^\circ=180^\circ$
$\Rightarrow 7\text{x}-20+3\text{x}=180$
$\Rightarrow 10\text{x}=200$
$\Rightarrow \text{x}=20$
$\therefore \angle \text{AOD}=(7\times 20-20)^\circ=120^\circ$
Now, $\angle \text{AOD}=\angle \text{BOC}=120^\circ$ [Vertically opposite angles]
$\therefore \text{y}=120$
Now,$ x + y = 20 + 120$
$= 140$
Hence, the correct answer is option $(b).$
Let the required angle be $x$
Now, supplementary of the required angle = $180^\circ$$- x$
Then,
$x = 2$($180^\circ$ $- x$)
$⇒ x =$ $360^\circ$ $- 2x$
$⇒ 3x =$ $360^\circ$
$⇒ x = $$120^\circ$
Hence, the correct answer is option $(b).$
Given that the ratio of two complementary angles is $13 : 5.$
Let the angles are $13x$ and $5x$ We know, $13x + 5x = 90^\circ $
$\Rightarrow x = 5$
Therefore, the angles are $65^\circ $ and $25^\circ .$
$\angle \text{BQF}=\angle \text{AQP}=(4\text{x})^\circ$ [Vertically opposite angles]
Since, $AB || CD$
$\therefore \angle \text{AQP}+ \angle \text{CPQ}=180^\circ$ [Angles on the same side of a transversal line are supplementary]
$\Rightarrow (4\text{x})^\circ+(5\text{x})^\circ=180^\circ$
$\Rightarrow 9\text{x}=180$
$\Rightarrow \text{x}=20$
$\therefore \angle \text{BQF}=(4\times20)^\circ=80^\circ$
Now, $\angle \text{BQF}=\angle \text{DPQ}=80^\circ$ [Corresponding angles]
Hence, the correct answer is option $(b).$
$\angle \text{AOD}+\angle \text{DOB}+\angle \text{BOC}=180^\circ [ \because AOC$ is a straight line$]$
$\Rightarrow 38^\circ+\text{x}+90^\circ=180^\circ$
$\Rightarrow \text{x}+128^\circ=180^\circ$
$\Rightarrow \text{x}=52^\circ$
Hence, the correct answer is option $(b).$
If the angle is $x^\circ ,$ then by hypothesis
$\Rightarrow x^\circ = 8 (90^\circ - x^\circ )$
$\Rightarrow 720 = 8x^\circ + x^\circ $
or $ 9x^\circ = 720^\circ $
$\therefore x = 80^\circ $
Let the angle be $x$
$\therefore$ complement $= (90 - x) x + (x + 20)$
$= 902x = 90 - 20$
$\text{x}=\frac{70}{2}$
$x = 35^\circ $
$\therefore$ other angle $= 90 - 35 = 55^\circ $
Let all the lines intersect at $O.$
$\angle \text{COF}=\angle \text{DOE}=4\text{x}^\circ$ [Vertically opposite angles]
$\angle \text{AOC}+\angle \text{COF}+\angle \text{BOF}=180^\circ [AOB$ is a straight line$]$
$\Rightarrow 2\text{x}^\circ+4\text{x}^\circ+3\text{x}^\circ=180^\circ$
$\Rightarrow 9\text{x}^\circ=180^\circ$
$\Rightarrow9\text{x}=180$
$\Rightarrow \text{x}=20$
Hence, the correct answer is option $(c).$
Sum of supplementary angles $= 180^\circ $
Given two supplementary angles are in the ratio $5 : 7.$
$1^\text{st} \text{angle}=\frac{5}{12}\times180^\circ=5\times15=75^\circ$
$2^\text{st} \text{angles}=\frac{7}{12}\times180^\circ=7\times15=105^\circ$
Since, $AB || CD$
$\therefore \angle \text{BPQ}= \angle \text{PQC}$ [Alternate interior angles]
$\Rightarrow(3\text{x}+34)^\circ=(5\text{x}-14)^\circ$
$\Rightarrow 3\text{x}+34=5\text{x}-14$
$\Rightarrow 48=2\text{x}$
$\Rightarrow \text{x}=24$
$\therefore \angle\text{BPQ}=(3\times 24+34)^\circ=106^\circ$
$\angle \text{BPQ}+\angle \text{BPE}=180^\circ$ [EF is a straight line]
$\Rightarrow 106^\circ+\angle \text{BPE}=180^\circ$
$\Rightarrow \angle \text{BPE}=74^\circ$
Hence, the correct answer is option $(c).$
If two angles are supplementary, then the sum of the angles is $180^\circ .$
If the ratio is $4 : 5,$ let angles are $4x$ and $5x$
Now we know, $4x + 5x = 180^\circ $
$\Rightarrow 9x = 180$
$\Rightarrow x = 20$
Therefore, angles are $100^\circ $ and $80^\circ $
For the lines $mm$ and $nn$ to be parallel corresponding angles should be equal, i.e,
$\angle1=\angle5$
$\Rightarrow26\text{x}-7^\circ=20\text{x}+17^\circ$
$\Rightarrow6\text{x}=24^\circ$
$\Rightarrow\text{x}=4^\circ$
Two supplementary angles differ by $44^\circ $
$\therefore x + (x + 44^\circ ) = 180^\circ $
$2x = 136^\circ $
$x = 68^\circ $
Other angle $= (x + 44^\circ ) = (68^\circ + 44^\circ ) = 112^\circ $
We have, $a : b = 3 : 2$ Let $a = 3x$ and $b = 2x.$
Since, $a$ and $b$ form a linear pair.
$\because \text{a}+\text{b} = 180^\circ$
$\Rightarrow 3\text{x} + 2\text{x} = 180^\circ$
$\Rightarrow 5\text{x} = 180^\circ [ \because$ sum of linear of angles is $180^\circ ]$
$\Rightarrow\text{x}=\frac{180°}{5}$
$\Rightarrow \text{x} = 36^\circ $
$\therefore \text {a}=3\text{x}\Rightarrow\text{a}=3\times36^\circ=108^\circ$
Now, $f = a [$Corresponding angles$]$
$\Rightarrow \text{f} = 108^\circ$
Supplementary angles are those whose sum is $180$ degrees.
Complementary angles are those angles whose sum is $90$ degrees.
hence they dont need to be adjacent.
$\text{x}=\frac{\text{x}-1}{\text{x}+1}$
Slope of tangent at
$\text{(x, y)}=\frac{\text{dy}}{\text{dx}}=\frac{2}{(\text{x}+1)^2}$
for tangent to be parallel to $y = 2x + 1$
$\frac{2}{(\text{x}+1)^2}=2$
$\Rightarrow{(\text{x}+1)}^2=1$
$\Rightarrow\text{x}=0$ or $\text{x}=-2$
$\Rightarrow $ corresponding points are $(0, -1)$ & amp; $(-2, 3)$
From them given figure, $\angle1=\angle5$ [Corresponding angle]
$\Rightarrow\angle5 = (2\text{a}+\text{b})^\circ$
$[\because\angle1=(2\text{a}+\text{b})^\circ,\text{given}]$
Also, $\angle5+\angle6=180^\circ$ [liner paire]
$\Rightarrow (2\text{a} + \text{b})^\circ + (3\text{a} - \text{b})^\circ = 180^\circ$
$[\because\angle6=(3\text{a}-\text{b})^\circ,\text{given}]$
$\Rightarrow(2\text{a} + 3\text{a}) + (\text{b} - \text{b}) = 180^\circ$
$\Rightarrow5\text{a} = 180^\circ$
$\Rightarrow\text{a}=\frac{180^\circ}{5}$
$\Rightarrow \text{a} = 36^\circ$
Now, $\angle1+\angle2=180^\circ$ [liner paier]
$\Rightarrow\angle2=180^\circ-\angle1$
$\Rightarrow\angle2=180^\circ-(2\text{a}+\text{b})^\circ$
$[\because\angle1=(2\text{a}+\text{b})^\circ,\text{given}]$
$\Rightarrow\angle2=180^\circ-2\text{a}-\text{b}$
$\Rightarrow\angle2=180^\circ-2\times36^\circ-\text{b}$
$[\because\ \text{a}=36^\circ]$
$\Rightarrow\angle2=180^\circ-72^\circ-\text{b}$
$\Rightarrow\angle2=(180^\circ-\text{b})^\circ$
Construction: Draw a line $PQ$ parallel to $AB$ which is also parallel to $CD$
Since, $PQ || AB$
$\therefore \angle \text{AME}+\angle \text{QEM}=180^\circ$ [Angles on the same side of a transversal line are supplementary]
$\Rightarrow 139^\circ+\angle \text{QEM}=180^\circ$
$\Rightarrow \angle \text{QEM}=41^\circ$
Now, $\angle \text{QEM}+\angle \text{DEQ}=\angle \text{MED}$
$\Rightarrow 41^\circ+\angle \text{DEQ}=67^\circ$
$\Rightarrow \angle \text{DEQ}=26^\circ$
Now, $\angle \text{PED}+\angle \text{DEQ}=180^\circ$ [Linear Pair angles]
$\Rightarrow \angle \text{PED}+26^\circ=180^\circ$
$\Rightarrow \angle \text{PED}=154^\circ$
Since, $ PQ || AB$
$\therefore \text{x}^\circ=\angle \text{PED}$ [Corresponding angles]
$\Rightarrow \text{x}^\circ-154^\circ$
$\Rightarrow \text{x}=154$
Hence, the correct answer is option $(a).$
Lets say an angle of a triangle is $\theta ,$ then after it is bisected each smaller angle will now be $\frac{\theta}{2}$.
If you take the external angle of $\theta ,$ it will be $180^\circ - \theta $ and if this is bisected the two new angles would be
$\frac{(180^\circ-\theta)}{2}$ which is equal to $\frac{90^\circ-\theta}{2}$ .
Now adding these two angles, we get $\frac{\theta}{2}+\frac{90^\circ-\theta}{2}=90^\circ.$
Hence the angle between the internal and external bisectors of an angle of a triangle is always $90^\circ .$
Let $x, y$ be any two supplementary angles and $x$ be the smaller angle.
$\therefore x + y = 180^\circ $
Also, $x = 4 (360^\circ - x)$
$⟹ x = 4 \times 360^\circ - 4x$
$⟹ 5x = 4 \times 360^\circ $
$⟹ x = 4 \times 72 = 288^\circ x + y = 180^\circ $
$⟹ y = -108^\circ = 252y = -108^\circ = 252^\circ $
$\therefore x − y = 288 - 252 = 36^\circ $
Let two angles be $x$ and $(180 - x)$
According to question,
$x = 4(90 − x)$
$\Rightarrow x = 360 − 4x$
$\Rightarrow 5x = 360$
$\Rightarrow x = 72^\circ $
$\therefore$ Angles are $72^\circ $ and $108^\circ $
Difference of these two angles $= 108^\circ - 72^\circ = 36^\circ .$
Two angles are called to be supplementary if the summation of both angles is $180^\circ $
say$\angle{\text{A}}$ and $\angle{\text{B}} $ are supplementary angles
$\Rightarrow\angle{\text{A}}+\angle{\text{B}}=180^\circ$
$\therefore\angle{\text{A}}=4\times\angle{\text{B}}$ (Given in question)
So, $4\times\angle{\text{B}}+\angle{\text{B}}=180^\circ$
$\Rightarrow\angle{\text{B}}=36^\circ$
Now, $\angle{\text{A}}=4\times\angle{\text{B}}$
$\Rightarrow\angle{\text{A}}=4\times36^\circ$
$\Rightarrow\angle{\text{A}}=144^\circ$
so these angles are $36^\circ , 144^\circ $
It is given that, $PA || BC$ and $AB$ is transversal.
$\therefore\angle\text{PAB}=\angle\text{ABC}$ [Alternate interior angles]
$\Rightarrow 50^\circ=\text{a}$
Also, $AB || DC$ and $BC$ is transversal.
$\therefore\angle\text{ABC}+\angle\text{DCB}=180^\circ$ [Consecutive interior angles]
$\Rightarrow\text{a}+\angle\text{DCB}=180^\circ$
$\Rightarrow\angle\text{DCB}=180^\circ-\text{a}$
$\Rightarrow\angle\text{DCB}=180^\circ-50^\circ$ $[\because\text{a}=50^\circ]$
$\Rightarrow\angle\text{DCB}=130^\circ$
Also, $BC || DT$ and $DC$ is transversal.
$\therefore\angle \text{CDT}=\text{DCB}$ [Alternate interior angles]
$\Rightarrow \text{b}= 130^\circ$ $[\because\angle\text{DCB}=130^\circ]$
Given, $2\angle\text{A}=3\angle\text{B}=6\angle\text{C},$
$2\angle\text{A}=6\angle\text{C}\angle\text{A}=3\angle\text{C}$
$3\angle\text{B}=6\angle\text{CB}=2\angle\text{C}$
Now, $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$3\angle\text{C}+2\angle\text{C}+\angle\text{C}=180^\circ$
$6\angle\text{C}=180^\circ$
$\angle\text{C}=30^\circ$
Small angle $= 30^\circ $
Two angles are called adjacent angles, if they have a common vertex and a common arm but no common interior points. It is not necessary that uncommon arms must be always
opposite rays.
Let one angle be $x^\circ .$
Then, another angle is $\frac{\text{x}^\circ}{3}.$
Thus, $\text{x}+\frac{\text{x}}{3}=180^\circ$
$4\text{x}=180\times3\text{x}=\frac{180\times3}{4}=135^\circ$
Thus, the required angles are $135^\circ , 45^\circ .$
Construction: Draw a line $PQ$ parallel to $AB$ which is also parallel to $CD$
Since, $PQ || CD$
$\therefore \angle \text{EFC}=\angle \text{FEQ}=37^\circ$ [Alternate angles]
Now, $\angle \text{AEQ}+\angle \text{FEQ}=\angle \text{AEF}$
$\Rightarrow \angle \text{AEQ}+37^\circ=95^\circ$
$\Rightarrow \angle \text{AEQ}=58^\circ$
Since, $PQ || AB$
$\therefore \angle \text{EAB}+\angle \text{AEQ}=180^\circ$ [Angles on the same side of a transversal line are supplementary]
$\Rightarrow \angle\text{EAB}+58^\circ=180^\circ$
$\Rightarrow \angle\text{EAB}=122^\circ$
$\angle\text{EAB}+\text{Reflex}\angle\text{EAB}=360^\circ$ [Complete angle]
$\therefore 122^\circ+(2\text{x})^\circ=360^\circ$
$\Rightarrow 2\text{x}=238$
$\Rightarrow \text{x}=119$
Hence, the correct answer is option $(d).$
cLet the required angle be $x$
Then,
$x = 3(180^\circ - x)$
$\Rightarrow x = 540^\circ - 3x$
$\Rightarrow 4x = 540^\circ $
$\Rightarrow x = 135^\circ $
Hence, the correct answer is option $(c).$
From the given figure, we can say that.
$\angle\text{BOC}=\angle\text{EOF} 40^\circ=\angle\text{EOF}$ [vertycally opposite angles]
$\Rightarrow40^\circ=\angle\text{EOF}$
Since, sum of all the angles on a straight line is $180^\circ $
$\therefore\angle\text{BOC} +\angle\text{FOE}+\angle\text{EOD}=180^\circ$
$\Rightarrow 90^\circ+40^\circ+5\text{a}=180^\circ$
$\Rightarrow130^\circ+5\text{a}=180^\circ\Rightarrow5\text{a}=180^\circ-130^\circ$
$\Rightarrow5\text{a}=50^\circ$
$\Rightarrow\text{a}=\frac{50^\circ}{5}=10^\circ$
We know that, the sum of all angles around a point is $360^\circ .$
$\therefore100^\circ+46^\circ+64^\circ+\text{x}=360^\circ$
$\Rightarrow210^\circ+\text{x}=360^\circ$
$\Rightarrow\text{x}=360^\circ-210^\circ$
$\Rightarrow\text{x}=150^\circ$
Construction: Draw a line $OE$ from the point $O$ parallel to $AB$ and $CD$
Since, $AB || OE$
$\therefore\angle \text{BAO}+\angle \text{AOE}=180^\circ$ [Angles on the same side of a transversal line are supplementary]
$\Rightarrow 150^\circ+\angle \text{AOE}=180^\circ$
$\Rightarrow \angle \text{AOE}=30^\circ$
Again, $CD || OE$
$\therefore \angle \text{DCO}+\angle \text{COE}=180^\circ$ [Angles on the same side of a transversal line are supplementary]
$\Rightarrow 120^\circ+\angle \text{COE}=180^\circ$
$\Rightarrow \angle \text{COE}=60^\circ$
Now, $\angle \text{AOC}=\angle \text{AOE}+\angle \text{COE}$
$=30^\circ+60^\circ$
$=90^\circ$
Hence, the correct answer is option $(b).$