Question 1: In the given figure, AB is a mirror. PQ is the incident ray and QR, the reflected ray. If \(\begin{array}{l}\angle PQR=112^{\circ}\end{array} \)
, find \(\begin{array}{l}\angle PQA\end{array} \)
.
                         Â
Ans:
We know that the angle of incidence = angle of reflection.
Hence, let \(\begin{array}{l}\angle PQA = \angle BQR = x^{o}\end{array} \)
Since, AQB is a straight line, we have
therefore, \(\begin{array}{l} \angle PQA + \angle PQR + \angle BQR = 180^{o}\end{array} \)
x + 112 + x = 180o
2x = 68
X = 34o
therefore, \(\begin{array}{l}\angle PQA = 34^{o}\end{array} \)
Question 2: If two straight lines intersect each other than prove that ray opposite to the bisector of one of the angles so formed bisect the vertically opposite angles.
Ans:
Let AB and CD be the two lines intersecting at a point O and let ray OE bisect \(\begin{array}{l}\angle AOC\end{array} \)
. Now draw a ray OF in the opposite direction of OE, such that EOF is a straight line.
Let \(\begin{array}{l}\angle COE = 1, \angle AOE = 2, \angle BOF = 3\ and\ \angle DOF = 4\end{array} \)
We know that vertically opposite angles are equal.
therefore, \(\begin{array}{l} \angle 1 = \angle 4\ and\ \angle 2 = \angle 3\end{array} \)
But, \(\begin{array}{l}\angle 1 = \angle 2\end{array} \)
[Since OE bisects \(\begin{array}{l}\angle AOC\end{array} \)
]
therefore, \(\begin{array}{l} \angle 4 = \angle 3\end{array} \)
Hence, the ray opposite the bisector of one of the angles so formed bisects the vertically-opposite angle.
Question 3: Prove that the bisector of two adjacent supplementary angles include a right angle.
Ans: Let AOB denote a straight line and let \(\begin{array}{l}\angle AOC\ and\ \angle BOC\end{array} \)
be the supplementary angles.
Thus, we have:
\(\begin{array}{l}\angle AOC = x^{o}\end{array} \)
and \(\begin{array}{l}\angle BOC = (180 -x)^{o}\end{array} \)
Let OE bisects \(\begin{array}{l}\angle AOC\end{array} \)
and OF bisect \(\begin{array}{l}\angle BOC\end{array} \)
Then, we have:
\(\begin{array}{l}\angle AOE = \angle COE = \frac{1}{2} x^{o}\end{array} \)
and
\(\begin{array}{l}\angle BOF = \angle FOC = \frac{1}{2}(180 – x)^{o}\end{array} \)
Therefore,
\(\begin{array}{l}\angle COE + \angle FOC = \frac{1}{2} x + \frac{1}{2} (180 – x)^{o}\end{array} \)
\(\begin{array}{l} = \frac{1}{2}(x + 180 – x)\end{array} \)
\(\begin{array}{l}= \frac{1}{2}(180^{o})\end{array} \)
= 90o
Q4) In the adjoining figure \(\begin{array}{l}AB\left | \right | CD\end{array} \)
are cut by a traversal t at E and F respectively. If L1 = \(\begin{array}{l}70^{\circ}\end{array} \)
, find the measure of each of the remaining marked angles.
Ans.
We have, \(\begin{array}{l}\angle 1=70^{\circ}\end{array} \)
. Then,
\(\begin{array}{l}\angle 1=\angle 5\end{array} \)
[Corresponding angle ]
Therefore, \(\begin{array}{l}\angle 5=70^{\circ}\end{array} \)
\(\begin{array}{l}\angle 1=\angle 3\end{array} \)
[Vertically-opposite angles]
Therefore, \(\begin{array}{l}\angle 3=70^{\circ}\end{array} \)
\(\begin{array}{l}\angle 5=\angle 7\end{array} \)
[Vertically-opposite angles]
Therefore, \(\begin{array}{l}\angle 7=70^{\circ}\end{array} \)
\(\begin{array}{l}\angle 1+\angle 2=180^{\circ}\end{array} \)
[Since AFB is a straight line ]
Therefore, \(\begin{array}{l}70^{\circ}+\angle 2=180^{\circ}\end{array} \)
à \(\begin{array}{l}\angle 2=110^{\circ}\end{array} \)
[Vertically-opposite angles]
à \(\begin{array}{l}\angle 4=110^{\circ}\end{array} \)
\(\begin{array}{l}\angle 2=\angle 6\end{array} \)
[Corresponding angles]
à \(\begin{array}{l}\angle 6=110^{\circ}\end{array} \)
\(\begin{array}{l}\angle 6=\angle 8\end{array} \)
[Vertically-opposite angles]
à \(\begin{array}{l}\angle 8=110^{\circ}\end{array} \)
Therefore, \(\begin{array}{l}\angle 1=70^{\circ}\end{array} \)
, \(\begin{array}{l}\angle 2=110^{\circ}\end{array} \)
, \(\begin{array}{l}\angle 3=70^{\circ}\end{array} \)
, \(\begin{array}{l}\angle 4=110^{\circ}\end{array} \)
, \(\begin{array}{l}\angle 5=70^{\circ}\end{array} \)
, \(\begin{array}{l}\angle 6=110^{\circ}\end{array} \)
, \(\begin{array}{l}\angle 7=70^{\circ}\end{array} \)
and \(\begin{array}{l}\angle 8=110^{\circ}\end{array} \)
.
Q5) In the adjoining figure, \(\begin{array}{l}AB\left | \right | CD\end{array} \)
are cut by a transversal t at E and F respectively. If L2:L1 = 5:4 , find the measure of each one of the marked angles.
Given \(\begin{array}{l}AB\left | \right | CD\end{array} \)
and a line t intersects them at E and F forming angles
L1 + L2 + L3 + L4 + L5 + L6 + L7 + L8
Ans.
Given, L2:L1 = 5:4
Let L2 = 5y and L1 = 4y
But L2 + L1 = \(\begin{array}{l}180^{\circ}\end{array} \)
[Linear pair]
à 5y + 4y = \(\begin{array}{l}180^{\circ}\end{array} \)
à Y = \(\begin{array}{l}\frac{180^{\circ}}{9}=20^{\circ}\end{array} \)
Therefore, L2 = 5y = 5 x \(\begin{array}{l}20^{\circ}\end{array} \)
= \(\begin{array}{l}100^{\circ}\end{array} \)
And L1 = 4y = 4 x \(\begin{array}{l}20^{\circ}\end{array} \)
= \(\begin{array}{l}80^{\circ}\end{array} \)
But L1 = L3 [vertically opp. Angles]
Therefore, L3 = \(\begin{array}{l}80^{\circ}\end{array} \)
Similarly, Since L2 = L4 [vertically opp. Angles]
Therefore, L4 = \(\begin{array}{l}100^{\circ}\end{array} \)
Since, L1 = L5 [corresponding angles]
Therefore, L5 = \(\begin{array}{l}80^{\circ}\end{array} \)
Since, L4 = L6 [Alternate angles]
Therefore, L6 = \(\begin{array}{l}100^{\circ}\end{array} \)
Since, L3 = L7 [Corresponding angles]
Therefore, L7 = \(\begin{array}{l}80^{\circ}\end{array} \)
Since, L4 = L8 [Corresponding angles]
Therefore, L8 = \(\begin{array}{l}100^{\circ}\end{array} \)
Hence, L3 = \(\begin{array}{l}80^{\circ}\end{array} \)
,L4 = \(\begin{array}{l}100^{\circ}\end{array} \)
, L5 = \(\begin{array}{l}80^{\circ}\end{array} \)
, L6 = \(\begin{array}{l}100^{\circ}\end{array} \)
, L7 = \(\begin{array}{l}80^{\circ}\end{array} \)
, L8 = \(\begin{array}{l}100^{\circ}\end{array} \)
.
Q6) In the adjoining fig. ABCD is a quadrilateral in which \(\begin{array}{l}AB\left | \right |DC\end{array} \)
and \(\begin{array}{l}AD\left | \right |BC\end{array} \)
.Prove that \(\begin{array}{l}\angle ADC=\angle ABC\end{array} \)
.
Ans.
Let \(\begin{array}{l}AD\parallel BC\end{array} \)
and CD is the transversal. Then,
\(\begin{array}{l}\angle ADC+\angle DCB=180^{\circ}\end{array} \)
 …(i) [Consecutive Interior angles]
Also,
\(\begin{array}{l}AB\parallel CD\end{array} \)
and BC is the transversal. Then,
\(\begin{array}{l}\angle DCB+\angle ABC=180^{\circ}\end{array} \)
  …(ii) [Consecutive Interior angles]
From (i) and (ii), we get:
\(\begin{array}{l}\angle ADC+\angle DCB=\angle DCB+\angle ABC\end{array} \)
à \(\begin{array}{l}\angle ADC=\angle ABC\end{array} \)
Q7) In each of the figure find the angle \(\begin{array}{l}AB\parallel CD\end{array} \)
. Find the value of x in each case.
(i)
Ans. (i)
In the fig. \(\begin{array}{l}AB\parallel CD\parallel EF\end{array} \)
Now, \(\begin{array}{l}AB\parallel EF\end{array} \)
and BE is the transversal. Then,
\(\begin{array}{l}\angle ABE=\angle BEF\end{array} \)
[Alternate interior angles]
à \(\begin{array}{l}\angle BEF=35^{\circ}\end{array} \)
Again, \(\begin{array}{l}EF\parallel CD\end{array} \)
and DE is the transversal.
Then,
\(\begin{array}{l}\angle DEF=\angle FED\end{array} \)
à \(\begin{array}{l}\angle FED=65^{\circ}\end{array} \)
Therefore, \(\begin{array}{l}x^{\circ}=\angle BEF+\angle FED\end{array} \)
= \(\begin{array}{l}(35+65)^{\circ}\end{array} \)
= \(\begin{array}{l}100^{\circ}\end{array} \)
Or, x = 100
(ii)
Draw \(\begin{array}{l}EO\parallel AB\parallel CD\end{array} \)
Then, \(\begin{array}{l}\angle EOB+\angle EOD=x^{\circ}\end{array} \)
Now, \(\begin{array}{l}EO\parallel AB\end{array} \)
and BO is the transversal.
Therefore, \(\begin{array}{l}\angle EOB+\angle ABO=180^{\circ}\end{array} \)
[Consecutive Interior angles]
à \(\begin{array}{l}\angle EOB+55^{\circ}=180^{\circ}\end{array} \)
à \(\begin{array}{l}\angle EOB=155^{\circ}\end{array} \)
Therefore,
\(\begin{array}{l}x^{\circ}=\angle EOB+\angle EOD\end{array} \)
= \(\begin{array}{l}(125+155)^{\circ}\end{array} \)
= \(\begin{array}{l}280^{\circ}\end{array} \)
Or, x = 280
(iii)
Draw \(\begin{array}{l}EF\parallel AB\parallel CD\end{array} \)
.
Then, \(\begin{array}{l}\angle AEF+\angle CEF=x^{\circ}\end{array} \)
Now, \(\begin{array}{l}EF\parallel AB\end{array} \)
and AE is the transversal.
Therefore, \(\begin{array}{l}\angle AEF+\angle BAE=180^{\circ}\end{array} \)
[Consecutive interior angles]
à \(\begin{array}{l}\angle AEF+116=180\end{array} \)
à \(\begin{array}{l}\angle AEF=64^{\circ} \end{array} \)
Again, \(\begin{array}{l}EF\parallel CD\end{array} \)
and CE is the transversal.
\(\begin{array}{l}\angle CEF+\angle ECD=180^{\circ}\end{array} \)
[Consecutive Interior angles]
à \(\begin{array}{l}\angle CEF+124=180\end{array} \)
à \(\begin{array}{l}\angle CEF=56^{\circ}\end{array} \)
Therefore,
\(\begin{array}{l}x^{\circ}=\angle AEF+\angle CEF\end{array} \)
= \(\begin{array}{l}(64+56)^{\circ}\end{array} \)
= \(\begin{array}{l}120^{\circ}\end{array} \)
Or, x = 120
Q8) In the given figure, \(\begin{array}{l}AB\parallel CD\parallel EF\end{array} \)
. Find the value of x.
Ans.
Given, \(\begin{array}{l}EF\parallel CD\end{array} \)
and CE is the transversal.
Then,
\(\begin{array}{l}\angle ECD+\angle CEF=180^{\circ}\end{array} \)
[Consecutive Interior angles]
à \(\begin{array}{l}\angle ECD+130^{\circ}=180^{\circ}\end{array} \)
à \(\begin{array}{l}\angle ECD=50^{\circ}\end{array} \)
Again, \(\begin{array}{l}AB\parallel CD\end{array} \)
and BC is the transversal.
Then,
\(\begin{array}{l}\angle ABC=\angle BCD\end{array} \)
[Alternate Interior Angles]
à \(\begin{array}{l}70^{\circ}=x+50^{\circ} ;Therefore, [\angle BCD=\angle BCE+\angle ECD]\end{array} \)
à \(\begin{array}{l}x=20^{\circ}\end{array} \)
Q9) In the given figure , \(\begin{array}{l}AB\parallel CD\end{array} \)
. Find the value of x.
Ans.
Draw \(\begin{array}{l}EF\parallel AB\parallel CD\end{array} \)
\(\begin{array}{l}EF\parallel CD\end{array} \)
and CE is the transversal.
Then,
\(\begin{array}{l}\angle ECD+\angle CEF=180^{\circ}\end{array} \)
[Anfles on the same side of a transversal are supplementary]
à \(\begin{array}{l}130^{\circ}+\angle CEF=180^{\circ}\end{array} \)
à \(\begin{array}{l}\angle CEF=50^{\circ}\end{array} \)
Again, \(\begin{array}{l}EF\parallel AB\end{array} \)
and AE is the transversal.
Then,
\(\begin{array}{l}\angle BAE+\angle AEF=180^{\circ}\end{array} \)
[Angles on the same side of a transversal line are supplementary]
à \(\begin{array}{l}x^{\circ}+20^{\circ}+50^{\circ}=180^{\circ}\;[\angle AEF=\angle AEC+\angle CEF]\end{array} \)
à \(\begin{array}{l}x^{\circ}+70^{\circ}=180^{\circ}\end{array} \)
à \(\begin{array}{l}x^{\circ}=110^{\circ}\end{array} \)
à x = 110
Q10) In the given figure, \(\begin{array}{l}AB\parallel CD\end{array} \)
, Prove that
\(\begin{array}{l}\angle BAE-\angle DCE=\angle AEC\end{array} \)
Ans.
Draw \(\begin{array}{l}EF\parallel AB\parallel CD\end{array} \)
through E.
Now, \(\begin{array}{l}EF\parallel AB\end{array} \)
and AE is the transversal.
Then,
\(\begin{array}{l}\angle BAE+\angle AEF=180^{\circ}\end{array} \)
[Angles on the same side of a transversal line are supplementary]
Again, \(\begin{array}{l}EF\parallel CD\end{array} \)
and CE is the transversal.
Then,
\(\begin{array}{l}\angle DCE+\angle CEF=180^{\circ}\end{array} \)
[Angles on the same side of a transversal line are supplementary]
à \(\begin{array}{l}\angle DCE+(\angle AEC+\angle AEF)=180^{\circ}\end{array} \)
à \(\begin{array}{l}\angle DCE+\angle AEC+180^{\circ}-\angle BAE=180^{\circ}\end{array} \)
à \(\begin{array}{l}\angle BAE-\angle DCE=\angle AEC\end{array} \)
Q11) In the given figure, \(\begin{array}{l}AB\parallel CD\; and \; CD\parallel EF\end{array} \)
. Find the value of x.
Ans.
We have, \(\begin{array}{l}AB\parallel CD\end{array} \)
and \(\begin{array}{l}BC\parallel ED\end{array} \)
.
\(\begin{array}{l}BD\parallel ED\end{array} \)
and CD is the transversal.
Then,
\(\begin{array}{l}\angle BCD+\angle CDE=180^{\circ}\end{array} \)
[Angles on the same side of a transversal line are supplementary]
à \(\begin{array}{l}\angle BCD+75=180\end{array} \)
à \(\begin{array}{l}\angle BCD=105^{\circ}\end{array} \)
\(\begin{array}{l}AB\parallel CD\end{array} \)
and BC is the transversal.
\(\begin{array}{l}\angle ABC=\angle BCD\end{array} \)
(alternate angles)
à \(\begin{array}{l}x^{\circ}=105^{\circ}\end{array} \)
à x = 105
Q12) In the given figure , \(\begin{array}{l}AB\parallel CD\end{array} \)
.Prove that P + q – r =  \(\begin{array}{l}180^{\circ}\end{array} \)
.
Ans.
Draw \(\begin{array}{l}PFQ\parallel AB\parallel CD\end{array} \)
Now, \(\begin{array}{l}PFQ\parallel AB\end{array} \)
and EF is the transversal.
Then,
\(\begin{array}{l}\angle AEF+\angle EFP=180^{\circ}\end{array} \)
   …(1)
[Angles on the same side of a transversal line are supplementary]
Also, \(\begin{array}{l}PFQ\parallel CD\end{array} \)
\(\begin{array}{l}\angle PFQ=\angle FGD=r^{\circ}\end{array} \)
[Alternate angles]
And \(\begin{array}{l}\angle EFP=\angle EFG-\angle PFG=q^{\circ}-r^{\circ}\end{array} \)
Putting the value of \(\begin{array}{l}\angle EFP\end{array} \)
in equ. (i)
We get,
\(\begin{array}{l}p^{\circ}+q^{\circ}-r^{\circ}=180^{\circ}\end{array} \)
à p + q – r = 180
Q13) In the given figure, \(\begin{array}{l}AB\parallel PQ\end{array} \)
. Find the value of x and y.
                                     Â
Ans.
Given \(\begin{array}{l}AB\parallel PQ\end{array} \)
Let CD be the transversal cutting AB and PQ at E and F, respectively.
Then,
\(\begin{array}{l}\angle CEB+\angle BEG+\angle GEF=180^{\circ}\end{array} \)
[Since CD is a straight line]
à \(\begin{array}{l}75^{\circ}+20^{\circ}+\angle GEF=180^{\circ}\end{array} \)
à \(\begin{array}{l}\angle GEF=85^{\circ}\end{array} \)
We know that the sum of angles of a triangle is \(\begin{array}{l}180^{\circ}\end{array} \)
therefore, \(\begin{array}{l} \angle GEF+\angle EGF+\angle EFG=180\end{array} \)
à \(\begin{array}{l}85^{\circ}+x+25^{\circ}=180^{\circ}\end{array} \)
à \(\begin{array}{l}110^{\circ}+x=180^{\circ}\end{array} \)
à x = \(\begin{array}{l}70^{\circ}\end{array} \)
And
\(\begin{array}{l}\angle FEG+\angle BEG=\angle DFQ\end{array} \)
[Corresponding angles]
à \(\begin{array}{l}85^{\circ}+20^{\circ}=\angle DFQ\end{array} \)
à \(\begin{array}{l}\angle DFQ=105^{\circ}\end{array} \)
\(\begin{array}{l}\angle EFG+\angle GFQ+\angle DFQ=180^{\circ}\end{array} \)
[Since CD is a straight line]
à \(\begin{array}{l}25^{\circ}+y+105^{\circ}=180^{\circ}\end{array} \)
à \(\begin{array}{l}y=50^{\circ}\end{array} \)
Therefore, \(\begin{array}{l} x=70^{\circ}\end{array} \)
and \(\begin{array}{l}y=50^{\circ}\end{array} \)
Q14) In the given figure, \(\begin{array}{l}AB\parallel CD\end{array} \)
. Find the value of x.
Ans.
\(\begin{array}{l}AB\parallel CD\end{array} \)
and AC is the transversal.
Then,
\(\begin{array}{l}\angle BAC+\angle ACD=180^{\circ}\end{array} \)
[Consecutive Interior angles]
à 75  + \(\begin{array}{l}\angle ACD=180\end{array} \)
à \(\begin{array}{l}\angle ACD=105^{\circ}\end{array} \)
And,
\(\begin{array}{l}\angle ACD=\angle ECF\end{array} \)
[Vertically –opposite angles]
à \(\begin{array}{l}\angle ECF=105^{\circ}\end{array} \)
We know that the sum of the angles of a triangle is \(\begin{array}{l}180^{\circ}\end{array} \)
\(\begin{array}{l}\angle ECF+\angle CFE+\angle CEF=180^{\circ}\end{array} \)
à \(\begin{array}{l}105^{\circ}+30^{\circ}+x=180^{\circ}\end{array} \)
à \(\begin{array}{l}135^{\circ}+x=180^{\circ}\end{array} \)
à \(\begin{array}{l}x=45^{\circ}\end{array} \)
Q15) In the given figure, \(\begin{array}{l}AB\parallel CD\end{array} \)
. Find the value of x.
Ans.
\(\begin{array}{l}AB\parallel CD\end{array} \)
and PQ is the transversal.
Then,
\(\begin{array}{l}\angle PEF=\angle EGH\end{array} \)
[Corresponding Angles]
à \(\begin{array}{l}\angle EGH=85^{\circ}\end{array} \)
And,
\(\begin{array}{l}\angle EGH+\angle QGH=180^{\circ}\end{array} \)
[Since PQ is a straight line]
à \(\begin{array}{l}85^{\circ}+\angle QGH=180^{\circ}\end{array} \)
à \(\begin{array}{l}\angle QGH=95^{\circ}\end{array} \)
Also,
\(\begin{array}{l}\angle CHQ+\angle GHQ=180^{\circ}\end{array} \)
[Since CD is a straight line]
à \(\begin{array}{l}115^{\circ}+\angle GHQ=180^{\circ}\end{array} \)
à \(\begin{array}{l}\angle GHQ=65^{\circ}\end{array} \)
We know that the sum of angles of a triangle is \(\begin{array}{l}180^{\circ}\end{array} \)
à \(\begin{array}{l}\angle QGH+\angle GHQ+\angle GQH=180^{\circ}\end{array} \)
à \(\begin{array}{l}95^{\circ}+65^{\circ}+x=180^{\circ}\end{array} \)
à x = \(\begin{array}{l}20^{\circ}\end{array} \)
Q16) In the given figure, \(\begin{array}{l}AB\parallel CD\end{array} \)
. Find the value of x,y and z.
Ans.
\(\begin{array}{l}\angle ADC= \angle DAB\end{array} \)
[Alternate interior angles]
à z = \(\begin{array}{l}75^{\circ}\end{array} \)
\(\begin{array}{l}\angle ABC=\angle BCD\end{array} \)
[Alternate Interior Angles]
à x = \(\begin{array}{l}75^{\circ}\end{array} \)
We know that the sum of the angles of triangle is \(\begin{array}{l}180^{\circ}\end{array} \)
à \(\begin{array}{l}35^{\circ}+y+75^{\circ}=180^{\circ}\end{array} \)
à y = \(\begin{array}{l}70^{\circ}\end{array} \)
Therefore, \(\begin{array}{l}x=35^{\circ},y=70^{\circ}\;and\;z=75^{\circ}\end{array} \)
Q17) In the given figure, \(\begin{array}{l}AB\parallel CD\end{array} \)
. Find the value of x,y and z.
Ans.
\(\begin{array}{l}AB\parallel CD\end{array} \)
and let EF and EG be the transversals.
Now, \(\begin{array}{l}AB\parallel CD\end{array} \)
and EF is the transversal.
Then,
\(\begin{array}{l}\angle AEF=\angle EFG\end{array} \)
[Alternate angles]
à \(\begin{array}{l}y^{\circ}=75^{\circ}\end{array} \)
à y = 75
Also,
\(\begin{array}{l}\angle EFC+\angle EFD=180^{\circ}\end{array} \)
[Since CFGD is a straight line]
à x + y =180
à x + 75 =180
à x = 105
And,
\(\begin{array}{l}\angle EGF+\angle EGD=180^{\circ}\end{array} \)
[Since CFGD is a straight line]
à \(\begin{array}{l}\angle EGF+125=180\end{array} \)
à \(\begin{array}{l}\angle EGF=55^{\circ}\end{array} \)
We know that the sum of angles of a triangle is \(\begin{array}{l}180^{\circ}\end{array} \)
\(\begin{array}{l}\angle EFG+\angle GEF+\angle EGF=180^{\circ}\end{array} \)
à \(\begin{array}{l}y+z+55=180\end{array} \)
à 75 + z + 55 = 180
à z = 50
Therefore, x = 105, y = 75 and z = 50
Q18) In the given figure, \(\begin{array}{l}AB\parallel CD\; and \; EF\parallel GH\end{array} \)
. Find the value of x,y,z and t.
Ans.
In the given figure,
x = \(\begin{array}{l}60^{\circ}\end{array} \)
[Vertically-opposite Angles]
\(\begin{array}{l}\angle PRQ=\angle SQR\end{array} \)
[Alternate angles]
y = \(\begin{array}{l}60^{\circ}\end{array} \)
\(\begin{array}{l}\angle APR=\angle PQS\end{array} \)
[Corresponding Angles]
à \(\begin{array}{l}110^{\circ}=\angle PQR+60^{\circ}\;because [\angle PQS=\angle PQR+\angle RQS]\end{array} \)
à \(\begin{array}{l}\angle PQR=50^{\circ}\end{array} \)
\(\begin{array}{l}\angle PQR+\angle RQS+\angle BQS=180^{\circ}\end{array} \)
[Since AB is straight line]
à \(\begin{array}{l}50^{\circ}+60^{\circ}+z=180^{\circ}\end{array} \)
à \(\begin{array}{l}110^{\circ}+z=180^{\circ}\end{array} \)
à \(\begin{array}{l}z=70^{\circ}\end{array} \)
\(\begin{array}{l}\angle DSH=z\end{array} \)
[Corresponding Angles]
à \(\begin{array}{l}\angle DSH=70^{\circ}\end{array} \)
therefore, \(\begin{array}{l} \angle DSH=t\end{array} \)
[Vertically-opposite Angles]
à t = \(\begin{array}{l}70^{\circ}\end{array} \)
Therefore,\(\begin{array}{l};x=60^{\circ},z=70^{\circ}\;and\;t=70^{\circ}\end{array} \)
Q19) For what value of x will the lines l and m be parallel to each other?
         Â
Ans.
For the lines l and m to be parallel
(i)
\(\begin{array}{l}\Leftrightarrow\end{array} \)
3x – 20 = 2x +10  [Corresponding angles]
\(\begin{array}{l}\Leftrightarrow\end{array} \)
x = 30
(ii)
\(\begin{array}{l}\Leftrightarrow\end{array} \)
3x + 5 + 4x = 180 [Consecutive Interior Angles]
\(\begin{array}{l}\Leftrightarrow\end{array} \)
7x = 175
\(\begin{array}{l}\Leftrightarrow\end{array} \)
x = 25
Q20) If two straight lines are perpendicular to the same line, prove that the lines are parallel to each other.
Ans:
Given: Two lines m and n are perpendicular to a given line l.
To Prove : \(\begin{array}{l}m\parallel n\end{array} \)
Proof : Since \(\begin{array}{l}m\perp l\end{array} \)
So, \(\begin{array}{l}\angle 1=90^{\circ}\end{array} \)
Again, Since \(\begin{array}{l}n\perp l\end{array} \)
\(\begin{array}{l}\angle 2=90^{\circ}\end{array} \)
therefore, \(\begin{array}{l} \angle 1=\angle 2=90^{\circ}\end{array} \)
But \(\begin{array}{l}\angle 1\end{array} \)
and \(\begin{array}{l}\angle 2\end{array} \)
are the corresponding angles made by the transversal l with lines m and n and they are proved to be equal.
Thus, \(\begin{array}{l}m\parallel n\end{array} \)
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