Exercise 4.1: Lines and Triangles

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
\(\begin{array}{l}\angle PQA\end{array} \)
.

                          

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.

                                      

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?

          

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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