**Question 1: In the given figure, AB is a mirror. PQ is the incident ray and QR, the reflected ray. If \(\angle PQR=112^{\circ}\), find \(\angle PQA\).**

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Ans:

We know that the angle of incidence = angle of reflection.

Hence, let \(\angle PQA = \angle BQR = x^{o}\)

Since, AQB is a straight line, we have

therefore, \( \angle PQA + \angle PQR + \angle BQR = 180^{o}\)

x + 112 + x = 180^{o}

2x = 68

X = 34^{o}

therefore, \(\angle PQA = 34^{o}\)

**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 \(\angle AOC\)

Let \(\angle COE = 1, \angle AOE = 2, \angle BOF = 3\ and\ \angle DOF = 4\)

We know that vertically opposite angles are equal.

therefore, \( \angle 1 = \angle 4\ and\ \angle 2 = \angle 3\)

But, \(\angle 1 = \angle 2\)

therefore,Â \( \angle 4 = \angle 3\)

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 \(\angle AOC\ and\ \angle BOC\)

Thus, we have:

\(\angle AOC = x^{o}\)

Let OE bisects \(\angle AOC\)

Then, we have:

\(\angle AOE = \angle COE = \frac{1}{2} x^{o}\)

\(\angle BOF = \angle FOC = \frac{1}{2}(180 – x)^{o}\)

Therefore,

\(\angle COE + \angle FOC = \frac{1}{2} x + \frac{1}{2} (180 – x)^{o}\)

\( = \frac{1}{2}(x + 180 – x)\)

\(= \frac{1}{2}(180^{o})\)

= 90^{o}

**Q4) In the adjoining figure \(AB\left | \right | CD\) are cut by a traversal t at E and F respectively. If L1 = \(70^{\circ}\) , find the measure of each of the remaining marked angles.**

**Ans.**

We have, \(\angle 1=70^{\circ}\)

\(\angle 1=\angle 5\)

Therefore, \(\angle 5=70^{\circ}\)

\(\angle 1=\angle 3\)

Therefore, \(\angle 3=70^{\circ}\)

\(\angle 5=\angle 7\)

Therefore, \(\angle 7=70^{\circ}\)

\(\angle 1+\angle 2=180^{\circ}\)

Therefore, \(70^{\circ}+\angle 2=180^{\circ}\)

Ã \(\angle 2=110^{\circ}\)

Ã \(\angle 4=110^{\circ}\)

\(\angle 2=\angle 6\)

Ã \(\angle 6=110^{\circ}\)

\(\angle 6=\angle 8\)

Ã \(\angle 8=110^{\circ}\)

Therefore, \(\angle 1=70^{\circ}\)

**Q5) In the adjoining figure, \(AB\left | \right | CD\) 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 \(AB\left | \right | CD\) 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 = \(180^{\circ}\)

Ã 5y + 4y = \(180^{\circ}\)

Ã Y = \(\frac{180^{\circ}}{9}=20^{\circ}\)

Therefore, L2 = 5y = 5 x \(20^{\circ}\)

And L1 = 4y = 4 x \(20^{\circ}\)

But L1 = L3 [vertically opp. Angles]

Therefore, L3 = \(80^{\circ}\)

Similarly, Since L2 = L4 [vertically opp. Angles]

Therefore, L4 = \(100^{\circ}\)

Since, L1 = L5 [corresponding angles]

Therefore, L5 = \(80^{\circ}\)

Since, L4 = L6 [Alternate angles]

Therefore, L6 = \(100^{\circ}\)

Since, L3 = L7 [Corresponding angles]

Therefore, L7 = \(80^{\circ}\)

Since, L4 = L8 [Corresponding angles]

Therefore, L8 = \(100^{\circ}\)

Hence, L3 = \(80^{\circ}\)

**Q6) In the adjoining fig. ABCD is a quadrilateral in which \(AB\left | \right |DC\) and \(AD\left | \right |BC\) .Prove that \(\angle ADC=\angle ABC\) .**

**Ans. **

Let \(AD\parallel BC\)

\(\angle ADC+\angle DCB=180^{\circ}\)

Also,

\(AB\parallel CD\)

\(\angle DCB+\angle ABC=180^{\circ}\)

From (i) and (ii), we get:

\(\angle ADC+\angle DCB=\angle DCB+\angle ABC\)

Ã \(\angle ADC=\angle ABC\)

**Q7) In each of the figure find the angle \(AB\parallel CD\). Find the value of x in each case.**

**(i)**

**Ans. (i)**

In the fig. \(AB\parallel CD\parallel EF\)

Now, \(AB\parallel EF\)

\(\angle ABE=\angle BEF\)

Ã \(\angle BEF=35^{\circ}\)

Again, \(EF\parallel CD\)

Then,

\(\angle DEF=\angle FED\)

Ã \(\angle FED=65^{\circ}\)

Therefore, \(x^{\circ}=\angle BEF+\angle FED\)

= \((35+65)^{\circ}\)

= \(100^{\circ}\)

Or, x = 100

**(ii)**

Draw \(EO\parallel AB\parallel CD\)

Then, \(\angle EOB+\angle EOD=x^{\circ}\)

Now, \(EO\parallel AB\)

Therefore, \(\angle EOB+\angle ABO=180^{\circ}\)

Ã \(\angle EOB+55^{\circ}=180^{\circ}\)

Ã \(\angle EOB=155^{\circ}\)

Therefore,

\(x^{\circ}=\angle EOB+\angle EOD\)

= \((125+155)^{\circ}\)

= \(280^{\circ}\)

Or, x = 280

**(iii)**

Draw \(EF\parallel AB\parallel CD\)

Then, \(\angle AEF+\angle CEF=x^{\circ}\)

Now, \(EF\parallel AB\)

Therefore, \(\angle AEF+\angle BAE=180^{\circ}\)

Ã \(\angle AEF+116=180\)

Ã \(\angle AEF=64^{\circ} \)

Again, \(EF\parallel CD\)

\(\angle CEF+\angle ECD=180^{\circ}\)

Ã \(\angle CEF+124=180\)

Ã \(\angle CEF=56^{\circ}\)

Therefore,

\(x^{\circ}=\angle AEF+\angle CEF\)

= \((64+56)^{\circ}\)

= \(120^{\circ}\)

Or, x = 120

**Q8) In the given figure, \(AB\parallel CD\parallel EF\). Find the value of x.**

**Ans. **

Given, \(EF\parallel CD\)

Then,

\(\angle ECD+\angle CEF=180^{\circ}\)

Ã \(\angle ECD+130^{\circ}=180^{\circ}\)

Ã \(\angle ECD=50^{\circ}\)

Again, \(AB\parallel CD\)

Then,

\(\angle ABC=\angle BCD\)

Ã \(70^{\circ}=x+50^{\circ} ;Therefore, [\angle BCD=\angle BCE+\angle ECD]\)

Ã \(x=20^{\circ}\)

**Q9) In the given figure , \(AB\parallel CD\) . Find the value of x.**

**Ans. **

Draw \(EF\parallel AB\parallel CD\)

\(EF\parallel CD\)

Then,

\(\angle ECD+\angle CEF=180^{\circ}\)

Ã \(130^{\circ}+\angle CEF=180^{\circ}\)

Ã \(\angle CEF=50^{\circ}\)

Again, \(EF\parallel AB\)

Then,

\(\angle BAE+\angle AEF=180^{\circ}\)

Ã \(x^{\circ}+20^{\circ}+50^{\circ}=180^{\circ}\;[\angle AEF=\angle AEC+\angle CEF]\)

Ã \(x^{\circ}+70^{\circ}=180^{\circ}\)

Ã \(x^{\circ}=110^{\circ}\)

Ã x = 110

**Q10) In the given figure, \(AB\parallel CD\) , Prove that**

**\(\angle BAE-\angle DCE=\angle AEC\)**

**Ans. **

Draw \(EF\parallel AB\parallel CD\)

Now, \(EF\parallel AB\)

Then,

\(\angle BAE+\angle AEF=180^{\circ}\)

Again, \(EF\parallel CD\)

Then,

\(\angle DCE+\angle CEF=180^{\circ}\)

Ã \(\angle DCE+(\angle AEC+\angle AEF)=180^{\circ}\)

Ã \(\angle DCE+\angle AEC+180^{\circ}-\angle BAE=180^{\circ}\)

Ã \(\angle BAE-\angle DCE=\angle AEC\)

**Q11) In the given figure, \(AB\parallel CD\; and \; CD\parallel EF\). Find the value of x.**

**Ans. **

We have, \(AB\parallel CD\)

\(BD\parallel ED\)

Then,

\(\angle BCD+\angle CDE=180^{\circ}\)

Ã \(\angle BCD+75=180\)

Ã \(\angle BCD=105^{\circ}\)

\(AB\parallel CD\)

\(\angle ABC=\angle BCD\)

Ã \(x^{\circ}=105^{\circ}\)

Ã x = 105

**Q12) In the given figure , \(AB\parallel CD\) .Prove that P + q â€“ r = Â \(180^{\circ}\) .**

**Ans. **

Draw \(PFQ\parallel AB\parallel CD\)

Now, \(PFQ\parallel AB\)

Then,

\(\angle AEF+\angle EFP=180^{\circ}\)

[Angles on the same side of a transversal line are supplementary]

Also, \(PFQ\parallel CD\)

\(\angle PFQ=\angle FGD=r^{\circ}\)

And \(\angle EFP=\angle EFG-\angle PFG=q^{\circ}-r^{\circ}\)

Putting the value of \(\angle EFP\)

We get,

\(p^{\circ}+q^{\circ}-r^{\circ}=180^{\circ}\)

Ã p + q â€“ r = 180

**Q13) In the given figure, \(AB\parallel PQ\). Find the value of x and y.**

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**Ans. **

Given \(AB\parallel PQ\)

Let CD be the transversal cutting AB and PQ at E and F, respectively.

Then,

\(\angle CEB+\angle BEG+\angle GEF=180^{\circ}\)

Ã \(75^{\circ}+20^{\circ}+\angle GEF=180^{\circ}\)

Ã \(\angle GEF=85^{\circ}\)

We know that the sum of angles of a triangle is \(180^{\circ}\)

therefore,Â \( \angle GEF+\angle EGF+\angle EFG=180\)

Ã \(85^{\circ}+x+25^{\circ}=180^{\circ}\)

Ã \(110^{\circ}+x=180^{\circ}\)

Ã x = \(70^{\circ}\)

And

\(\angle FEG+\angle BEG=\angle DFQ\)

Ã \(85^{\circ}+20^{\circ}=\angle DFQ\)

Ã \(\angle DFQ=105^{\circ}\)

\(\angle EFG+\angle GFQ+\angle DFQ=180^{\circ}\)

Ã \(25^{\circ}+y+105^{\circ}=180^{\circ}\)

Ã \(y=50^{\circ}\)

Therefore, \( x=70^{\circ}\)

**Q14) In the given figure, \(AB\parallel CD\). Find the value of x.**

**Ans. **

\(AB\parallel CD\)

Then,

\(\angle BAC+\angle ACD=180^{\circ}\)

Ã 75 Â + \(\angle ACD=180\)

Ã \(\angle ACD=105^{\circ}\)

And,

\(\angle ACD=\angle ECF\)

Ã \(\angle ECF=105^{\circ}\)

We know that the sum of the angles of a triangle is \(180^{\circ}\)

\(\angle ECF+\angle CFE+\angle CEF=180^{\circ}\)

Ã \(105^{\circ}+30^{\circ}+x=180^{\circ}\)

Ã \(135^{\circ}+x=180^{\circ}\)

Ã \(x=45^{\circ}\)

**Q15) In the given figure, \(AB\parallel CD\). Find the value of x.**

**Ans. **

\(AB\parallel CD\)

Then,

\(\angle PEF=\angle EGH\)

Ã \(\angle EGH=85^{\circ}\)

And,

\(\angle EGH+\angle QGH=180^{\circ}\)

Ã \(85^{\circ}+\angle QGH=180^{\circ}\)

Ã \(\angle QGH=95^{\circ}\)

Also,

\(\angle CHQ+\angle GHQ=180^{\circ}\)

Ã \(115^{\circ}+\angle GHQ=180^{\circ}\)

Ã \(\angle GHQ=65^{\circ}\)

We know that the sum of angles of a triangle is \(180^{\circ}\)

Ã \(\angle QGH+\angle GHQ+\angle GQH=180^{\circ}\)

Ã \(95^{\circ}+65^{\circ}+x=180^{\circ}\)

Ã x = \(20^{\circ}\)

**Q16) In the given figure, \(AB\parallel CD\). Find the value of x,y and z.**

**Ans. **

\(\angle ADC= \angle DAB\)

Ã z = \(75^{\circ}\)

\(\angle ABC=\angle BCD\)

Ã x = \(75^{\circ}\)

We know that the sum of the angles of triangle is \(180^{\circ}\)

Ã \(35^{\circ}+y+75^{\circ}=180^{\circ}\)

Ã y = \(70^{\circ}\)

Therefore, \(x=35^{\circ},y=70^{\circ}\;and\;z=75^{\circ}\)

**Q17) In the given figure, \(AB\parallel CD\). Find the value of x,y and z.**

**Ans. **

\(AB\parallel CD\)

Now, \(AB\parallel CD\)

Then,

\(\angle AEF=\angle EFG\)

Ã \(y^{\circ}=75^{\circ}\)

Ã y = 75

Also,

\(\angle EFC+\angle EFD=180^{\circ}\)

Ã x + y =180

Ã x + 75 =180

Ã x = 105

And,

\(\angle EGF+\angle EGD=180^{\circ}\)

Ã \(\angle EGF+125=180\)

Ã \(\angle EGF=55^{\circ}\)

We know that the sum of angles of a triangle is \(180^{\circ}\)

\(\angle EFG+\angle GEF+\angle EGF=180^{\circ}\)

Ã \(y+z+55=180\)

Ã 75 + z + 55 = 180

Ã z = 50

Therefore, x = 105, y = 75 and z = 50

**Q18) In the given figure, \(AB\parallel CD\; and \; EF\parallel GH\). Find the value of x,y,z and t.**

**Ans. **

In the given figure,

x = \(60^{\circ}\)

\(\angle PRQ=\angle SQR\)

y = \(60^{\circ}\)

\(\angle APR=\angle PQS\)

Ã \(110^{\circ}=\angle PQR+60^{\circ}\;because [\angle PQS=\angle PQR+\angle RQS]\)

Ã \(\angle PQR=50^{\circ}\)

\(\angle PQR+\angle RQS+\angle BQS=180^{\circ}\)

Ã \(50^{\circ}+60^{\circ}+z=180^{\circ}\)

Ã \(110^{\circ}+z=180^{\circ}\)

Ã \(z=70^{\circ}\)

\(\angle DSH=z\)

Ã \(\angle DSH=70^{\circ}\)

therefore,Â \( \angle DSH=t\)

Ã t = \(70^{\circ}\)

Therefore,\(;x=60^{\circ},z=70^{\circ}\;and\;t=70^{\circ}\)

**Q19) For what value of x will the lines l and m be parallel to each other?**

**Â Â Â Â Â Â Â Â Â Â **

**Ans. **

For the lines l and m to be parallel

**(i)**

\(\Leftrightarrow\)

\(\Leftrightarrow\)

(ii)

\(\Leftrightarrow\)

\(\Leftrightarrow\)

\(\Leftrightarrow\)

**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 : **\(m\parallel n\)

Proof : Since \(m\perp l\)

So, \(\angle 1=90^{\circ}\)

Again, Since \(n\perp l\)

\(\angle 2=90^{\circ}\)

therefore, \( \angle 1=\angle 2=90^{\circ}\)

But \(\angle 1\)

Thus, \(m\parallel n\)