Finish Line & Beyond - हिन्दी ... · PDF file9/10/2009 · Finish Line &...

Finish Line & Beyond

LINES AND ANGLES

• Basic Terms and Definitions• Intersecting Lines and Non-intersecting Lines• Pairs of Angles• Parallel Lines And A Transversal• Lines Parallel To The Same Line• Angle Sum Property of A Triangle

(a) Segment: - A part of line with two end points is called a line-segment.

A line segment is denoted by AB and its length is is denoted by AB.

(b) Ray: - A part of a line with one end-point is called a ray.

A ray is denoted by AB.

We can denote a line-segment AB, a ray AB and length AB and line AB by the same symbol AB.

(c) Collinear points: - If three or more points lie on the same line, then they are called collinear points, otherwise they are called non-collinear points.

(c) Angle: - An angle is formed by two rays originating from the same end point.

The rays making an angle are called the arms of the angle and the end-points are called the vertex of the angle.

(d)Types of Angles:-

(i) Acute angle: - An angle whose measure lies between 0° and 90°, is called an acute angle.

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(ii) Right angle: - An angle, whose measure is equal to 90°, is called a right angle.

(iii) Obtuse angle: - An angle, whose measure lies between 90° and 180°, is called an obtuse angle.




(iv) Straight angle: - The measure of a straight angle is 180°.

(v) Reflex angle: - An angle which is greater than 180° and less than 360°,

is called the reflex angle.

(vi) Complimentary angle: - Two angles, whose sum is 90°, are called complimentary angle.

(vii) Supplementary angle: - Two angles whose sum is 180º, are called

supplementary angle.




(viii) Adjacent angle: - Two angles are adjacent, if they have a common vertex, a common vertex, common arm and their non-common arms are on different sides of the common arm.

In the above figure ABD∠ and DBC∠ are adjacent angle. Ray BD is their common arm and point B is their common vertex. Ray BA and ray BC are non-common arms. When the two angles are adjacent, then their sum is always equal to the angle formed by the two non-common arms.

Thus, ABC∠ = ABD∠ + DBC∠ .

Here we can observe that ABC∠ and DBC∠ are not adjacent angles, because their non-common arms BD and AB lie on the same side of the common arm BC.

(ix) Linear pair of angles: - If the sum of two adjacent angles is 180º, then their non-common lines are in the same straight line and two adjacent angles form a linear pair of angles.




In the fig. ABD∠ and CBD∠ form a linear pair of angles because

ABD∠ + CBD∠ = 180º.

(x) Vertically opposite angles: - When two lines AB and CD intersect at a point O, the vertically opposite angles are formed.

Here are two pairs of vertically opposite angles. One pair is AOD∠ and BOC∠and the second pair is AOC∠ and BOD∠The vertically opposite angles are always equal.

So, AOD∠ = BOC∠ and AOC∠ = BOD∠

(e) Intersecting lines and non-intersecting lines: - Two lines are intersecting if they have one point in common. We have observed in the above figure that lines AB and CD are intersecting lines, intersecting at O, their point of intersection.




Parallel lines: - If two lines do not meet at a point if extended to both directions, such lines are called parallel lines.

Lines PQ and RS are parallel lines.

The length of the common perpendiculars at different points on these parallel lines is same. This equal length is called the distance between two parallel lines.

Axiom 1. If a ray stands on a line, then the sum of two adjacent angles so formed is 180º. Conversely if the sum of two adjacent angles is 180º, then a ray stands on a line (i.e., the non-common arms form a line).

Axiom 2. If the sum of two adjacent angles is 180º, then the non-common arms of the angles form a line. It is called Linear Pair Axiom.

(f) Theorem 1. If two lines intersect each other, then the vertically opposite angles are equal.

Sol. Given: Two lines AB and CD intersect each other at O. To Prove: -

AOC∠ = BOD∠and AOD∠ = BOC∠


P Q

R S



Proof: -

Ray OA stands on line CD.

∴ AOC∠ + AOD∠ = 180º ………….equation (i) {Linear Pair Axiom}

Again ray OD stands on line AB.

∴ AOD∠ + BOD∠ = 180º …………..equation (ii)

From equation (i) and (ii),

AOC∠ + AOD∠ = AOD∠ + BOD∠

⇒ AOC∠ + AOD∠ - AOD∠ = BOD∠

⇒ AOC∠ = BOD∠

Now, Again

Ray OB stands on line CD.

∴ BOC∠ + BOD∠ = 180º ………….equation (iii) {Linear Pair Axiom}

Again ray OD stands on line AB.

∴ AOD∠ + BOD∠ = 180º …………..equation (iv)

From equation (iii) and (iv),

⇒ BOC∠ + BOD∠ = AOD∠ + BOD∠




⇒ BOC∠ + BOD∠ - BOD∠ = AOD∠

⇒ BOC∠ = AOD∠

Hence Proved.

Parallel Lines And A Transversal

In the above figure m and n are two parallel lines and l is the transversal, which intersect the parallel line m and n at points P and Q respectively.

Here Exterior Angles are: - 1∠ , 2∠ , 7∠ and 8∠ Interior Angles are: - 3∠ , 4∠ , 5∠ and 6∠

Corresponding Angles are: -

(i) 1∠ and 5∠ (ii) 2∠ and 6∠ (iii) 4∠ and 8∠ (iv) 3∠ and 7∠




If a transversal intersects two parallel lines, then each pair of corresponding angle is equal. (Corresponding Angles Axiom) – Axiom 3.

Axiom 4. If a transversal intersects two lines such that a pair of corresponding angles is equal, then the two lines are parallel to each other.

Thus, (i) 1∠ = 5∠ (ii) 2∠ = 6∠ (iii) 4∠ = 8∠ (iv) 3∠ = 7∠

Alternate Interior Angles: - (i) 4∠ and 6∠ (ii) 3∠ and 5∠

Alternate Exterior Angles: - (i) 1∠ and 7∠ (ii) 2∠ and 8∠

If a transversal intersects two parallel lines, then each pair of alternate interior and exterior angles are equal.

Thus, Alternate Interior Angles: - (i) 4∠ = 6∠ (ii) 3∠ = 5∠

Alternate Exterior Angles: - (i) 1∠ = 7∠ (ii) 2∠ = 8∠

Interior angles on the same side of the transversal line are called the consecutive interior angles or allied angles or co-interior angles.

(i) 4∠ and 5∠ (ii) 3∠ and 6∠

Theorem 2. If a transversal intersects two parallel lines, then each pair of alternate interior angles is equal.

Sol. Given: Let PQ and RS are two parallel lines and AB be the transversal which intersects them on L and M respectively.

To Prove: - PLM∠ = SML∠And LMR∠ = MLQ∠




Proof: - PLM∠ = RMB∠ ……….equation (i) {corresponding angle}

And RMB∠ = SML∠ ……….equation (ii) {vertically opposite angle are equal}

From equation (i) and (ii),

PLM∠ = SML∠

Similarly,

LMR∠ = ALP∠ ……….equation (iii) {corresponding angle}

And ALP∠ = MLQ∠ ……….equation (iv) {vertically opposite angle are equal}




From equation (iii) and (iv),

LMR∠ = MLQ∠

Hence Proved.

Theorem 3. If a transversal intersects two lines such that a pair of alternate interior angles is equal, then the two lines are parallel.

Sol. Given: - A transversal AB intersects two lines PQ and RS such that

PLM∠ = SML∠ .

To Prove: - PQ║RS

Figure same as in theorem 2.

Proof: - PLM∠ = SML∠ …………equation (i) {given}

SML∠ = RMB∠ …………equation (ii) {vertically opposite angles}

From equations (i) & (ii),

PLM∠ = RMB∠ But these are corresponding angles.We know that if a transversal intersects two lines such that a pair of corresponding angles is equal, then the two lines ate parallel to each other.

Hence, PQ║RS Proved.




Theorem 4. If a transversal intersects two parallel lines, then each pair of interior angles on the same side of the transversal is supplementary.

Solution:

Given: - Transversal EF intersects two parallel lines AB and CD at G and H respectively.

To Prove: - 1∠ + 4∠ = 180º and 2∠ + 3∠ = 180º.

Proof: - 2∠ + 5∠ = 180º …………….equation (i) {Linear Pair}

But 5∠ = 3∠ …………equation (ii) {corresponding angles}





2∠ + 3∠ = 180º

Also, 3∠ + 4∠ = 180º …………….equation (iii) {Linear Pair}

But 3∠ = 1∠ …………equation (iv) {Alt. Interior angles}

From equations (iii) & (iv),

1∠ + 4∠ = 180º

Hence, 1∠ + 4∠ = 180º and 2∠ + 3∠ = 180º. Proved.

Theorem 5. If a transversal intersects two lines such that a pair of interior angles on the same side of the transversal is supplementary, then the two lines are parallel.

Sol.




Given: - A transversal EF intersects two lines AB and CD at P and Q respectively.

1∠ + 2∠ = 180º

To Prove: AB║CD

Proof: - 1∠ + 2∠ = 180º ………….equation (i) {Given}

1∠ + 3∠ = 180º ………….equation (ii) {Linear Pair}


1∠ + 2∠ = 1∠ + 3∠




⇒ 1∠ + 2∠ - 1∠ = 3∠

⇒ 2∠ = 3∠

But these are alternate interior angles. We know that if a transversal intersects two lines such that the pair of alternate interior angles are equal, then the lines are parallel.

Hence, AB║CD Proved.

Theorem 6. Lines which are parallel to the same line are parallel to each other.

Sol.




Given: - Three lines AB, CD and EF are such that AB║CD, CD║EF.

To Prove: - AB║EF.

Construction: - Let us draw a transversal GH which intersects the lines AB, CD and EF at P, Q and R respectively.

Proof: - Since, AB║CD and GH is the transversal. Therefore,

1∠ = 2∠ ………..equation (i) {corresponding angles}

Similarly, CD║ EF and GH is the transversal, Therfore,

2∠ = 3∠ ……… equation (ii) {corresponding angles} From equations (i) & (ii),

1∠ = 3∠

But these are corresponding angles.We know that if a transversal intersects two lines such that a pair of corresponding angles is equal, then the two lines ate parallel to each other.

Hence, AB║ EF Proved.

Angle Sum Property of Triangle: -

Theorem 7. The sum of the angles of a triangle is 180º.

Sol.




Given: - A ∆ ABC.

To Prove: - 1∠ + 2∠ + 3∠ = 180º.

Construction: - Let us draw a line m through A, parallel to BC.

Proof: - BC║m and AB and AC are its transversal.

∴ 1∠ = 4∠ …………..equation (i) {alternate interior angles}

2∠ = 5∠ …………..equation (ii) {alternate interior angles}

By adding equation (i) & (ii),

1∠ + 2∠ = 4∠ + 5∠ ………..equation (iii)

Now by adding 3∠ to both sides of equation (iii), we get

1∠ + 2∠ + 3∠ = 4∠ + 5∠ + 3∠

4∠ + 5∠ + 3∠ = 180º {Linear Pair}

∴ 1∠ + 2∠ + 3∠ = 180º

Hence Proved.




Theorem 8. If a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two interior opposite angles.

Sol.

Given: - A ∆ ABC in which side BC is produced to D forming exterior angle ACD∠ of ∆ ABC.

To Prove: - 4∠ = 1∠ + 2∠ .

Proof: - 1∠ + 2∠ + 3∠ = 180º ………equation (i) {Angle Sum Property of a ∆ }

3∠ + 4∠ = 180º ………equation (ii) {Linear Pair}


1∠ + 2∠ + 3∠ = 3∠ + 4∠

⇒ 1∠ + 2∠ + 3∠ - 3∠ = 4∠

⇒ 1∠ + 2∠ = 4∠

Hence, 4∠ = 1∠ + 2∠ Proved.




EXERCISE 1

Q1. In the Fig. lines AB and CD intersect at O. If AOC∠ + BOE∠ = 70º AND BOD∠ = 40º, find BOE∠ and reflex COE∠ .

Sol. Given: - AOC∠ + BOE∠ = 70º …………equation (i)

And BOD∠ = 40º.

Now, AOC∠ = BOD∠ {vertically opposite angles}

∴ AOC∠ = 40º. ……equation (ii) { BOD∠ = 40º, given}

Now, putting the value of equation (ii) in equation (i),

AOC∠ + BOE∠ = 70º

⇒ 40º + BOE∠ = 70º

⇒ BOE∠ = 70º - 40º

⇒ BOE∠ = 30º

∴ BOE∠ = 30º




Now, AOC∠ + BOE∠ + COE∠ = 180º {Angles at a common point on a line}

⇒ 70º + COE∠ = 180º {from equation (i)}

⇒ COE∠ = 180º - 70º

⇒ COE∠ = 110º

Reflex COE∠ = 360º - 110º = 250º

Hence, BOE∠ = 30º

And Reflex COE∠ = 250º

Q2. In the following figure, lines XY and MN intersect at O. If POY∠ = 90º and a : b = 2 : 3, find c.

Sol. Given: - POY∠ = 90º

And a : b = 2 : 3.

∴ba

= 32

⇒ a = 32

b ………….equation (i)

Now, POX∠ + POY∠ = 180º




⇒ POX∠ + 90º = 180º

⇒ POX∠ = 180º - 90º

⇒ POX∠ = 90º

⇒ a + b = 90º { POX∠ = a + b}

⇒32

b + b = 90º

⇒3

32 bb + = 90º

⇒ 2b + 3b = 90º × 3

⇒ 5b = 270º

⇒ b = 5

270°

⇒ b = 54º

Putting the value of b in equation (i)

a = 32

b

⇒ a = 32

x 54º

⇒ a = 2 x 18º

⇒ a = 36º

Now, b + c = 180º {Angles at a common point on a line}

⇒ 54º + c = 180º

⇒ c = 180º- 54º

⇒ c = 126º

So, c = 126º.




Q3. In the figure, PQR∠ = PRQ∠ , then prove that PRTPQS ∠=∠ .

Sol. Given: - PQR∠ = PRQ∠

To Prove: - PRTPQS ∠=∠

Proof: - PQR∠ + PQS∠ = 180º …………equation (i) {Linear Pair}

And PRQ∠ + PRT∠ = 180º ………...equation (ii) {Linear Pair}


PQR∠ + PQS∠ = PRQ∠ + PRT∠

⇒ PQR∠ + PQS∠ = PQR∠ + PRT∠ { PQR∠ = PRQ∠ given}

⇒ PQR∠ + PQS∠ - PQR∠ = PRT∠




⇒ PQS∠ = PRT∠

Hence, PQS∠ = PRT∠ Proved.

Q4. In the figure if x + y = w + z, then prove that AOB is a line.

Given: - x + y = w + z

To Prove: - AOB is a straight line

Proof: - x + y = w + z ………equation (i) {given}

But, x + y + w + z = 360º {Angles around a point}




⇒ (x + y) + (w + z) = 360º

⇒ (x + y) + (x + y) = 360º {from equation (i)}

⇒ x + y + x + y = 360º

⇒ 2x + 2y = 360º

⇒ 2(x + y) = 360º

⇒ x + y = 2

360°

⇒ x + y = 180º

Therefore, x and y form a linear pair.

Hence, AOB is a straight line. Proved.

Q5. In the given figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR.

Prove that ROS∠ = 21

( QOS∠ - POS∠ )

P Q O

S

Sol. Given: - POQ is a straight line.

OR ⊥ PQ

Ray OS meets line PQ at O.

To Prove: - ROS∠ = 21

( QOS∠ - POS∠ )

Proof: - QOS∠ = ROS∠ + ROQ∠ …………….equation (i)




POS∠ = POR∠ - ROS∠ …………….equation (ii)

Subtracting equation (ii) from equation (i),

QOS∠ - POS∠ = ( ROS∠ + ROQ∠ ) – ( POR∠ - ROS∠ )

⇒ QOS∠ - POS∠ = ROS∠ + ROQ∠ – POR∠ + ROS∠

⇒ QOS∠ - POS∠ = 2 ROS∠ + ROQ∠ – POR∠

⇒ QOS∠ - POS∠ = 2 ROS∠ + 90º - 90º { OR ⊥ PQ }

⇒ QOS∠ - POS∠ = 2 ROS∠

⇒21

( QOS∠ - POS∠ ) = ROS∠

⇒ ROS∠ = 21

( QOS∠ - POS∠ )

Hence, ROS∠ = 21

( QOS∠ - POS∠ ) Proved.

Q6. It is given that XYZ∠ = 64º and XY is produced to a point P. Draw a figure from the given information. If ray YQ bisects ZYP∠ , find XYQ∠ and reflex QYP∠ .

Sol.

Given: - XYZ∠ = 64º

Now, XYZ∠ + ZYP∠ = 180º {Linear Pair}

⇒ 64º + ZYP∠ = 180º




⇒ ZYP∠ = 180º - 64º

⇒ ZYP∠ = 116º

Since, YQ bisects ZYP∠ .

∴ ZYQ∠ = PYQ∠ = 21

ZYP∠ = 2

116° = 58°

∴ XYQ∠ = XYZ∠ + ZYQ∠ = 64º + 58° = 122º

Now reflex QYP∠ = 360º - 58° = 302º.

EXERCISE 2

1. In the following figure find the value of x and y, the show that AB║CD.

Answer: It is clear that °=∠+∠ 180APNAPM ( Angles on the same side of a line)⇒ °=°−°=∠ 13050180APN = x ----------------------(1)




Now, DQMCQN ∠=∠ (Opposite Angles)yDQM =°=∠⇒ 130 ------------------------------------(2)

As we have seen CQNAPN ∠=∠So by the theorem of corresponding angles on one side of the transversal it is clear thatAB║CD

2. In the following figure if AB║CD and CD║EF and y:z=3:7, find the value of x

Answer: As CD and EF are parallel lines, so DPOFQP ∠=∠ (corresponding angles)

Now °=∠+∠ 180CPODPO°=+⇒ 18073 xx

°=⇒ 18010x°=⇒ 18x

Putting the value of x in the given ratio we get following values:




°=∠ 126DPO°=∠ 54CPO

Now it is given that AB║CD, So, xAOPDPO =°=∠=∠ 126

3. In the following figure if AB║CD , EF ⊥ CD and °=∠ 126GED , find FGEandGEFAGE ∠∠∠ ,,

Answer: FEDGEDGEF ∠−∠=∠⇒ °=°−°=∠ 3690126GEF

As AB║CD, So, °=∠=∠ 90FEDEFG

°=°+°−°=∠⇒ 54)3690(180FGE

Now, °=∠+∠ 180FGEAGE (Angles on the same side of a line)°=°−°=∠⇒ 12654180AGE

4. In the following figure PQ║ST, values of °=∠ 110PQR and °=∠ 130RST , find the

value of QRS∠ .




Answer: Let us draw another line AB which is parallel to PQ and ST.

Now, °=∠+∠ 180BRSRSTAnd, °=∠+∠ 180ARQPQR

Because, internal angles on the one side of the transversal are complementary angles.Hence, °=°−°=∠ 50130180BRS

°=°−°=∠ 70110180ARQNow, it is clear that °=∠+∠+∠ 180BRSQRSARQ

°=°+∠+°⇒ 1805070 QRS°=∠⇒ 60QRS

5. In the following figure AB║CD, °=∠ 50APQ and °=∠ 127PRD , find values of x and y.

Answer: °=∠+∠ 180PRDBPR (Internal Angles on one side of transversal)°=°−°=∠⇒ 53127180BPR




On the line CD °=+∠ 180PRQPRD°=°−°=∠⇒ 53127180PRQ

On the line AB, °=∠+∠+∠ 180BPRQPRAPQ°=°+°−°=∠⇒ 77)5350(180QPR

In ΔPQR, °=∠+∠+∠ 180PRQQPRPQR (Sum of angles of a Triangle)

°=°+°−°=∠⇒ 50)5377(180PQR°=⇒ 50x and °= 77y

6. In the given figure PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ, the reflected ray moves along the path BC and strikes the mirror RS. The second mirror reflects the ray along CD. Prove that AB║CD.

Answer: From the theory of reflection in Physics we know that angle of incidence is equal to angle of reflection. Here, In the case of mirror PQ, Angle of incidence ABPi ∠=

And angle of reflection QBCr ∠=

In the case of mirror RS, Angle of incidence BCRi ∠= And angle of reflection SCDr ∠=

Required evidence to prove AB║CDWe need to check if BCDABC ∠=∠ (Alternate angles)

On line PQ, °=∠+∠+∠ 180QBCABCABP°=+∠+⇒ 180rABCi

°=∠++⇒ 180ABCii ………………………..(1)Similarly, on line RS it can be observed that

°+∠++ 180BCDii ……………………………(2)

From the question it is given that PQ║RSHence, BCRQBC ∠=∠ (Alternate Angles)




Hence, values of angles of incidence for both mirrors are same.

Correlating this finding with equations (1) and (2) it is clear that BCDABC ∠=∠

So, AB║CD Proved.

Angle Sum Property of a Triangle

Theorem: The sum of the angles of a triangle is 180º.

Construction: Let us draw a triangle PQR and draw a line XY║QR so that it touches the vertex P of the triangle. For convenience let us name angles as 1, 2, 3, 4 and 5.

Proof: 41 ∠=∠ (Alternate angles) 53 ∠=∠ (Alternate angles)

°=∠+∠+∠ 180321

Substituting the values of 1∠ and 2∠ we get °=∠+∠+∠ 180524

So, it is proved that the sum of angles of a triangle is 180º

Theorem: If a side of a triangle is produced, then the exterior angle soformed is equal to the sum of the two interior opposite angles.

Construction: Let us construct a triangle PQR and extend it base to S. Let us name angles as 1, 2, 3 and 4 for convenience.




Required Proof: 214 ∠+∠=∠

Evidence: From earlier theorem we know that,°=∠+∠+∠ 180321 (Sum of angles of a triangle) ………………..(1)

On line QS,°=∠+∠ 18043 (Angles on the same side of a line) …………………..(2)

From equation (1) and (2) it is clear,214 ∠+∠=∠

Exercise 3:

1. In the following figure, sides QP and RQ of ΔPQR are produced to points S and T respectively. If °=∠ 135SPR and °=∠ 110PQT , find the value of PRQ∠

Answer: On line QS, °=∠+∠ 180SPRQPR ⇒ °=°−°=∠ 45135180QPR

Similarly on line TR, °=+∠ 180PQRTQP




⇒ °=°−°=∠ 70110180PQR

Now we have values of two angles of the given triangle so value of the third angle can be calculated as follows:

°=°+°−°=∠ 65)4570(180PRQ

2. In the following figure °=∠ 62X and °=∠ 54XYZ . If YO and ZO are the bisectors of XYZ∠ and XZY∠ respectively of ΔXYZ, find values of OZY∠ and YOZ∠

Answer: In ΔXYZ, °=∠+∠+∠ 180XZYYXZXYZ

⇒ °=°+°−°=∠ 64)5462(180XZY

As per the question YO and ZO are bisectors of XYZ∠ and XZY∠ respectively

Hence, °=÷=∠=∠ 2725421 XYZOYZ

And, °=÷=∠=∠ 3226421 XZYOZY

Now, for ΔOYZ,)(180 OZYOYZYOZ ∠+∠−°=∠

=180°-(27°+32°)=121°Requires answers are 32° and 121°

3. In the following figure AB║DE, °=∠ 35BAC and °=∠ 53CDE , find the value of DCE∠




Answer: °=∠=∠ 35CEDBAC (Alternate angles)

In ΔDCE, °=∠+∠+∠ 180CEDCDEDCE (Sum of angles of a triangle)Hence, °=°+°−°=∠ 92)3553(180DCE

4. In the following figure lines PQ and RS intersect at point T, such that °=∠ 40PRT, °=∠ 95RPT and °=∠ 75TSQ . Find the value of SQT∠ .

Answer: In ΔPRT,°=∠+∠+∠ 180PTRRPTPRT

⇒ °=°+°−°=∠ 45)4095(180PTR

As we know opposite angles are equal so °=∠=∠ 45STQPTR

Now, in ΔQST,




°=∠+∠+∠ 180SQTSTQQST⇒ °=°+°−°=∠ 60)4575(180SQT

5. In the following figure, PQ ⊥ PS, PQ║SR, °=∠ 28SQR and °=∠ 65QRT . Find the values of x and y.

Answer: On the line ST, °=∠+∠ 180QRTQRS

⇒ °=°−°=∠ 11565180QRS

In ΔQRS, °=∠+∠+∠ 180QRSSQRQSR

°=°+°−°=∠ 27)11528(180QSR

Now, PSRSPQ ∠=∠ (Complementary Angles on the inner side of transversal)

∴ °=°−°=∠ 632790PSQ

In ΔSPQ,)(180 PSQSPQPQS ∠+∠−°=∠

=180°-(90°+63°) = 27°

So, x=27° Y=63°

6. In the following figure, the side QR of ΔPQR is produced to a point S. If the

bisectors of PQR∠ and PRS∠ meet at point T, then prove that QPRQTR ∠=∠21




Answer: In Δ PQR

PQRQPRSRP ∠+∠=∠

∴ PQRQPRSRP ∠+∠=∠21

21

21

or, PQRSRPQPR ∠−∠=∠21

21

21

…………………………..(1)

In Δ TQRTQRQTRSRT ∠+∠=∠

Or, PQRQTRSRP ∠+∠=∠21

21

Or, PQRSRPQTR ∠−∠=∠21

21

……………………………..(2)

As RHS of both equations are same So, following can be written:

QPRQTR ∠=∠21



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