Course / Category - Foundation program for Railways

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Foundation program for Railways

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Maths / Lines and Angles / Alternate Angles

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Alternate Angles

Two angles, formed when a line crosses two other lines, that lie on opposite sides of the transversal line and on opposite relative sides of the other lines. If the two lines crossed are parallel, the alternate angles are equal.

Alternate interior angles: Alternate interior angles are the pair of angles lying in the region between the two lines (intersected by a transversal) and on the opposite sides of the transversal but one below the transversal and the other above the transversal. In the adjoining figure $large (angle 4&angle 5)$ and $(angle 3&angle 6)$ are pair of alternate interior angles.

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

Given: m and n are parallel lines and the transversal l cuts m and n

To Prove : $large angle 3 = angle 6 ;and; angle 4= angle 5$

Proof: Corresponding angles are equal as m and n are parallel lines and l is the transversal

$large angle 1 = angle 5 ;and; angle 2= angle 6;;;;;(1)$

Vertically opposite angles are equal when m and l intersect.

$large angle 1 = angle 4 ;and; angle 2= angle 3 ;;; .......(2)$

From Eq. (1) and (2)

$large angle 5 = angle 4 ;and; angle 6= angle 3$

Hence Proved that alternate interior angles are equal.

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

Given : m and n are lines and the transversal l cuts m and n and $angle 3 = angle 6$

To Prove : m and n are parallel lines

Proof: Vertically opposite angles are equal when m and l intersect.

$angle 2= angle 3 ;;; .......(1)$

and $angle 3 = angle 6;;;;.....(2)$

From Eq. (1) and (2)

$angle 2 = angle 6$

But they are Corresponding angles as m and n are lines and l is the transversal and as they are equal the lines are parallel.

Hence Proved that m and n are parallel lines.

Same is the case with exterior alternate angles

Alternate exterior angles: Alternate exterior angles are the pair of angles lying outside the region between the two lines (intersected by a transversal) and on the opposite sides of the transversal but one below the transversal and the other above the transversal. In the given figure $(angle 1&angle 8)$ and $(angle 2&angle 7)$ are pair of alternate exterior angles.

Illustration: Find the angle which is alternate to : $angle 7,angle 6;and ;angle 1$

Solution:

Angle alternate to $angle 7$ = $angle 2$ , these are alternate exterior angles

Angle alternate to $angle 6$ = $angle 3$ , these are alternate interior angles

Angle alternate to $angle 1$ = $angle 8$ , these are alternate exterior angles

Illustration: In the given figure m || n and $large angle 1:angle 2$ = 5:3. Find the value of $large angle 3$ .

Solution: $angle 1 : angle 2 = 5: 3$

Let the common factor be x

$angle 1 = 5x; and ; angle 2 = 3x$

$angle 1 + angle 2 = 180^0 [ Linear; Pair]$

$5x + 3x = 180^0$

$8x = 180^0$

$x = frac{180^0}{8} = frac{45^0}{2}$

$angle 2 =3frac{45^0}{2}= frac{135^0}{2}$

$angle 2 = angle 3; [ Alternate ;angle ]$

$angle 3 = frac{135^0}{2}$

Illustration: In the given figure, PQ and RS are two mirrors placed parallel to each other . An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path and strikes the mirror RS at C and again reflects back along CD. Prove that AB || CD.

Solution:

Construction: Draw BB' $small perp$ PQ and CC' $small perp$ RS.

Since BB' and CC' are normals, therefore $small angle 1=angle 2$ and $small angle 3=angle 4$ [Angle of incidence = angle of reflection]

Since PQ || RS, therefore BB' || CC'

Now, BB' || CC' and BC is a transversal

$small therefore angle 2=angle 3$

Multiply by 2 on both sides, we have

$small Rightarrow 2 angle 2=2angle 3$

$small Rightarrow angle 2+angle 2=angle 3+angle 3$

$small Rightarrow angle 1+angle 2=angle 3+angle 4$ $small left [ because angle 1=angle 2;and;angle 3=angle 4 right ]$

$small Rightarrow angle ABC=angle BCD$

But, $small angle ABC;and;angle BCD$ are alternate angles formed by transversal BC with AB and CD.

Now, using the theorem "If a transversal intersects two lines such that a pair of alternate interior angles is equal, then the lines are parallel".

Hence, AB || CD.

Hence the proof.

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Alternate Angles

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