Alternate Interior Angle

Alternate Interior angle

Q-5 In the figure, if A B C D , ∠ A P Q = 50 ° and ∠ P R D = 127 ° , find x and y.

Triangle poly1 Triangle poly1: Polygon Q, P, R Angle Î± Angle Î±: Angle between F, P, Q Angle Î± Angle Î±: Angle between F, P, Q Angle Î± Angle Î±: Angle between F, P, Q Angle Î² Angle Î²: Angle between P, R, H Angle Î² Angle Î²: Angle between P, R, H Angle Î² Angle Î²: Angle between P, R, H Angle Î³ Angle Î³: Angle between P, Q, R Angle Î³ Angle Î³: Angle between P, Q, R Angle Î³ Angle Î³: Angle between P, Q, R Angle Î´ Angle Î´: Angle between Q, P, R Angle Î´ Angle Î´: Angle between Q, P, R Angle Î´ Angle Î´: Angle between Q, P, R Segment c Segment c: Segment [Q, P] of Triangle poly1 Segment a Segment a: Segment [P, R] of Triangle poly1 Segment b Segment b: Segment [R, Q] of Triangle poly1 Segment f Segment f: Segment [A, B] Segment g Segment g: Segment [C, D] Point Q Q = (-2.5, 0.52) Point Q Q = (-2.5, 0.52) Point Q Q = (-2.5, 0.52) Point P P = (-0.8, 3.25) Point P P = (-0.8, 3.25) Point P P = (-0.8, 3.25) Point R R = (1.46, 0.52) Point R R = (1.46, 0.52) Point R R = (1.46, 0.52) Point A A = (-3.7, 3.28) Point A A = (-3.7, 3.28) Point A A = (-3.7, 3.28) Point B B = (2.28, 3.28) Point B B = (2.28, 3.28) Point B B = (2.28, 3.28) Point C C = (-3.69, 0.53) Point C C = (-3.69, 0.53) Point C C = (-3.69, 0.53) Point D D = (2.71, 0.55) Point D D = (2.71, 0.55) Point D D = (2.71, 0.55) Point E Point E: Point on g Point E Point E: Point on g Point F Point F: Point on f Point F Point F: Point on f Point G Point G: Point on f Point G Point G: Point on f Point H Point H: Point on g Point H Point H: Point on g the AB Parallel CD Also Angle of X and Y

The AB parallel CD also angle of x and y

Solution:

Given,

A B C D , ∠ A P Q = 50 ° and ∠ P R D = 127 °

Now,

x = 50 ° (Alternate interior angles)

∠ P R D + ∠ R P B = 180 ° (Angles on the same side of transversal) ⇒ 127 ° + ∠ R P B = 180 ° ⇒ ∠ R P B = 53 °

Now, y + 50 ° + ∠ R P B = 180 ° (AB is a straight line) ⇒ y + 50 ° + 53 ° = 180 °

⇒ y + 103 ° = 180 ° ⇒ y = 77 °

Q-6 In the 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 BC and strikes the mirror RS at C and again reflects back along CD. Prove that A D C D .

Quadrilateral poly1 Quadrilateral poly1: Polygon C_2, D, E, F Quadrilateral poly2 Quadrilateral poly2: Polygon G, H, I, J Segment e Segment e: Segment [E, F] of Quadrilateral poly1 Segment i Segment i: Segment [I, J] of Quadrilateral poly2 Segment k Segment k: Segment [K, B] Segment k Segment k: Segment [K, B] Segment k Segment k: Segment [K, B] Segment l Segment l: Segment [B, C] Segment l Segment l: Segment [B, C] Segment l Segment l: Segment [B, C] Vector u Vector u: Vector[C, N] Vector u Vector u: Vector[C, N] Point E E = (10.46, 3.04) Point E E = (10.46, 3.04) Point F F = (5.5, 3.04) Point F F = (5.5, 3.04) Point P Point P: Point on e Point P Point P: Point on e Point P Point P: Point on e Point Q Point Q: Point on e Point Q Point Q: Point on e Point Q Point Q: Point on e Point I I = (10.76, 0.1) Point I I = (10.76, 0.1) Point J J = (5.6, 0.1) Point J J = (5.6, 0.1) Point R Point R: Point on i Point R Point R: Point on i Point R Point R: Point on i Point S Point S: Point on i Point S Point S: Point on i Point S Point S: Point on i Point B Point B: Point on e Point B Point B: Point on e Point B Point B: Point on e Point C Point C: Point on i Point C Point C: Point on i Point C Point C: Point on i D text1 = "D" A text3 = "A" Line of PQ and RS It Is a Parallel

Line of PQ and RS it is a parallel the mirror PQ at B,the reflected ray moves along the path BC and strikes the mirror RSat C and again reflects back along CD

Solution:

Draw ray B L ⊥ P Q a n d C M ⊥ R S

Quadrilateral poly1 Quadrilateral poly1: Polygon C_2, D, E, F Quadrilateral poly2 Quadrilateral poly2: Polygon G, H, I, J Angle Î± Angle Î±: Angle between K, B, L Angle Î± Angle Î±: Angle between K, B, L Angle Î² Angle Î²: Angle between L, B, C Angle Î² Angle Î²: Angle between L, B, C Angle Î³ Angle Î³: Angle between B, C, M Angle Î³ Angle Î³: Angle between B, C, M Angle Î´ Angle Î´: Angle between M, C, N Angle Î´ Angle Î´: Angle between M, C, N Segment e Segment e: Segment [E, F] of Quadrilateral poly1 Segment i Segment i: Segment [I, J] of Quadrilateral poly2 Segment k Segment k: Segment [K, B] Segment k Segment k: Segment [K, B] Segment k Segment k: Segment [K, B] Segment l Segment l: Segment [B, C] Segment l Segment l: Segment [B, C] Segment l Segment l: Segment [B, C] Vector u Vector u: Vector[C, N] Vector u Vector u: Vector[C, N] Vector v Vector v: Vector[B, L] Vector v Vector v: Vector[B, L] Vector w Vector w: Vector[C, M] Vector w Vector w: Vector[C, M] Point E E = (10.46, 3.04) Point E E = (10.46, 3.04) Point F F = (5.5, 3.04) Point F F = (5.5, 3.04) Point P Point P: Point on e Point P Point P: Point on e Point P Point P: Point on e Point Q Point Q: Point on e Point Q Point Q: Point on e Point Q Point Q: Point on e Point I I = (10.76, 0.1) Point I I = (10.76, 0.1) Point J J = (5.6, 0.1) Point J J = (5.6, 0.1) Point R Point R: Point on i Point R Point R: Point on i Point R Point R: Point on i Point S Point S: Point on i Point S Point S: Point on i Point S Point S: Point on i Point B Point B: Point on e Point B Point B: Point on e Point B Point B: Point on e Point C Point C: Point on i Point C Point C: Point on i Point C Point C: Point on i D text1 = "D" A text3 = "A" L text2 = "L" M text4 = "M" Two Line PQ and RS at a Parallel

Two line PQ and RS at a parallel

B L ⊥ P Q a n d C M ⊥ R S and P Q R S

∴ B L ∥ C M

∠ L B C = ∠ M C B ………..equation(1) (Alternate interiror Angle)

∠ A B L = ∠ L B C ………….equation(2) (Angle of incidence = Angle of reflection)

∠ M C B = ∠ M C D …………..equation (3) (Angle of incidence = Angle of reflection)

From (1),(2) and (3),we get ∠ A B L = ∠ M C D ………..equation(4)

Adding (1)and(4),we get

∠ L B C + ∠ A B L = ∠ M C B + ∠ M C D

s o , ∠ A B C = ∠ B C D ( ∵ ∠ A B C = ∠ L B C + ∠ A B L a n d ∠ B C D = ∠ M C B + ∠ M C D )

But these are alternate interior angle and they are equal. So, A D C D