Solved Examples

Solved Examples 1. Find the supplementary angles of (a) 80o (b) 90o (c) 100o (d) 150o Sol: (a) Let x o be the supplementary angle of 80o Then x o + 80o = 180o (Since sum of the supplementary angles = 180o ) or x o = 180o − 80o (By transposing 80o to RHS) or x o = 100o Thus, supplementary angle of 80o is 100o (b) Let x o be the supplementary angle of 90o Then x o + 90o = 180o (Since sum of the supplementary angles = 180o ) or x o = 180o − 90o (By transposing 90o to RHS) or x o = 90o Thus, supplementary angle of 90o is 90o (c) Let x o be the supplementary angle of 100o Then x o + 100o = 180o (Since sum of the supplementary angles = 180o ) or x o = 180o − 100o (By transposing 100o to RHS) or x o = 80o Thus, supplementary angle of 100o is 80o (d) Let x o be the supplementary angle of 150o Then x o + 150o = 180o (Since sum of the supplementary angles = 180o )

or x o = 180o − 150o (By transposing 150o to RHS) or x o = 30o Thus, supplementary angle of 150o is 30o 2. Find an angle which is equal to (a) One-half of its complement (b) One-fourth of its supplement Sol: (a) Let x o be the required angle. Then its complement = 90o − x o As the angle is equal to 1 of the complement 2 1 x o = (90o − x o ) 2

By crossing multiplying, we get 2 x o = (90o − x o )

or 2 x o + x o = 90o

(By transposing x o to LHS)

or 3x o = 90o or x o =

90o 3

or x o = 30o Therefore Degree measure of the required angle is 30o (b) Let x o be the required angle. Then its supplement = 180o − x o As the angle is equal to 1 of the supplement 4

1 x o = (180o − x o ) 4 By crossing multiplying, we get

4 x o = (180o − x o ) or 4 x o + x o = 180o

(By transposing x o to LHS)

or 5 x o = 180o or x o =

180o 5

or x o = 36o Therefore Degree measure of the required angle is 36o 3. In figure 3-35, line l II line m and n is a transversal. If ∠2 = 50o . Find ∠1 ?

Sol: In the figure, ∠2 and ∠3 form a linear pair.

Therefore ∠2 + ∠3 = 180o (Linear pair angles are supplementary) or 50o + ∠3 = 180o or ∠3 = 180o − 50o (Transposing 50o to RHS) or ∠3 = 130o Now, angles ∠1 and ∠3 are corresponding angles, and corresponding angles are equal.

∠1 = 130o 4. In figure 3-36, PQ II RS and PT II RU. ∠URV = 90o and ∠QPR = 115o . Find the values of x, y and z . Sol:

Since PQ II RS and ∠115o and ∠x o are consecutive interior angles.

115o + x o = 180o (Consecutive interior angles are supplementary) or ∠x o = 180o − ∠115o or x o = 65o Now SRV is a straight line.

Therefore x o + y o + 90o = 180o (Sum of the angles formed on a straight line is 180o ) or 65o + y o + 90o = 180o or y o + 155o = 180o or y o = 180o − 155o (Transposing 155o to RHS) or y o = 25o

Also PT II RU

Therefore y o + z o = 180o ( y o and z o are consecutive interior angle) or 25o + z o = 180o (Q y o = 25o , found above) or z o = 180o − 25o or z o = 155o Hence z o = 65o , z o = 25o and z o = 155o 5. In figure 3-37 l II m . Find x o ? Sol:

Construct line n parallel to line m . Mark angles ∠a and ∠b as shown in figure. Since l II m and m II n , ∴ l II n (Transitive property) Now l II n , Therefore ∠a = 40o (Alternate angles) Also m II n , Therefore ∠b = 50o (Alternate angles) But ∠BAC = ∠a + ∠b = 40o + 500 = 90o Therefore, Reflex angle BAC = x o = 360o − ∠BAC = 360o − 90o = 270o

6. In figure 3-38, ∠1 = 75o and ∠8 = 95o . Is line l II line m ? Sol:

Since ∠1 and ∠2 form a linear pair

∠1 + ∠2 = 180o or 75o + ∠2 = 180o ( ∠1 = 75o given) or ∠2 = 180o − 75o or ∠2 = 105o Now, ∠2 and ∠8 form a pair of alternate angles, and if l II m , then ∠2 should be equal to ∠8 . But ∠2 = 105o and ∠8 = 95o

Therefore ∠2 ≠ ∠8 Hence, line l

line m (or, lines l and m are not parallel)

7. In figure 3-39, PQ parallel to BC. Find x o ? Sol: PQ parallel to BC and AB is the transversal.

∠PAB = 70o (Alternate angles) Also, PQ parallel to BC and AC is the transversal.

∠QAC = 50o (Alternate angles) Now, angles ∠PAB , x o and ∠QAC form a straight angle at A.

Therefore ∠PAB + x o + ∠QAC = 180o or 70o + x o + 50o = 180o or x o + 120o = 180o or x o = 180o − 120o or x o = 60o 8. In figure 3-40, Find the value of x Sol: Since l parallel to m, Therefore ∠a = 80o (alternate angles) Also, ∠a and ∠x are linear pair angles. Therefore ∠ a + ∠ x = 180 o

or 80o + ∠x = 180o or ∠x = 180o − 80o = 100o

uuur uuur 9. In figure 3-41, OA and OB are opposite rays. If x = 55o . Find the values of y ?

Sol:

Since x = 55o , therefore (2 x)o = 2 × 55o = 110o

Now angles ∠AOC and ∠BOC form a linear pair.

Therefore ∠AOC + ∠BOC = 180o or (2 x)o + ∠BOC = 180o or 110o + ∠BOC = 180o or ∠BOC = 180o − 110o = 70o or ∠BOC = 70o Now, ∠BOC = (3 y + 10)o

Therefore (3 y + 10)o = 70o or 3 y = 70o − 10o or 3 y = 60o or y =

60o 3

or y = 20o Therefore, the value of y is 20o

u uur u uur 10. In figure 3-42, T U p a r a l l e l t o P Q and u uur uuur P Q p a r a l l e l t o R S ∠T = 75o and ∠R = 140o . Find the value of x ? Sol:

uuur uuur Since TU PQ

Therefore ∠UTP and ∠TPQ form a alternate angles. Thus ∠UTP = ∠TPU

∠UTP = ∠x + ∠RPQ ( Since ∠x and ∠RPQ are adjacent angles) or 75o = ∠x + ∠RPQ

… (1)

u uuur u uur Now, R S p a r a l l e l t o P Q and PR is the transversal. ∠SRP + ∠RPQ = 180o

(Interior angels on the same side of the transversal are supplementary)

or 140o + ∠RPQ = 180o or ∠RPQ = 180o − 140o or ∠RPQ = 40o Substituting the value of ∠RPQ = 40o in equation (1), we get

75o = ∠x + 40o or 75o − 40o = ∠x (Transposing 40o to LHS) or 35o = ∠x Therefore, the value of x is 35o

LET US REVISE 1. An angle is formed when two distinct rays originate from a common point. 2. Two angles are called adjacent angles if they have (a) common vertex, (b) a common arm and (c) the other two arms are on the opposite sides of the common arm. 3. When the sum of two adjacent angles equals 180o , they form a linear pair. 4. Two angles are called complementary if their sum equals 90o and supplementary if their sum equals 180o . 5. When two lines intersect, they form four angles and the pair of angles that are not adjacent to each other are called vertically opposite angles. Vertically opposite angels are equal in magnitude. 6. A transversal is a line that intersects two or more lines. 7. If a transversal intersects to parallel lines, then (a) The pairs of corresponding angles are equal. (b) Alternate angles have equal measures. (c) Sum of the measures of consecutive interior angles (interior angles on the same side of the transversal) is equal to 180o . 8. The measure of an acute angle is greater than 0o but less than 90o . 9. The measure of a right angle is 90o . 10. The measure of an obtuse angle is greater than 90o but less than 180o . 11. The measure of a straight line is 180o . 12. The measure of a reflex angle is greater than 180o but less than 360o . 13. The measure of a complete angle is 360o . 14. The sum of angles formed about a point equals to 360o . 15. When two lines intersect, they form four angles and the pair of angles that are not adjacent to each other are called vertically opposite angles. Vertically opposite angels are equal in magnitude. _______________________________________________________