Match the following:
(1) (c) Explanation: The given triangle has a right angle. (2) (d) Explanation: The given triangle has an obtuse angle. (3) (a) Explanation: All the sides in the given triangle are equal. (4) (b) Explanation: All the angles in the given triangle are acute. Page No 254:
Based on the sides, classify the following triangles (figures not drawn to the scales): 
(i) It is a scalene triangle since all three sides have different lengths. (ii) It is a scalene triangle since all three sides have different lengths. (iii) It is a scalene triangle since all three sides have different lengths. (iv) It is an isosceles triangle since two of its sides have the same length. (v) It is a scalene triangle since all three sides have different lengths. (vi) It is a scalene triangle since all three sides have different lengths. (vii) It is a scalene triangle since all three sides have different lengths. (viii) It is an equilateral triangle since all of its sides have the same length. (ix) It is an isosceles triangle since two of its sides have the same length. (x) It is an isosceles triangle since two of its sides have the same length. Page No 259:
In a triangle ABC, if ∠A = 55° and ∠B = 40°, find ∠C.
Given: ∠A = 55° and ∠B = 40° To Find: ∠C Solution: In ΔABC, ∠A + ∠B + ∠C = 180° (Sum of the angles of a triangle) 55° + 40° + ∠C = 180° ∠C = 180° − 95° ∴∠C = 85° Page No 259:
In a right angled triangle, if one of the other two angles is 35°, find the remaining angle.
Given: ∠B = 90° and ∠C = 35° To Find: ∠A Solution: In ΔABC, ∠A + ∠B + ∠C = 180° (Sum of the angles of a triangle) ∠A + 90° + 35° = 180° ∠A = 180° − 125° ∴∠C = 55° Page No 259:
If the vertex angle of an isosceles triangle is 50°, find the other angles.
Given: ∠A = 50° and AB = AC To Find: ∠B and ∠C Proof: AB = AC (Given) ∠B = ∠C (Angles opposite to equal sides are equal) In ΔABC, ∠A + ∠B + ∠C = 180° (Sum of the angles of a triangle) 50° + ∠C + ∠C = 180° 2 ∠C = 180° − 50° 2∠C = 130° ∴∠C = 65° ⇒∠B = 65° Page No 259:
The angles of a triangle are in the ratio 1:2:3. Determine the three angles.
Given: ∠A: ∠B: ∠C = x: 2x: 3x To Find: ∠A, ∠B and ∠C Solution: In ΔABC, ∠A + ∠B + ∠C = 180° (Sum of the angles of a triangle) x + 2x + 3x = 180° 6x = 180° ∴x = 30° ∠A = x = 30° ∠B = 2x = 60° ∠C = 3x = 90° Page No 259:
In the adjacent triangle ABC, find the value of x and calculate the measure of all the angles of the triangle.
Given: ∠A = x + 15°, ∠B = x − 15°, ∠C = x + 30° To Find: ∠A, ∠B and ∠C Solution: In ΔABC, ∠A + ∠B + ∠C = 180° (Sum of the angles of a triangle) x +15° + x − 15° + x + 30° = 180° 3x + 30° = 180° 3x = 180° − 30° 3x = 150° ∴ x = 50° ∠A = x + 15° = 50° + 15° = 65° ∠B = x − 15° = 50° − 15° = 35° ∠C = x + 30° = 50° + 30° = 80° Page No 259:
The angles of a triangle are arranged in ascending order of their magnitude. If the difference between two consecutive angles is 10°, find the three angles.
Given: ∠A = x, ∠B = x + 10°, ∠C = x + 20° To Find: ∠A, ∠B and ∠C Solution: In ΔABC, ∠A + ∠B + ∠C = 180° (Sum of the angles of a triangle) x + x +10° + x + 20° = 180° 3x + 30° = 180° 3x = 180° − 30° 3x = 150° ∴ x = 50° ∠A = x = 50° ∠B = x + 10° = 50° +10° = 60° ∠C = x + 20° = 50° + 20° = 70° Page No 262:
The exterior angles obtained on producing the base of a triangle both ways are 104° and 136°. Find the angles of the triangle.
Given: ∠ABD = 104° and ∠ACE = 136° To find: ∠A, ∠B and ∠C Solution: We know that an exterior angle is equal to the sum of its interior opposite angles. ∠2 + ∠3 = ∠ABD ∠2 + ∠3 = 104° ... (1) ∠2 + ∠1 = ∠ACE ∠2 + ∠1 = 136° … (2) Also, ∠1 + ∠2 + ∠3 = 180° (Sum of angles of triangle) 136° + ∠3 = 180° (Using (2)) ∠3 = 180° − 136° ∴∠3 = 44° We have ∠2 + ∠3 = 104° ∠2 + 44° = 104° ∠2 = 104° − 44° ∴∠2 = 60° We have ∠2 + ∠1 = 136° 60° + ∠1 = 136° ∠1 = 136° − 60° ∴∠1 = 76° Thus, the angles of the triangle are Page No 262:
Sides BC, CA and AB of a triangle ABC are produced in an order, forming exterior angles ∠ACD, ∠BAE and ∠CBF. Show that ∠ACD + ∠BAE + ∠CBF = 360°.
Given: ∠ACD, ∠BAE and ∠CBF are exterior angles. To Show: ∠ACD + ∠BAE + ∠CBF = 360° Proof: We know that an exterior angle is equal to the sum of its interior opposite angles. ∠1 + ∠2 = ∠ACD … (1) ∠2 + ∠3 = ∠BAE … (2) ∠3 + ∠1 = ∠CBF … (3) Adding (1), (2) and (3), we get: ∠1 + ∠2 + ∠2 + ∠3 + ∠3 + ∠1 = ∠ACD + ∠BAE + ∠CBF 2(∠1 + ∠2 + ∠3) = ∠ACD + ∠BAE + ∠CBF 2 × 180° = ∠ACD + ∠BAE + ∠CBF (Sum of the angles of a triangle) ∴ ∠ACD + ∠BAE + ∠CBF = 360° Page No 263:
Fill up the blanks to make the following statements true: (a) Sum of the angles of triangle is ___________. (b) An exterior angle of a triangle is equal to the sum of ___________ opposite angles. (c) An exterior angle of a triangle is always ___________ than either of the interior opposite angles. (d) A triangle cannot have more than ___________ right angle. (e) A triangle cannot have more than ___________ obtuse angle.
(a) Sum of the angles of triangle is 180° (b) An exterior angle of a triangle is equal to the sum of interior opposite angles. (c) An exterior angle of a triangle is always greater than either of the interior opposite angles. (d) A triangle cannot have more than one right angle. (e) A triangle cannot have more than one obtuse angle. Page No 263:
Compute the value of x in each of the following figures:
(i) AB = AC (Given) ∠1 = 50° (Angles opposite to equal sides are equal) Now, we have ∠1 + x = 180° (Linear angles) x = 180° − 50° ∴ x = 130° (ii) ∠1 + 106° = 180° (Linear angles) ∠1 = 180° − 106° ∠1 = 74° … (1) ∠2 + 130° = 180° (Linear angles) ∠2 = 180° − 130° ∠2 = 50° … (2) ∠1 + ∠2 + x = 180° (Sum of the angles of a triangle) 74° + 50° + x = 180° (From (1) and (2)) x = 180° − 124° ∴ x = 56° (iii) ∠2 + 100° = 180° (Linear angles) ∠2 = 180° − 100° ∠2 = 80° … (1) ∠1 = 65° … (2) (Vertically opposite angles) ∠1 + ∠2 + x = 180° (Sum of the angles of a triangle) 65° + 80° + x = 180° (From (1) and (2)) x = 180° − 145° ∴ x = 35° (iv) ∠2 + 112° = 180° (Linear angles) ∠2 = 180° − 112° ∠2 = 68° … (1) ∠1 + 120° = 180° (Linear angles) ∠1 = 180° − 120° ∠1 = 60° … (2) ∠1 + ∠2 + x = 180° (Sum of the angles of a triangle) 60° + 68° + x = 180° (From (1) and (2)) x = 180° − 128° ∴ x = 52° (v) AB = AC (Given) ∠1 = 20° (Angles opposite to equal sides are equal) In ΔABC, ∠A + ∠B + ∠C = 180° (Sum of the angles of a triangle) 20° + ∠2 + 20° = 180° (Sum of the angles of a triangle) ∠2 = 180° − 40° ∠2 = 140° Also, ∠2 + x = 180° (Linear angles) x = 180° − 140° ∴ x = 40° Page No 263:
In the figure, QT ⊥ PR, ∠TQR = 40° and ∠SPR = 30°. Find ∠TRS and ∠PSQ
Given: ∠SPR = 30°, ∠TQR = 40° and QT ⊥ PR i.e., ∠QTR = 90° To Find: ∠TRS and ∠PSQ Solution: In ΔQTR, ∠QTR + ∠TQR + ∠TRQ = 180° (Sum of the angles of a triangle) 90° + 40° + ∠TRQ = 180° ∠TRQ = 180° − 130° ∠TRQ = 50° ∴∠TRS = 50° In ΔPSR, ∠SPR + ∠PRS + ∠RSP = 180° (Sum of the angles of a triangle) 30° + 50° + ∠RSP = 180° ∠RSP = 180° − 80° ∠RSP = 100° ∠RSP + ∠PSQ = 180° (Linear angles) 100° + ∠PSQ = 180° ∠PSQ = 180° − 100° ∴∠PSQ = 80° Page No 263:
An exterior angle of a triangle is 120° and one of the interior opposite angles is 30°. Find the other angles of the triangle.
Given: ∠ACD = 120° and ∠ABC = 30° To Find: ∠A and ∠C We know that an exterior angle is equal to the sum of its interior opposite angles. ∠A + ∠B = ∠ACD ∠A + 30° = 120° ∠A = 120° − 30° ∴∠A = 90° Also, ∠ACD + ∠ACB = 180° (Linear angles) 120° + ∠ACB = 180° ∠ACB = 180° − 120° ∴∠ACB = 60° Page No 263:
Find the sum of all the angles at the five vertices of the adjoining star.
The given figure of a star is shown below. To Find: ∠A + ∠B + ∠C + ∠D + ∠E In triangle BEP, ∠B + ∠D = ∠QPD [Exterior angle property of triangles] … (1) In triangle ACQ, ∠A + ∠C = ∠PQD [Exterior angle property of triangles] … (2) Now, in triangle PQD, ∠QPD + ∠PQD + ∠D = 180° [Angle sum property of triangles] ⇒ (∠B + ∠E) + (∠A + ∠C) + ∠D = 180° [Using (1) and (2)] ∴ ∠A + ∠B + ∠C + ∠D + ∠E = 180° Page No 264:
Choose the correct answer from the given alternative: In a triangle ABC, ∠A = 80° and AB = AC, then ∠B is _____________. A. 50° B. 60° C. 40° D. 70°
In a triangle ABC, if ∠A = 80° and AB = AC, then ∠B is 50° Since AB = AC, we get In a triangle ABC, we have: Correct Option: A Page No 264:
Choose the correct answer from the given alternative: In right angled triangle, ∠A is right angle and ∠B = 35°, then ∠C is ___________. A. 65° B. 55° C. 75° D. 45°
In a right-angled triangle, if ∠A is a right angle and ∠B measures 35°, then ∠C is 55° In triangle ABC, we have: Correct Option: B Page No 264:
Choose the correct answer from the given alternative: In a triangle ABC, ∠B = ∠C = 45°, then the triangle is ____________. A. right triangle B. acute angled triangle C. obtuse angle triangle D. equilateral triangle
In a triangle ABC, if ∠B = ∠C = 45°, then the triangle is a right triangle In triangle ABC, we have: Correct Option: A Page No 264:
Choose the correct answer from the given alternative: In an equilateral triangle, each exterior angle is ____________. A. 60° B. 90° C. 120° D. 150°
In an equilateral triangle, each exterior angle is 120° Correct Option: C Page No 264:
Choose the correct answer from the given alternative: Sum of the three exterior angles of a triangle is _____________. A. two right angles B. three right angles C. one right angle D. four right angles
The sum of three exterior angles of triangle is four right angles Correct Option: D Page No 264:
In a triangle ABC, ∠B = 70°, Find ∠A + ∠C.
In ΔABC, ∠A + ∠B + ∠C = 180° (Sum of angles of a triangle) ∠A + 70° + ∠C = 180° (Given ∠B = 70°) ∠A + ∠C = 180° − 70° ∴∠A + ∠C = 110° Page No 264:
In a triangle ABC, ∠A = 110° and AB = AC. Find ∠B and ∠C.
AB = AC (Given) ∠B = ∠C (Angles opposite to equal sides are equal) In ΔABC, ∠A + ∠B + ∠C = 180° (Sum of the angles of a triangle) 110° + ∠B + ∠C = 180° (Given ∠B = 70°) 2∠C = 180° − 110° 2∠C = 70° ∠C = 35° ∴∠B = ∠C = 35° Page No 264:
If three angles of a triangle are in the ratio 2: 3: 5, determine three angles.
Given: ∠A: ∠B: ∠C = 2x: 3x: 5x To Find: ∠A, ∠B and ∠C Solution: In ΔABC, ∠A + ∠B + ∠C = 180° (Sum of the angles of a triangle) 2x + 3x + 5x = 180° 10x = 180° ∴x = 18° ∠A = 2x = 36° ∠B = 3x = 54° ∠C =5x = 90° Page No 264:
The angles of triangle are arranged in ascending order of magnitude. If the difference between two consecutive angles is 15°, find the three angles.
Given: ∠A = x, ∠B = x + 15°, ∠C = x + 30° To Find: ∠A, ∠B and ∠C Solution: In ΔABC, ∠A + ∠B + ∠C = 180° (Sum of the angles of a triangle) x + x +15° + x + 30° = 180° 3x + 45° = 180° 3x = 180° − 45° 3x = 135° ∴x = 45° ∠A = x = 45° ∠B = x + 15° = 45° +15° = 60° ∠C = x + 20° = 45° + 30° = 75° Page No 264:
The sum of two angles of a triangle is equal to its third angle. Determine the measure of the third angle.
Given: ∠A + ∠B = ∠C To Find: ∠C Solution: In ΔABC, ∠A + ∠B + ∠C = 180° (Sum of the angles of a triangle) ∠C + ∠C = 180° 2∠C = 180° ∴∠C = 90° Page No 264:
In a triangle ABC, if 2∠A = 3∠B = 6∠C, determine ∠A, ∠B and ∠C.
Given: 2∠A = 3∠B = 6∠C ⇒ ∠B = To Find: ∠A, ∠B and ∠C Solution: In ΔABC, ∠A + ∠B + ∠C = 180° (Sum of the angles of a triangle) ∠A + 6∠A = 540° ∴∠A = 90° ∠B = ∠B = ∴ ∠B = 60° ∠C = ∠C = ∴∠C = 30° Page No 264:
The angles of triangle are x − 40°, x − 20° and
In ΔABC, ∠A + ∠B + ∠C = 180° (Sum of the angles of a triangle) x − 40° + x − 20° + 2x + 2x + 2x + 2x + 5x = 2 × 225° 5x = 450° x = 90° Page No 265:
In triangle ABC, ∠A − ∠B = 15° and ∠B − ∠C = 30°, find ∠A, ∠B and ∠C.
Subtracting (2) from (1), we get: Subtracting (4) from (3), we get: Page No 265:
The sum of two angles of a triangle is 80° and their difference is 20°. Find the angles of the triangle.
Let the angles of the triangle be x, y and z. Adding the two equations, we get: Page No 265:
In a triangle ABC, ∠B = 60° and ∠C = 80°. Suppose the bisector of ∠B and ∠C meet at I. Find ∠BIC.
According to the given information, triangle ABC can be drawn as shown below. In Page No 265:
In a triangle, each of the smaller angles is half the largest angle. Find the angles.
Let the smaller angles be x. Then the largest angle will be 2x. The measures of the angles are 45°, 45° and 90°. Page No 265:
In a triangle, each of the bigger angles is twice the third angle. Find the angles.
Let each of the bigger angles measure 2x. Then the measure of the smallest angle will be x. The angles measure 36°, 36° and 72°. Page No 265:
In a triangle ABC, ∠B = 50° and ∠A = 60°. Suppose BC is extended to D. Find ∠ACD.
According to the given information, triangle ABC can be drawn as shown below. Page No 265:
In an isosceles triangle, he vertex angle is twice the sum of the base angles. Find the angles of the triangle.
The base angles of an isosceles triangle are equal. Let the measure of each base angle be x. Then the measure of the vertex angle will be 2(x + x) = 2(2x) = 4x. The angles measure 30°, 30° and 120°.
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Is the largest angle in an isosceles triangle is 80 What is the measure of each of the remaining angles?
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