Chapter 15: Properties of Triangles Exercise – 15.2
Question: 1
Two angles of a triangle are of measures 150° and 30°. Find the measure of the third angle.
Solution:
Let the third angle be x
Sum of all the angles of a triangle = 180°
105° + 30° + x = 180°
135° + x = 180°
x = 180° – 135°
x = 45°
Therefore the third angle is 45°
Question: 2
One of the angles of a triangle is 130°, and the other two angles are equal. What is the measure of each of these equal angles?
Solution:
Let the second and third angle be x
Sum of all the angles of a triangle = 180°
130° + x + x = 180°
130° + 2x = 180°
2x = 180°– 130°
2x = 50°
x = 50/2
x = 25°
Therefore the two other angles are 25° each
Question: 3
The three angles of a triangle are equal to one another. What is the measure of each of the angles?
Solution:
Let the each angle be x
Sum of all the angles of a triangle =180°
x + x + x = 180°
3x = 180°
x = 180/3
x = 60°
Therefore angle is 60° each
Question: 4
If the angles of a triangle are in the ratio 1: 2: 3, determine three angles.
Solution:
If angles of the triangle are in the ratio 1: 2: 3 then take first angle as ‘x’, second angle as ‘2x’ and third angle as ‘3x’
Sum of all the angles of a triangle=180°
x + 2x + 3x = 180°
6x = 180°
x = 180/6
x = 30°
2x = 30° × 2 = 60°
3x = 30° × 3 = 90°
Therefore the first angle is 30°, second angle is 60° and third angle is 90°
Question: 5
The angles of a triangle are (x − 40) °, (x − 20) ° and (1/2 − 10) °. Find the value of x.
Solution:
Question: 6
The angles of a triangle are arranged in ascending order of magnitude. If the difference between two consecutive angles is 10°. Find the three angles.
Solution:
Let the first angle be x
Second angle be x + 10°
Third angle be x + 10° + 10°
Sum of all the angles of a triangle = 180°
x + x + 10° + x + 10° +10° = 180°
3x + 30 = 180
3x = 180 - 30
3x = 150
x = 150/3
x = 50°
First angle is 50°
Second angle x + 10° = 50 + 10 = 60°
Third angle x + 10° +10° = 50 + 10 + 10 = 70°
Question: 7
Two angles of a triangle are equal and the third angle is greater than each of those angles by 30°. Determine all the angles of the triangle
Solution:
Let the first and second angle be x
The third angle is greater than the first and second by 30° = x + 30°
The first and the second angles are equal
Sum of all the angles of a triangle = 180°
x + x + x + 30° = 180°
3x + 30 = 180
3x = 180 - 30
3x = 150
x = 150/3
x = 50°
Third angle = x + 30° = 50° + 30° = 80°
The first and the second angle is 50° and the third angle is 80°
Question: 8
If one angle of a triangle is equal to the sum of the other two, show that the triangle is a right triangle.
Solution:
One angle of a triangle is equal to the sum of the other two
x = y + z
Let the measure of angles be x, y, z
x + y + z = 180°
x + x = 180°
2x = 180°
x = 180/2
x = 90°
If one angle is 90° then the given triangle is a right angled triangle
Question: 9
If each angle of a triangle is less than the sum of the other two, show that the triangle is acute angled.
Solution:
Each angle of a triangle is less than the sum of the other two
Measure of angles be x, y and z
x > y + z
y < x + z
z < x + y
Therefore triangle is an acute triangle
Question: 10
In each of the following, the measures of three angles are given. State in which cases the angles can possibly be those of a triangle:
(i) 63°, 37°, 80°
(ii) 45°, 61°, 73°
(iii) 59°, 72°, 61°
(iv) 45°, 45°, 90°
(v) 30°, 20°, 125°
Solution:
(i) 63°, 37°, 80° = 180°
Angles form a triangle
(ii) 45°, 61°, 73° is not equal to 180°
Therefore not a triangle
(iii) 59°, 72°, 61° is not equal to 180°
Therefore not a triangle
(iv) 45°, 45°, 90° = 180°
Angles form a triangle
(v) 30°, 20°, 125° is not equal to 180°
Therefore not a triangle
Question: 11
The angles of a triangle are in the ratio 3: 4: 5. Find the smallest angle
Solution:
Given that
Angles of a triangle are in the ratio: 3: 4: 5
Measure of the angles be 3x, 4x, 5x
Sum of the angles of a triangle =180°
3x + 4x + 5x = 180°
12x = 180°
x = 180/12
x = 15°
Smallest angle = 3x
=3 × 15°
= 45°
Question: 12
Two acute angles of a right triangle are equal. Find the two angles.
Solution:
Given acute angles of a right angled triangle are equal
Right triangle: whose one of the angle is a right angle
Measured angle be x, x, 90°
x + x + 180°= 180°
2x = 90°
x = 90/2
x = 45°
The two angles are 45° and 45°
Question: 13
One angle of a triangle is greater than the sum of the other two. What can you say about the measure of this angle? What type of a triangle is this?
Solution:
Angle of a triangle is greater than the sum of the other two
Measure of the angles be x, y, z
x > y + z or
y > x + z or
z > x + y
x or y or z > 90° which is obtuse
Therefore triangle is an obtuse angle
Question: 14
AC, AD and AE are joined. Find
∠FAB + ∠ABC + ∠BCD + ∠CDE + ∠DEF + ∠EFA
∠FAB + ∠ABC + ∠BCD + ∠CDE + ∠DEF + ∠EFA
Solution:
We know that sum of the angles of a triangle is 180°
Therefore in ∆ABC, we have
∠CAB + ∠ABC + ∠BCA = 180° — (i)
In ∆ACD, we have
∠DAC + ∠ACD + ∠CDA = 180° — (ii)
In ∆ADE, we have
∠EAD + ∠ADE + ∠DEA =180° — (iii)
In ∆AEF, we have
∠FAE + ∠AEF + ∠EFA = 180° — (iv)
Adding (i), (ii), (iii), (iv) we get
∠CAB + ∠ABC + ∠BCA + ∠DAC + ∠ACD + ∠CDA + ∠EAD + ∠ADE + ∠DEA + ∠FAE + ∠AEF +∠EFA = 720°
Therefore ∠FAB + ∠ABC + ∠BCD + ∠CDE + ∠DEF + ∠EFA = 720°
Question: 15
Find x, y, z (whichever is required) from the figures given below:
Solution:
(i) In ∆ABC and ∆ADE we have:
∠ADE = ∠ABC (corresponding angles)
x = 40°
∠AED = ∠ACB (corresponding angles)
y = 30°
We know that the sum of all the three angles of a triangle is equal to 180°
x + y + z = 180° (Angles of ∆ADE)
Which means: 40° + 30° + z = 180°
z = 180° - 70°
z = 110°
Therefore, we can conclude that the three angles of the given triangle are 40°, 30° and 110°.
(ii) We can see that in ∆ADC, ∠ADC is equal to 90°.
(∆ADC is a right triangle)
We also know that the sum of all the angles of a triangle is equal to 180°.
Which means: 45° + 90° + y = 180° (Sum of the angles of ∆ADC)
135° + y = 180°
y = 180° – 135°.
y = 45°.
We can also say that in ∆ABC, ∠ABC + ∠ACB + ∠BAC is equal to 180°.
(Sum of the angles of ∆ABC)
40° + y + (x + 45°) = 180°
40° + 45° + x + 45° = 180° (y = 45°)
x = 180° –130°
x = 50°
Therefore, we can say that the required angles are 45° and 50°.
(iii) We know that the sum of all the angles of a triangle is equal to 180°.
Therefore, for △ABD:
∠ABD +∠ADB + ∠BAD = 180° (Sum of the angles of ∆ABD)
50° + x + 50° = 180°
100° + x = 180°
x = 180° – 100°
x = 80°
For ∆ABC:
∠ABC + ∠ACB + ∠BAC = 180° (Sum of the angles of ∆ABC)
50° + z + (50° + 30°) = 180°
50° + z + 50° + 30° = 180°
z = 180° – 130°
z = 50°
Using the same argument for ∆ADC:
∠ADC + ∠ACD + ∠DAC = 180° (Sum of the angles of ∆ADC)
y +z + 30° =180°
y + 50° + 30° = 180° (z = 50°)
y = 180° – 80°
y = 100°
Therefore, we can conclude that the required angles are 80°, 50° and 100°.
(iv) In ∆ABC and ∆ADE we have:
∠ADE = ∠ABC (Corresponding angles)
y = 50°
Also, ∠AED = ∠ACB (Corresponding angles)
z = 40°
We know that the sum of all the three angles of a triangle is equal to 180°.
Which means: x + 50° + 40° = 180° (Angles of ∆ADE)
x = 180° – 90°
x = 90°
Therefore, we can conclude that the required angles are 50°, 40° and 90°.
Question: 16
If one angle of a triangle is 60° and the other two angles are in the ratio 1: 2, find the angles
Solution:
We know that one of the angles of the given triangle is 60°. (Given)
We also know that the other two angles of the triangle are in the ratio 1: 2.
Let one of the other two angles be x.
Therefore, the second one will be 2x.
We know that the sum of all the three angles of a triangle is equal to 180°.
60° + x + 2x = 180°
3x = 180° – 60°
3x = 120°
x = 120/3
x = 40°
2x = 2 × 40
2x = 80°
Hence, we can conclude that the required angles are 40° and 80°.
Question: 17
It one angle of a triangle is 100° and the other two angles are in the ratio 2: 3. find the angles.
Solution:
We know that one of the angles of the given triangle is 100°.
We also know that the other two angles are in the ratio 2: 3.
Let one of the other two angles be 2x.
Therefore, the second angle will be 3x.
We know that the sum of all three angles of a triangle is 180°.
100° + 2x + 3x = 180°
5x = 180° – 100°
5x = 80°
x = 80/5
2x = 2 ×16
2x = 32°
3x = 3×16
3x = 48°
Thus, the required angles are 32° and 48°.
Question: 18
In △ABC, if 3∠A = 4∠B = 6∠C, calculate the angles.
Solution:
We know that for the given triangle, 3∠A = 6∠C
∠A = 2∠C — (i)
We also know that for the same triangle, 4∠B = 6∠C
∠B = (6/4)∠C — (ii)
We know that the sum of all three angles of a triangle is 180°.
Therefore, we can say that:
∠A + ∠B + ∠C = 180° (Angles of ∆ABC) — (iii)
On putting the values of ∠A and ∠B in equation (iii), we get:
2∠C + (6/4)∠C +∠C = 180°
(18/4) ∠C = 180°
∠C = 40°
From equation (i), we have:
∠A = 2∠C = 2 × 40
∠A = 80°
From equation (ii), we have:
∠B = (6/4)∠C = (6/4) × 40°
∠B = 60°
∠A = 80°, ∠B = 60°, ∠C = 40°
Therefore, the three angles of the given triangle are 80°, 60°, and 40°.
Question: 19
Is it possible to have a triangle, in which
(i) Two of the angles are right?
(ii) Two of the angles are obtuse?
(iii) Two of the angles are acute?
(iv) Each angle is less than 60°?
(v) Each angle is greater than 60°?
(vi) Each angle is equal to 60°
Solution:
Give reasons in support of your answer in each case.
(i) No, because if there are two right angles in a triangle, then the third angle of the triangle must be zero, which is not possible.
(ii) No, because as we know that the sum of all three angles of a triangle is always 180°. If there are two obtuse angles, then their sum will be more than 180°, which is not possible in case of a triangle.
(iii) Yes, in right triangles and acute triangles, it is possible to have two acute angles.
(iv) No, because if each angle is less than 60°, then the sum of all three angles will be less than 180°, which is not possible in case of a triangle.
Proof:
Let the three angles of the triangle be ∠A, ∠B and ∠C.
As per the given information,
∠A < 60° … (i)
∠B< 60° … (ii)
∠C < 60° … (iii)
On adding (i), (ii) and (iii), we get:
∠A + ∠B + ∠C < 60°+ 60°+ 60°
∠A + ∠B + ∠C < 180°
We can see that the sum of all three angles is less than 180°, which is not possible for a triangle.
Hence, we can say that it is not possible for each angle of a triangle to be less than 60°.
(v) No, because if each angle is greater than 60°, then the sum of all three angles will be greater than 180°, which is not possible.
Proof:
Let the three angles of the triangle be ∠A, ∠B and ∠C. As per the given information,
∠A > 60° … (i)
∠B > 60° … (ii)
∠C > 60° … (iii)
On adding (i), (ii) and (iii), we get:
∠A + ∠B + ∠C > 60°+ 60°+ 60°
∠A + ∠B + ∠C > 180°
We can see that the sum of all three angles of the given triangle are greater than 180°, which is not possible for a triangle.
Hence, we can say that it is not possible for each angle of a triangle to be greater than 60°.
(vi) Yes, if each angle of the triangle is equal to 60°, then the sum of all three angles will be 180° , which is possible in case of a triangle.
Proof:
Let the three angles of the triangle be ∠A, ∠B and ∠C. As per the given information,
∠A = 60° … (i)
∠B = 60° …(ii)
∠C = 60° … (iii)
On adding (i), (ii) and (iii), we get:
∠A + ∠B + ∠C = 60°+ 60°+ 60°
∠A + ∠B + ∠C =180°
We can see that the sum of all three angles of the given triangle is equal to 180°, which is possible in case of a triangle. Hence, we can say that it is possible for each angle of a triangle to be equal to 60°.
Question: 20
In ∆ABC, ∠A = 100°, AD bisects ∠A and AD perpendicular BC. Find ∠B
Solution:
Consider ∆ABD
∠BAD = 100/2 (AD bisects ∠A)
∠BAD = 50°
∠ADB = 90° (AD perpendicular to BC)
We know that the sum of all three angles of a triangle is 180°.
Thus,
∠ABD + ∠BAD + ∠ADB = 180° (Sum of angles of ∆ABD)
Or,
∠ABD + 50° + 90° = 180°
∠ABD =180° – 140°
∠ABD = 40°
Question: 21
In ∆ABC, ∠A = 50°, ∠B = 100° and bisector of ∠C meets AB in D. Find the angles of the triangles ADC and BDC
Solution:
We know that the sum of all three angles of a triangle is equal to 180°.
Therefore, for the given △ABC, we can say that:
∠A + ∠B + ∠C = 180° (Sum of angles of ∆ABC)
50° + 70° + ∠C = 180°
∠C= 180° –120°
∠C = 60°
∠ACD = ∠BCD =∠C2 (CD bisects ∠C and meets AB in D. )
∠ACD = ∠BCD = 60/2= 30°
Using the same logic for the given ∆ACD, we can say that:
∠DAC + ∠ACD + ∠ADC = 180°
50° + 30° + ∠ADC = 180°
∠ADC = 180°– 80°
∠ADC = 100°
If we use the same logic for the given ∆BCD, we can say that
∠DBC + ∠BCD + ∠BDC = 180°
70° + 30° + ∠BDC = 180°
∠BDC = 180° – 100°
∠BDC = 80°
Thus,
For ∆ADC: ∠A = 50°, ∠D = 100° ∠C = 30°
∆BDC: ∠B = 70°, ∠D = 80° ∠C = 30°
Question: 22
In ∆ABC, ∠A = 60°, ∠B = 80°, and the bisectors of ∠B and ∠C, meet at O. Find
(i) ∠C
(ii) ∠BOC
Solution:
We know that the sum of all three angles of a triangle is 180°.
Hence, for △ABC, we can say that:
∠A + ∠B + ∠C = 180° (Sum of angles of ∆ABC)
60° + 80° + ∠C= 180°.
∠C = 180° – 140°
∠C = 140°.
For △OBC,
∠OBC = ∠B2 = 80/2 (OB bisects ∠B)
∠OBC = 40°
∠OCB =∠C2 = 40/2 (OC bisects ∠C)
∠OCB = 20°
If we apply the above logic to this triangle, we can say that:
∠OCB + ∠OBC + ∠BOC = 180° (Sum of angles of ∆OBC)
20° + 40° + ∠BOC = 180°
∠BOC = 180° – 60°
∠BOC = 120°
Question: 23
The bisectors of the acute angles of a right triangle meet at O. Find the angle at O between the two bisectors.
Solution:
We know that the sum of all three angles of a triangle is 180°.
Hence, for ∆ABC, we can say that:
∠A + ∠B + ∠C = 180°
∠A + 90° + ∠C = 180°
∠A + ∠C = 180° – 90°
∠A + ∠C = 90°
For ∆OAC:
∠OAC = ∠A2 (OA bisects LA)
∠OCA = ∠C2 (OC bisects LC)
On applying the above logic to △OAC, we get:
∠AOC + ∠OAC + ∠OCA = 180° (Sum of angles of ∆AOC)
∠AOC + ∠A2 + ∠C2 = 180°
∠AOC + ∠A + ∠C2 = 180°
∠AOC + 90/2 = 180°
∠AOC = 180° – 45°
∠AOC = 135°
Question: 24
In ∆ABC, ∠A = 50° and BC is produced to a point D. The bisectors of ∠ABC and ∠ACD meet at E. Find ∠E.
Solution:
In the given triangle,
∠ACD = ∠A + ∠B. (Exterior angle is equal to the sum of two opposite interior angles.)
We know that the sum of all three angles of a triangle is 180°.
Therefore, for the given triangle, we can say that:
Question: 25
In ∆ABC, ∠B = 60°, ∠C = 40°, AL perpendicular BC and AD bisects ∠A such that L and D lie on side BC. Find ∠LAD
Solution:
We know that the sum of all angles of a triangle is 180°
Therefore, for ∆ABC, we can say that:
Question: 26
Line segments AB and CD intersect at O such that AC perpendicular DB. It ∠CAB = 35° and ∠CDB = 55°. Find ∠BOD.
Solution:
We know that AC parallel to BD and AB cuts AC and BD at A and B, respectively.
∠CAB = ∠DBA (Alternate interior angles)
∠DBA = 35°
We also know that the sum of all three angles of a triangle is 180°.
Hence, for △OBD, we can say that:
∠DBO + ∠ODB + ∠BOD = 180°
35° + 55° + ∠BOD = 180° (∠DBO = ∠DBA and ∠ODB = ∠CDB)
∠BOD = 180° - 90°
∠BOD = 90°
Question: 27
In Figure, ∆ABC is right angled at A, Q and R are points on line BC and P is a point such that QP perpendicular to AC and RP perpendicular to AB. Find ∠P
Solution:
In the given triangle, AC parallel to QP and BR cuts AC and QP at C and Q, respectively.
∠QCA = ∠CQP (Alternate interior angles)
Because RP parallel to AB and BR cuts AB and RP at B and R, respectively,
∠ABC = ∠PRQ (alternate interior angles).
We know that the sum of all three angles of a triangle is 180°.
Hence, for ∆ABC, we can say that:
∠ABC + ∠ACB + ∠BAC = 180°
∠ABC + ∠ACB + 90° = 180° (Right angled at A)
∠ABC + ∠ACB = 90°
Using the same logic for △PQR, we can say that:
∠PQR + ∠PRQ + ∠QPR = 180°
∠ABC + ∠ACB + ∠QPR = 180° (∠ABC = ∠PRQ and ∠QCA = ∠CQP)
Or,
90°+ ∠QPR =180° (∠ABC+ ∠ACB = 90°)
∠QPR = 90°