∠CBX is an exterior angle of ∆ABC at B. Name

(i) the interior adjacent angle

(ii) the interior opposite angles to exterior ∠CBX

Also, name the interior opposite angles to an exterior angle at A.

(i) ∠ABC

(ii) ∠BAC and ∠ACB

Also the interior angles opposite to exterior are ∠ABC and ∠ACB

In the fig, two of the angles are indicated. What are the measures of ∠ACX and ∠ACB?

In ∆ABC, ∠A = 50° and ∠B = 55°

Because of the angle sum property of the triangle, we can say that

∠A + ∠B + ∠C = 180°

50°+ 55°+ ∠C = 180°

Or

∠C = 75°

∠ACB = 75°

∠ACX = 180°− ∠ACB = 180°−75° = 105°

In a triangle, an exterior angle at a vertex is 95° and its one of the interior opposite angles is 55°. Find all the angles of the triangle.

We know that the sum of interior opposite angles is equal to the exterior angle.

Hence, for the given triangle, we can say that:

∠ABC+ ∠BAC = ∠BCO

55° + ∠BAC = 95°

Or,

∠BAC= 95°– 95°

= ∠BAC = 40°

We also know that the sum of all angles of a triangle is 180°.

Hence, for the given △ABC, we can say that:

∠ABC + ∠BAC + ∠BCA = 180°

55° + 40° + ∠BCA = 180°

Or,

∠BCA = 180° –95°

= ∠BCA = 85°

One of the exterior angles of a triangle is 80°, and the interior opposite angles are equal to each other. What is the measure of each of these two angles?

Let us assume that A and B are the two interior opposite angles.

We know that ∠A is equal to ∠B.

We also know that the sum of interior opposite angles is equal to the exterior angle.

Hence, we can say that:

∠A + ∠B = 80°

Or,

∠A +∠A = 80° (∠A= ∠B)

2∠A = 80°

∠A = 40/2 =40°

∠A= ∠B = 40°

Thus, each of the required angles is of 40°.

The exterior angles, obtained on producing the base of a triangle both ways are 104° and 136°. Find all the angles of the triangle.

In the given figure, ∠ABE and ∠ABC form a linear pair.

∠ABE + ∠ABC =180°

∠ABC = 180°– 136°

∠ABC = 44°

We can also see that ∠ACD and ∠ACB form a linear pair.

∠ACD + ∠ACB = 180°

∠AUB = 180°– 104°

∠ACB = 76°

We know that the sum of interior opposite angles is equal to the exterior angle.

Therefore, we can say that:

∠BAC + ∠ABC = 104°

∠BAC = 104°– 44° = 60°

Thus,

∠ACE = 76° and ∠BAC = 60°

In Fig, the sides BC, CA and BA of a ∆ABC have been produced to D, E and F respectively. If ∠ACD = 105° and ∠EAF = 45°; find all the angles of the ∆ABC

In a ∆ABC, ∠BAC and ∠EAF are vertically opposite angles.

Hence, we can say that:

∠BAC = ∠EAF = 45°

Considering the exterior angle property, we can say that:

∠BAC + ∠ABC = ∠ACD = 105°

∠ABC = 105°– 45° = 60°

Because of the angle sum property of the triangle, we can say that:

∠ABC + ∠ACS +∠BAC = 180°

∠ACB = 75°

Therefore, the angles are 45°, 65° and 75°.

In Figure, AC perpendicular to CE and C ∠A: ∠B: ∠C= 3: 2: 1. Find the value of ∠ECD.

In the given triangle, the angles are in the ratio 3: 2: 1.

Let the angles of the triangle be 3x, 2x and x.

Because of the angle sum property of the triangle, we can say that:

3x + 2x + x = 180°

6x = 180°

Or,

x = 30° … (i)

Also, ∠ACB + ∠ACE + ∠ECD = 180°

x + 90° + ∠ECD = 180° (∠ACE = 90°)

∠ECD = 60° [From (i)]

A student when asked to measure two exterior angles of ∆ABC observed that the exterior angles at A and B are of 103° and 74° respectively. Is this possible? Why or why not?

Here,

Internal angle at A + External angle at A = 180°

Internal angle at A + 103° =180°

Internal angle at A = 77°

Internal angle at B + External angle at B = 180°

Internal angle at B + 74° = 180°

Internal angle at B = 106°

Sum of internal angles at A and B = 77° + 106° =183°

It means that the sum of internal angles at A and B is greater than 180°, which cannot be possible.

In Figure, AD and CF are respectively perpendiculars to sides BC and AB of ∆ABC. If ∠FCD = 50°, find ∠BAD

We know that the sum of all angles of a triangle is 180°

Therefore, for the given ∆FCB, we can say that:

∠FCB + ∠CBF + ∠BFC = 180°

50° + ∠CBF + 90°= 180°

Or,

∠CBF = 180° – 50°– 90° = 40° … (i)

Using the above rule for ∆ABD, we can say that:

∠ABD + ∠BDA + ∠BAD = 180°

∠BAD = 180° – 90°– 40° = 50° [from (i)]

In Figure, measures of some angles are indicated. Find the value of x.

Here,

∠AED + 120° = 180° (Linear pair)

∠AED = 180°– 120° = 60°

We know that the sum of all angles of a triangle is 180°.

Therefore, for △ADE, we can say that:

∠ADE + ∠AED + ∠DAE = 180°

60°+ ∠ADE + 30° =180°

Or,

∠ADE = 180°– 60°– 30° = 90°

From the given figure, we can also say that:

∠FDC + 90° = 180° (Linear pair)

∠FDC = 180°– 90° = 90°

Using the above rule for △CDF, we can say that:

∠CDF + ∠DCF + ∠DFC = 180°

90° + ∠DCF + 60° =180°

∠DCF = 180°−60°− 90°= 30°

Also,

∠DCF + x = 180° (Linear pair)

30° + x = 180°

Or,

x = 180°– 30° = 150°

In Figure, ABC is a right triangle right angled at A. D lies on BA produced and DE perpendicular to BC intersecting AC at F. If ∠AFE = 130°, find

(i) ∠BDE

(ii) ∠BCA

(iii) ∠ABC

(i) Here,

∠BAF + ∠FAD = 180° (Linear pair)

∠FAD = 180°- ∠BAF = 180°– 90° = 90°

Also,

∠AFE = ∠ADF + ∠FAD (Exterior angle property)

∠ADF + 90° = 130°

∠ADF = 130°− 90° = 40°

(ii) We know that the sum of all the angles of a triangle is 180°.

Therefore, for ∆BDE, we can say that:

∠BDE + ∠BED + ∠DBE = 180°.

∠DBE = 180°– ∠BDE ∠BED = 180°− 90°− 40°= 50° — (i)

Also,

∠FAD = ∠ABC + ∠ACB (Exterior angle property)

90° = 50° + ∠ACB

Or,

∠ACB = 90°– 50° = 40°

(iii) ∠ABC = ∠DBE = 50° [From (i)]

ABC is a triangle in which ∠B = ∠C and ray AX bisects the exterior angle DAC. If ∠DAX = 70°. Find ∠ACB.

Here,

∠CAX = ∠DAX (AX bisects ∠CAD)

∠CAX =70°

∠CAX +∠DAX + ∠CAB =180°

70°+ 70° + ∠CAB =180°

∠CAB =180° –140°

∠CAB =40°

∠ACB + ∠CBA + ∠CAB = 180° (Sum of the angles of ∆ABC)

∠ACB + ∠ACB+ 40° = 180° (∠C = ∠B)

2∠ACB = 180°– 40°

∠ACB = 140/2

∠ACB = 70°

The side BC of ∆ABC is produced to a point D. The bisector of ∠A meets side BC in L. If ∠ABC= 30° and ∠ACD = 115°, find ∠ALC

∠ACD and ∠ACL make a linear pair.

∠ACD+ ∠ACB = 180°

115° + ∠ACB =180°

∠ACB = 180°– 115°

∠ACB = 65°

We know that the sum of all angles of a triangle is 180°.

Therefore, for ∆ABC, we can say that:

∠ABC + ∠BAC + ∠ACB = 180°

30° + ∠BAC + 65° = 180°

Or,

∠BAC = 85°

∠LAC = ∠BAC/2 = 85/2

Using the above rule for ∆ALC, we can say that:

∠ALC + ∠LAC + ∠ACL = 180°

D is a point on the side BC of ∆ABC. A line PDQ through D, meets side AC in P and AB produced at Q. If ∠A = 80°, ∠ABC = 60° and ∠PDC = 15°, find

(i) ∠AQD

(ii) ∠APD

∠ABD and ∠QBD form a linear pair.

∠ABC + ∠QBC =180°

60° + ∠QBC = 180°

∠QBC = 120°

∠PDC = ∠BDQ (Vertically opposite angles)

∠BDQ = 75°

In ∆QBD:

∠QBD + ∠QDB + ∠BDQ = 180° (Sum of angles of ∆QBD)

120°+ 15° + ∠BQD = 180°

∠BQD = 180°– 135°

∠BQD = 45°

∠AQD = ∠BQD = 45°

In ∆AQP:

∠QAP + ∠AQP + ∠APQ = 180° (Sum of angles of ∆AQP)

80° + 45° + ∠APQ = 180°

∠APQ = 55°

∠APD = ∠APQ

Explain the concept of interior and exterior angles and in each of the figures given below. Find x and y

The interior angles of a triangle are the three angle elements inside the triangle.

The exterior angles are formed by extending the sides of a triangle, and if the side of a triangle is produced, the exterior angle so formed is equal to the sum of the two interior opposite angles.

Using these definitions, we will obtain the values of x and y.

(I) From the given figure, we can see that:

∠ACB + x = 180° (Linear pair)

75°+ x = 180°

Or,

x = 105°

We know that the sum of all angles of a triangle is 180°.

Therefore, for ∆ABC, we can say that:

∠BAC+ ∠ABC +∠ACB = 180°

40°+ y +75° = 180°

Or,

y = 65°

(ii) x + 80°= 180° (Linear pair)

= x = 100°

In ∆ABC:

x+ y+ 30° = 180° (Angle sum property)

100° + 30° + y = 180°

= y = 50°

(iii) We know that the sum of all angles of a triangle is 180°.

Therefore, for ∆ACD, we can say that:

30° + 100° + y = 180°

Or,

y = 50°

∠ACB + 100° = 180°

∠ACB = 80° … (i)

Using the above rule for ∆ACD, we can say that:

x + 45° + 80° = 180°

= x = 55°

(iv) We know that the sum of all angles of a triangle is 180°.

Therefore, for ∆DBC, we can say that:

30° + 50° + ∠DBC = 180°

∠DBC = 100°

x + ∠DBC = 180° (Linear pair)

x = 80°

And,

y = 30° + 80° = 110° (Exterior angle property)

Compute the value of x in each of the following figures

(i) From the given figure, we can say that:

∠ACD + ∠ACB = 180° (Linear pair)

Or,

∠ACB = 180°– 112° = 68°

We can also say that:

∠BAE + ∠BAC = 180° (Linear pair)

Or,

∠BAC = 180°– 120° = 60°

We know that the sum of all angles of a triangle is 180°.

Therefore, for ∆ABC:

x + ∠BAC + ∠ACB = 180°

x = 180°– 60°– 68° = 52°

= x = 52°

(ii) From the given figure, we can say that:

∠ABC + 120° = 180° (Linear pair)

∠ABC = 60°

We can also say that:

∠ACB+ 110° = 180° (Linear pair)

∠ACB = 70°

We know that the sum of all angles of a triangle is 180°.

Therefore, for ∆ABC:

x + ∠ABC + ∠ACB = 180°

x = 50°

(iii) From the given figure, we can see that:

∠BAD = ∠ADC = 52° (Alternate angles)

We know that the sum of all the angles of a triangle is 180°.

Therefore, for ∆DEC:

x + 40°+ 52° = 180°

= x = 88°

(iv) In the given figure, we have a quadrilateral whose sum of all angles is 360°.

Thus,

35° + 45° + 50° + reflex ∠ADC = 360°

Or,

reflex ∠ADC = 230°

230° + x = 360° (A complete angle)

= x = 130°

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