myhelper: RD Sharma solution class 7 chapter 15 Properties of triangles Exercise 15.3

Exercise 15.3

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Question 1:

In Fig., ∠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.

Answer 1:

(i) The interior angle adjacent to exterior ∠ CBX is ∠ ABC . (ii) The interior angles opposite to exterior ∠ CBX are ∠ BAC and ∠ ACB . Also, the interior angles opposite to exterior ∠ BAY are ∠ ABC and ∠ ACB .

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Question 2:

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

Answer 2:

$In ∆ ABC, ∠ A = 50 ° and ∠ B = 55 ° . Because of the angle sum property of the triangle, we can say that : ∠ A + ∠ B + ∠ C = 180 ° \Rightarrow 50 ° + 55 ° + ∠ C = 180 ° Or, ∠ C = 75 ° i . e . ∠ ACB = 75 ° ∠ ACX = 180 ° - ∠ ACB = 180 ° - 75 ° = 105 ° (Linear pair)$

Question 3:

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.

Answer 3:

$We know that the sum of interior opposite angles is equal to the exterior angle . Hence, for the given triangle, we can say that : ∠ ABC + ∠ BAC = ∠ BCO \Rightarrow 55 ° + ∠ BAC = 95 ° Or, ∠ BAC = 95 ° - 55 ° = ∠ BAC = 40 ° We also know that the sum of all angles of a triangle is 180 ° . Hence, for the given △ ABC, we can say that : ∠ ABC + ∠ BAC + ∠ BCA = 180 ° \Rightarrow 55 ° + 40 ° + ∠ BCA = 180 ° Or, ∠ BCA = 180 ° - 95 ° = ∠ BCA = 85 °$

Question 4:

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?

Answer 4:

$Let us assume that A and B are the two interior opposite angles . We know that ∠ A is equal to ∠ B . We also know that the sum of interior opposite angles is equal to the exterior angle . Hence, we can say that : ∠ A + ∠ B = 80 ° Or, ∠ A + ∠ A = 80 ° (∵ ∠ A = ∠ B) 2 ∠ A = 80 ° ∠ A = \frac{80 °}{2} = 40 ° ∠ A = ∠ B = 40 ° Thus, each of the required angles is of 40 ° .$

Question 5:

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

Answer 5:

$In the given figure, ∠ ABE and ∠ ABC form a linear pair . ∴ ∠ ABE + ∠ ABC = 180 ° ∠ ABC = 180 ° - 136 ° ∠ ABC = 44 ° We can also see that ∠ ACD and ∠ ACB form a linear pair . ∴ ∠ ACD + ∠ ACB = 180 ° ∠ ACB = 180 ° - 104 ° ∠ ACB = 76 ° We know that the sum of interior opposite angles is equal to the exterior angle . Therefore, we can say that : ∠ BAC + ∠ ABC = 104 ° ∠ BAC = 104 ° - 44 = 60 ° Thus, ∠ ACB = 76 ° and ∠ BAC = 60 °$

Question 6:

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.

Answer 6:

$In ∆ ABC, ∠ BAC and ∠ EAF are vertically opposite angles . Hence, we can say that : ∠ BAC = ∠ EAF = 45 ° Considering the exterior angle property, we can say that : ∠ BAC + ∠ ABC = ∠ ACD = 105 ° \Rightarrow ∠ ABC = 105 ° - 45 ° = 60 ° Because of the angle sum property of the triangle, we can say that : ∠ ABC + ∠ ACB + ∠ BAC = 180 ° ∠ ACB = 75 ° Therefore, the angles are 45 °, 60 ° and 75 ° .$

Question 7:

In Fig., AC ⊥ CE and ∠A :∠B : ∠C = 3 : 2 : 1, find the value of ∠ECD.

Answer 7:

$In the given triangle, the angles are in the ratio 3 : 2 : 1 . Let the angles of the triangle be 3 x, 2 x and x . Because of the angle sum property of the triangle, we can say that : 3 x + 2 x + x = 180 ° \Rightarrow 6 x = 180 ° Or, x = 30 ° . . . (i) Also, ∠ ACB + ∠ ACE + ∠ ECD = 180 ° x + 90 ° + ∠ ECD = 180 ° (∠ ACE = 90 °) ∠ ECD = 60 ° [From (i)]$

Question 8:

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?

Answer 8:

$Here, Internal angle at A + External angle at A = 180 ° Internal angle at A + 103 ° = 180 ° Internal angle at A = 77 ° Internal angle at B + External angle at B = 180 ° Internal angle at B + 74 ° = 180 ° Internal angle at B = 106 ° Sum of internal angles at A and B = 77 ° + 106 ° = 183 ° It means that the s um of internal angle s at A and B is greater than 180 °, which cannot be possible .$

Question 9:

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

Answer 9:

$We know that the sum of all angles of a triangle is 180 ° . Therefore, for the given ∆ FCB, we can say that : ∠ FCB + ∠ CBF + ∠ BFC = 180 ° \Rightarrow 50 ° + ∠ CBF + 90 ° = 180 ° Or, ∠ CBF = 180 ° - 50 ° - 90 ° = 40 ° . . . (i) Using the above rule for ∆ ABD, we can say that : ∠ ABD + ∠ BDA + ∠ BAD = 180 ° \Rightarrow ∠ BAD = 180 ° - 90 ° - 40 ° = 50 ° [From (i)]$

Question 10:

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

Answer 10:

$Here, ∠ AED + 120 ° = 180 ° (Linear pair) \Rightarrow ∠ AED = 180 ° - 120 ° = 60 ° We know that the sum of all angles of a triangle is 180 ° . Therefore, for ∆ ADE, we can say that : ∠ ADE + ∠ AED + ∠ DAE = 180 ° \Rightarrow 60 ° + ∠ ADE + 30 ° = 180 ° Or, ∠ ADE = 180 ° - 60 ° - 30 ° = 90 ° From the given figure, we can also say that : ∠ FDC + 90 ° = 180 ° (Linear pair) \Rightarrow ∠ FDC = 180 ° - 90 ° = 90 ° Using the above rule for ∆ CDF, we can say that : ∠ CDF + ∠ DCF + ∠ DFC = 180 ° \Rightarrow 90 ° + ∠ DCF + 60 ° = 180 ° ∠ DCF = 180 ° - 60 ° - 90 ° = 30 ° Also, ∠ DCF + x = 180 ° (Linear pair) \Rightarrow 30 ° + x = 180 ° Or, x = 180 ° - 30 ° = 150 °$

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Question 11:

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

(i) ∠BDE
(ii) ∠BCA
(iii) ∠ABC

Answer 11:

$(i) Here, ∠ BAF + ∠ FAD = 180 ° (Linear pair) \Rightarrow ∠ FAD = 180 ° - ∠ BAF = 180 ° - 90 ° = 90 ° Also, ∠ AFE = ∠ ADF + ∠ FAD (Exterior angle property) ∠ ADF + 90 ° = 130 ° ∠ ADF = 130 ° - 90 ° = 40 ° (ii) We know that the sum of all the angles of a triangle is 180 ° . Therefore, for ∆ BDE, we can say that : ∠ BDE + ∠ BED + ∠ DBE = 180 ° . \Rightarrow ∠ DBE = 180 ° - ∠ BDE - ∠ BED = 180 ° - 90 ° - 40 ° = 50 ° . . . (i) Also, ∠ FAD = ∠ ABC + ∠ ACB (Exterior angle property) \Rightarrow 90 ° = 50 ° + ∠ ACB Or, ∠ ACB = 90 ° - 50 ° = 40 ° (iii) ∠ ABC = ∠ DBE = 50 ° [From (i)]$

Question 12:

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

Answer 12:

$Here, ∠ CAX = ∠ DAX (∵ AX bisects ∠ CAD) \Rightarrow ∠ CAX = 70 ° ∠ CAX + ∠ DAX + ∠ CAB = 180 ° 70 ° + 70 ° + ∠ CAB = 180 ° ∠ CAB = 180 ° - 140 ° ∠ CAB = 40 ° ∠ ACB + ∠ CBA + ∠ CAB = 180 ° (Sum of the angles of △ ABC) ∠ ACB + ∠ ACB + 40 ° = 180 ° (∵ ∠ C = ∠ B) 2 ∠ ACB = 180 ° - 40 ° ∠ ACB = \frac{140 °}{2} \Rightarrow ∠ ACB = 70 °$

Question 13:

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.

Answer 13:

$∠ ACD and ∠ ACL make a linear pair . ∴ ∠ ACD + ∠ ACB = 180 ° \Rightarrow 115 ° + ∠ ACB = 180 ° ∠ ACB = 180 ° - 115 ° ∠ ACB = 65 ° We know that the sum of all angles of a triangle is 180 ° . Therefore, for △ ABC, we can say that : ∠ ABC + ∠ BAC + ∠ ACB = 180 ° \Rightarrow 30 ° + ∠ BAC + 65 ° = 180 ° O r, ∠ BAC = 85 ° = ∠ LAC = \frac{∠ BAC}{2} = \frac{85 °}{2} Using the above rule for △ ALC, we can say that : ∠ ALC + ∠ LAC + ∠ ACL = 180 ° \Rightarrow ∠ ALC + \frac{85 °}{2} + 65 ° = 180 ° (∵ ∠ ACL = ∠ ACB) O r, ∠ ALC = 180 ° - \frac{85 °}{2} - 65 ° = ∠ ALC = \frac{145 °}{2} = 72 \frac{1}{2} ° Thus, ∠ A L C = 72 \frac{1}{2} °$

Question 14:

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.

Answer 14:

$∠ ABD and ∠ QBD form a linear pair . ∴ ∠ ABC + ∠ QBC = 180 ° \Rightarrow 60 ° + ∠ QBC = 180 ° ∠ QBC = 120 ° ∠ PDC = ∠ BDQ (Vertically opposite angles) \Rightarrow ∠ BDQ = 15 ° In △ QBD : ∠ QBD + ∠ QDB + ∠ BQD = 180 ° (Sum of angles of △ QBD) 120 ° + 15 ° + ∠ BQD = 180 ° ∠ BQD = 180 ° - 135 ° ∠ BQD = 45 ° ∠ AQD = ∠ BQD = 45 ° In △ AQP : ∠ QAP + ∠ AQP + ∠ APQ = 180 ° (Sum of angles of △ AQP) 80 ° + 45 ° + ∠ APQ = 180 ° ∠ APQ = 55 ° ∠ APD = ∠ APQ$

Question 15:

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

Answer 15:

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)
$From the given figure, we can see that : ∠ ACB + x = 180 ° (Linear pair) \Rightarrow 75 ° + x = 180 ° Or, x = 105 ° We know that the sum of all angles of a triangle is 180 ° . Therefore, for ∆ ABC, we can say that : ∠ BAC + ∠ ABC + ∠ ACB = 180 ° \Rightarrow 40 ° + y + 75 ° = 180 ° Or, y = 65 °$

(ii)
$x + 80 ° = 180 ° (Linear pair) = x = 100 ° In ∆ ABC : x + y + 30 ° = 180 ° (Angle sum property) 100 ° + 30 ° + y = 180 ° = y = 50 °$

$(iii) We know that the sum of all angles of a triangle is 180 ° . Therefore, for ∆ ACD, we can say that : 30 ° + 100 ° + y = 180 ° Or, y = 50 ° ∠ ACB + 100 ° = 180 ° ∠ ACB = 80 ° . . . (i) Using the above rule for ∆ ACB, we can say that : x + 45 ° + 80 ° = 180 ° . = x = 55 ° (iv) We know that the sum of all angles of a triangle is 180 ° . Therefore, for ∆ DBC, we can say that : 30 ° + 50 ° + ∠ DBC = 180 ° ∠ DBC = 100 ° x + ∠ DBC = 180 ° (Linear pair) x = 80 ° And, y = 30 ° + 80 ° = 110 ° (Exterior angle property)$

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Question 16:

Compute the value of x in each of the following figures:

Answer 16:

$(i) From the given figure, we can say that : ∠ ACD + ∠ ACB = 180 ° (Linear pair) Or, ∠ ACB = 180 ° - 112 ° = 68 ° . . . (i) We can also say that : ∠ BAE + ∠ BAC = 180 ° (Linear pair) Or, ∠ BAC = 180 ° - 120 ° = 60 ° . . . (ii) We know that the sum of all angles of a triangle is 180 ° . Therefore, for ∆ ABC : x + ∠ BAC + ∠ ACB = 180 ° \Rightarrow x = 180 ° - 60 ° - 68 ° = 52 ° = x = 52 ° (ii) From the given figure, we can say that : ∠ ABC + 120 ° = 180 ° (Linear pair) \Rightarrow ∠ ABC = 60 ° We can also say that : ∠ ACB + 110 ° = 180 ° (Linear pair) \Rightarrow ∠ ACB = 70 ° We know that the sum of all angles of a triangle is 180 ° . Therefore, for ∆ ABC : x + ∠ ABC + ∠ ACB = 180 ° = x = 50 ° (iii) From the given figure, we can see that : ∠ BAD = ∠ ADC = 52 ° (Alternate angles) We know that the sum of all the angles of a triangle is 180 ° . Therefore, for ∆ DEC : x + 40 ° + 52 ° = 180 ° = x = 88 °$

$(iv) In the given figure, we have a quadrilateral whose sum of all angles is 360 ° . Thus, 35 ° + 45 ° + 50 ° + reflex ∠ ADC = 360 ° Or, reflex ∠ ADC = 230 ° 230 ° + x = 360 ° (A complete angle) = x = 130 °$

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RD Sharma solution class 7 chapter 15 Properties of triangles Exercise 15.3