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

Exercise 15.3

Page-15.19

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:

$$\begin{gathered} \left(\mathrm{i}\right) \mathrm{The} \mathrm{interior} \mathrm{angle} \mathrm{adjacent} \mathrm{to} \mathrm{exterior} \angle \mathrm{CBX} \mathrm{is} \angle \mathrm{ABC}. \\ \left(\mathrm{ii}\right) \mathrm{The} \mathrm{interior} \mathrm{angles} \mathrm{opposite} \mathrm{to} \mathrm{exterior} \angle \mathrm{CBX} \mathrm{are} \angle \mathrm{BAC} \mathrm{and} \angle \mathrm{ACB}. \\ \mathrm{Also}, \mathrm{the} \mathrm{interior} \mathrm{angles} \mathrm{opposite} \mathrm{to} \mathrm{exterior} \angle \mathrm{BAY} \mathrm{are} \angle \mathrm{ABC} \mathrm{and} \angle \mathrm{ACB}. \end{gathered}$$

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Question 2:

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

Answer 2:

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

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:

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

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:

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

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:

$$\begin{gathered} \mathrm{In} \mathrm{the} \mathrm{given} \mathrm{figure}, \angle \mathrm{ABE} \mathrm{and} \angle \mathrm{ABC} \mathrm{form} \mathrm{a} \mathrm{linear} \mathrm{pair}. \\ \therefore \angle \mathrm{ABE} + \angle \mathrm{ABC}=180° \\ \angle \mathrm{ABC}=180°-136° \\ \mathbf{\angle }\mathbf{ABC}\mathbf{=}\mathbf{44}\mathbf{°} \\ \mathrm{We} \mathrm{can} \mathrm{also} \mathrm{see} \mathrm{that} \angle \mathrm{ACD} \mathrm{and} \angle \mathrm{ACB} \mathrm{form} \mathrm{a} \mathrm{linear} \mathrm{pair}. \\ \therefore \angle \mathrm{ACD} + \angle \mathrm{ACB}=180° \\ \angle \mathrm{ACB}=180°-104° \\ \mathbf{\angle }\mathbf{ACB}\mathbf{=}\mathbf{76}\mathbf{°} \\ \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{interior} \mathrm{opposite} \mathrm{angles} \mathrm{is} \mathrm{equal} \mathrm{to} \mathrm{the} \mathrm{exterior} \mathrm{angle}. \\ \mathrm{Therefore}, \mathrm{we} \mathrm{can} \mathrm{say} \mathrm{that}: \\ \angle \mathrm{BAC}+\angle \mathrm{ABC}=104° \\ \mathbf{\angle }\mathbf{BAC}\mathbf{=}\mathbf{104}\mathbf{°}\mathbf{-}\mathbf{44}\mathbf{=}\mathbf{60}\mathbf{°} \\ \mathrm{Thus}, \\ \angle \mathrm{ACB} = 76° \\ \mathrm{and} \\ \angle \mathrm{BAC}\hspace{0.17em}= 60° \end{gathered}$$

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:

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

Question 7:

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

Answer 7:

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

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:

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

Question 9:

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

Answer 9:

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

Question 10:

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

Answer 10:

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

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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:

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

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:

$$\begin{gathered} \mathrm{Here}, \\ \angle \mathrm{CAX} =\angle \mathrm{DAX} (\because \mathrm{AX} \mathrm{bisects} \angle \mathrm{CAD}) \\ \Rightarrow \angle \mathrm{CAX}=70° \\ \angle \mathrm{CAX} +\angle \mathrm{DAX} + \angle \mathrm{CAB} =180° \\ 70° +70°+ \angle \mathrm{CAB} =180° \\ \angle \mathrm{CAB} =180°-140° \\ \angle \mathrm{CAB} =40° \\ \angle \mathrm{ACB} +\angle \mathrm{CBA} + \angle \mathrm{CAB} =180° (\mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{angles} \mathrm{of} △\mathrm{ABC}) \\ \angle \mathrm{ACB} +\angle \mathrm{ACB}+ 40° =180° (\because \angle \mathrm{C}=\angle \mathrm{B}) \\ 2\angle \mathrm{ACB}=180°-40° \\ \angle \mathrm{ACB}=\frac{140°}{2} \\ \mathbf{\Rightarrow }\mathbf{\angle }\mathbf{ACB}\mathbf{=}\mathbf{70}\mathbf{°} \end{gathered}$$

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:

$$\begin{gathered} \angle \mathrm{ACD} \mathrm{and} \angle \mathrm{ACL} \mathrm{make} \mathrm{a} \mathrm{linear} \mathrm{pair}. \\ \therefore \angle \mathrm{ACD} + \angle \mathrm{ACB}=180° \\ \Rightarrow 115°+ \angle \mathrm{ACB}=180° \\ \angle \mathrm{ACB}=180°-115° \\ \angle \mathrm{ACB}=65° \\ \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{is} 180°. \\ \mathrm{Therefore}, \mathrm{for} △\mathrm{ABC}, \mathrm{we} \mathrm{can} \mathrm{say} \mathrm{that}: \\ \angle \mathrm{ABC}+\angle \mathrm{BAC}+\angle \mathrm{ACB}=180° \\ \Rightarrow 30°+\angle \mathrm{BAC}+65°=180° \\ Or, \\ \angle \mathrm{BAC}=85° \\ =\angle \mathrm{LAC}=\frac{\angle \mathrm{BAC}}{2}=\frac{85°}{2} \\ \mathrm{Using} \mathrm{the} \mathrm{above} \mathrm{rule} \mathrm{for} △\mathrm{ALC}, \mathrm{we} \mathrm{can} \mathrm{say} \mathrm{that}: \\ \angle \mathrm{ALC}+\angle \mathrm{LAC}+\angle \mathrm{ACL}=180° \\ \Rightarrow \angle \mathrm{ALC}+\frac{85°}{2}+65°=180° (\because \angle \mathrm{ACL}=\angle \mathrm{ACB}) \\ Or, \\ \angle \mathrm{ALC}=180°-\frac{85°}{2}-65° \\ = \mathbf{\angle }\mathbf{ALC}\mathbf{=}\frac{\mathbf{145}\mathbf{°}}{\mathbf{2}}\mathbf{=}\mathbf{72}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{°} \\ \mathrm{Thus}, \\ \angle ALC = 72\frac{1}{2}° \end{gathered}$$

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:

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

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

(ii)
$$\begin{gathered} x + 80° =180° (\mathrm{Linear} \mathrm{pair}) \\ =\mathit{ }x = 100° \\ \mathrm{In} ∆\mathrm{ABC}: \\ x+\mathit{ }y+ 30° = 180° (\mathrm{Angle} \mathrm{sum} \mathrm{property}) \\ 100°+30°+y = 180° \\ = y = 50° \end{gathered}$$

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

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Question 16:

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

Answer 16:

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

$$\begin{gathered} \left(\mathrm{iv}\right) \mathrm{In} \mathrm{the} \mathrm{given} \mathrm{figure}, \mathrm{we} \mathrm{have} \mathrm{a} \mathrm{quadrilateral} \mathrm{whose} \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{angles} \mathrm{is} 360°. \\ \mathrm{Thus}, \\ 35° + 45° + 50° + \mathrm{reflex} \angle \mathrm{ADC} = 360° \\ \mathrm{Or}, \\ \mathrm{reflex} \angle \mathrm{ADC} = 230° \\ 230°+ \mathrm{x}= 360° (\mathrm{A} \mathrm{complete} \mathrm{angle}) \\ = \mathrm{x} = 130° \end{gathered}$$

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