c61Question 41: In the given figure, AB is the diameter of the circle with center O . DO \parallel CB and \angle DCB = 120^o . Calculate (i) \angle DAB (ii) \angle DBA (iii) \angle DBC (iv) \angle ADC . Show that \triangle AOD is an equilateral triangle.

Answer:

(i) ABCD is a cyclic quadrilateral

\therefore \angle DAB + \angle DCB = 180^o

\Rightarrow \angle DAB = 60^o

(ii) \angle ADB = 90^o (angle in the  semicircle)

In \triangle ADO ,

\angle ADO = \angle AOD = 60^o

\therefore \angle ODB = 30^o

In \triangle \angle ODB, OD = OB 

\Rightarrow \angle DBA = 30^o

(iii) \angle ODB = 30^o

\therefore \angle DBC = 30^o (alternate angles)

(iv) \angle ABC = 60^o

\angle ADC + \angle ABC = 180^o (cyclic quadrilateral)

\angle ADC = 180^o-60^o = 120^o

In \triangle AOD ,

OD = OA (Radius of the same circle)

\therefore \angle ADO = \angle AOD = 60^o

Therefore all angles are 60^o . Hence \triangle AOD is an equilateral triangle.

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c62Question 42: In the given figure I is the incenter of the \triangle ABC . BI when produced meets the circumcenter of the the \triangle ABC at D . Given \angle BAC = 55^o and \angle ACB = 65^o . Calculate: (i) \angle DCA (ii) \angle DAC (iii) \angle DCI (iv) \angle AIC

Answer:

(i) IB is bisector of  \angle ABC

\angle ABD = \frac{1}{2} \angle ABC

\angle ABC = 180^o-55^o-65^o=60^o

\therefore \angle ABD = 30^o

(ii) \angle DAC = \angle CBD = 30^o (angles in the same segment)

(iii) \angle ACI = \frac{1}{2} \angle ACB

CI \ bisects \  \angle ACB

\therefore \angle ACI = \frac{65}{2} = 32.5^o

(iv) \angle IAC = \frac{1}{2} \angle BAC  = \frac{55}{2} = 27.5^o

AI bisects \angle BAC

\therefore \angle AIC = 180^o - \angle IAC- \angle ICA = 180^o-27.5^o-32.5^o = 120^o

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c63Question 43: A \triangle ABC is inscribed in a circle. The bisector of \angle BAC, \angle ABC and \angle ACB meet the circumference of the triangle at points P, Q and R respectively. Prove that (i) \angle ABC = 2 \angle APQ (ii) \angle ACB = 2 \angle APR (iii) \angle QPR = 90^o- \frac{1}{2} \angle BAC

Answer:

(i) BQ bisects \angle ABC

\therefore \angle ABQ = \frac{1}{2}  \angle ABC

\angle APQ = \angle ABQ (angle in same segment)

\therefore \angle ABC = 2 \angle ABQ = 2 \angle APQ  … … … … (i)

(ii) CR bisects \angle ACB

\therefore \angle ACR = \frac{1}{2} \angle ABC

and \angle ACR = \angle APR (angles in the same segment)

\therefore \angle ACB = 2 \angle APR … … … … (ii)

(iii) Adding (i) and (ii)  we get

\angle ABC + \angle ACB = 2 (\angle APQ + \angle APR) = 2 \angle QPR

\Rightarrow 180^o - \angle BAC = 2 \angle QPR

\Rightarrow \angle QPR = 90^o-\frac{1}{2} \angle BAC

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c64Question 44: Calculate angles x, y, z if: \frac{x}{3}= \frac{y}{4}=\frac{z}{5}

Answer:

\frac{x}{3}= \frac{y}{4}=\frac{z}{5} = k \ (say)

\Rightarrow x = 3k, y = 4k \ and \ z = 5k

\angle ABC + \angle ADC = 180^o 

\angle PBC = 180^o-3k-4k = 180^o-7k

\therefore ABC = 7k

Similarly \angle ADC = 180^o -(180^o-3k-5k) = 8k

\therefore \angle ABC + \angle ADC = 7k+8k = 15k

15 = 180^o \Rightarrow k = 12

Therefore x = 36^o, y = 48^o \ and \ z = 60^o

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c65Question 45: In the given figure AB = AC = CD and \angle ADC = 38^o , calculate (i) \angle ABC (ii) \angle BEC [1995]

Answer:

(i) AC = CD

\angle CAD = \angle CDA = 38^o

\angle ACD = 180^o-38^o - 38^o = 104^o

\angle ACB = 180^o-104^o = 76^o

AB = AC

\angle ABC = \angle ACB = 76^o

(ii) \angle BAC = 180^o-76^o-76^o = 38^o

\angle BAC = \angle BEC = 38^o (angles in the same segment)

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c66Question 46: In the given figure AC is the diameter of the circle with center O . Chord BD \perp AC . Write down the angles p, q and r in terms of x . [1996]

Answer:

\angle AOB = 2 \angle ACB = 2 \angle ADB (angles subtended by an chord on the center is double that subtended by the same chord on the circumference)

x = 2q \Rightarrow q=\frac{x}{2}

\angle ADB = \frac{x}{2}

\angle ADC = 90^o (angles in a semicircle)

r +\frac{x}{2} = 90^o \Rightarrow r = 90-\frac{x}{2}

Now \angle DAC = \angle DBC (angles in the same segment)

p = 90^o-q \Rightarrow p = 90^o-\frac{x}{2}

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c67Question 47:  In the given figure AC is the diameter of the circle with center O . $CD \parallel BE &s=0$. \angle AOB = 80^o and \angle ACE = 110^o . Calculate (i) \angle BEC (ii) \angle BCD (iii) \angle CED [1998]

Answer:

(i) \angle BOC = 180^o-80^o=100^o (Since AC is a straight line)

\angle BOC = 2 \angle BEC   (angles subtended by an chord on the center is double that subtended by the same chord on the circumference)

\Rightarrow \angle BEC = \frac{100}{2} = 50^o

(ii) DC \parallel EB

\angle DCE = \angle BEC = 50^o

\angle AOB = 80^o (given)

\Rightarrow \angle ACB = \frac{1}{2} \angle AOB = 40^o  (angles subtended by an chord on the center is double that subtended by the same chord on the circumference)

\angle BCD = \angle ACB + \angle ACE + \angle DCE = 40^o+10^o+50^o = 100^o

(iii) \angle BED = 180^o-\angle BCD = 180^o-100^o = 80^o (cyclic quadrilateral)

\angle CED + 50^o = 80^o \Rightarrow \angle CED = 30^o

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c68Question 48: In the given figure, AE is the diameter of the circle. Write down the numerical value of \angle ABC + \angle CDE . Give reasons for your answer. [1998]

Answer:

\angle AOC = \frac{180}{2} = 90^o

\angle AOC = 2 \angle AEC  (angles subtended by an chord on the center is double that subtended by the same chord on the circumference)

\Rightarrow \angle AEC = \frac{90}{2} = 45^o

ABCE is a cyclic quadrilateral

\therefore \angle ABC + \angle AEC = 180 ^o

\Rightarrow \angle ABC = 180-45 = 135^o

Similarly, \angle CDE = 135^o

\therefore \angle ABC + \angle CDE = 135^o + 135^o = 270^o

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c69.jpgQuestion 49: In the given figure AOC is the diameter and AC \parallel ED . If \angle CBE = 64^o , calculate \angle DEC . [1991]

Answer:

\angle ABC = 90^o (angle in a semi circle)

\angle ABE = 90^o-64^o = 26^o

\angle ABE = \angle ACE = 26^o (angles in the same segment)

AC \parallel ED

\therefore \angle DEC = \angle ACE = 26^o (alternate angles)

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c610Question 50: Use the given figure below to find (i) \angle BAD (ii) \angle DQB  [1987]

Answer:

(i) In \triangle ADP 

\angle BAD = 180^o - 85^o - 40^o = 55^o 

(ii) \angle ABC = 180^o - \angle ADC = 180^o - 85^o = 95^o 

\angle AQB = 180^o - 95^o - 55^o = 30^o 

\Rightarrow \angle DQB = 30^o 

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