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Introduction to Trigonometry : Exercise - 8.1 (Mathematics NCERT Class 10th)


               CLICK HERE to watch the second part

 

Q.1     In \Delta ABC, right angled at B, AB = 24 cm, BC = 7 cm. Determine :
           (i) sin A, cos A               (ii) sin C, cos C
Sol.       Let us draw a right \Delta ABC.
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             By using the Pythagoras theorem, we have
             A{C^2} = A{B^2} + B{C^2} = {(24)^2} + {(7)^2}
                                            = 576 + 49 = 625
              \Rightarrow AC = \sqrt {625} = 25cm
             (i) \sin A = {{BC} \over {AC}} = {7 \over {25}}, [Since, \sin \theta = {P \over H}]
              \cos A = {{AB} \over {AC}} = {{24} \over {25}}     [Since,\cos \theta = {B \over H}]

             (ii) \sin C\, = {{AB} \over {AC}} = {{24} \over {25}},
             \cos C\, = {{BC} \over {AC}} = {7 \over {25}}.


Q.2     In adjoining figure, find tan P – cot R.

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Sol.      By using the Pythagoras theorem, we have
            P{R^2} = P{Q^2} + Q{R^2}
             \Rightarrow {13^2} = {12^2} + Q{R^2}
             \Rightarrow Q{R^2} = {13^2}-{12^2} = 169 - 144 = 25
             \Rightarrow QR = \sqrt {25} = 5
            Therefore, \tan P\, = {P \over B} = {{QR} \over {PQ}} = {5 \over {12}} and \cot R\, = {B \over P} = {{QR} \over {PQ}} = {5 \over {12}}
            Hence, tan P – cot R = {5 \over {12}} - {5 \over {12}} = 0


Q.3     If \sin A\, = {3 \over 4}, calculate cos A and tan A.
Sol.       Consider a \Delta ABC in which \angle B = 90^\circ .
              For \angle A, we have
              Base = AB, Perp = BC and Hyp = AC
100
              Therefore, \sin A\, = {P \over H}, = {{BC} \over {AC}} = {3 \over 4}
              Let BC = 3k and AC = 4k.
              Then, AB = \sqrt {A{C^2} - B{C^2}} = \sqrt {{{\left( {4k} \right)}^2} - {{\left( {3k} \right)}^2}}
                             = \sqrt {16{k^2} - 9{k^2}} = \sqrt {7{k^2}} = \sqrt {7k}
              Therefore, \cos A = {B \over H} = {{AB} \over {AC}} = {{\sqrt {7k} } \over {4k}} = {{\sqrt 7 } \over 4}
              \tan A = {P \over B} = {{BC} \over {AB}} = {{3k} \over {\sqrt {7k} }} = {3 \over {\sqrt 7 }}


Q.4     Given 15 cot A = 8, find sin A and sec A.
Sol.       Consider a \Delta ABC in which \angle B = 90^\circ .
              For \angle A, we have
              Base = AB, Perp = BC
              and Hyp = AC

104                
              [Since, 15 cot A = 8  \Rightarrow \cot A = {8 \over {15}} ]
              Let AB = 8k and BC = 15k.
              Then, AC = \sqrt {A{B^2} + B{C^2}} = \sqrt {{{\left( {8k} \right)}^2} + {{\left( {15k} \right)}^2}}
                               = \sqrt {64{k^2} + 225{k^2}} = \sqrt {289{k^2}} = 17k
              Therefore, \sin A = {P \over H} = {{BC} \over {AC}} = {{15k} \over {17k}} = {{15} \over {17}}
              and, \sec A = {H \over B} = {{AC} \over {AB}} = {{17k} \over {8k}} = {{17} \over 8}


Q.5     Given \sec \theta = {{13} \over {12}}, calculate all other trigonometric ratios. 
Sol.       Consider a \Delta ABC in which \angle A = \theta ^\circ and \angle B = 90^\circ
             Then, Base = AB, Perp = BC and Hyp = AC
             Therefore, \sec \theta = {H \over B} = {{AC} \over {AB}} = {{13} \over {12}} 
             Let AC = 13k and AB = 12k. Then,

             BC = \sqrt {A{C^2} - A{B^2}} = \sqrt {{{\left( {13k} \right)}^2} - {{\left( {12k} \right)}^2}}
                    = \sqrt {169{k^2} - 144{k^2}} = \sqrt {25{k^2}} = 5k
             Therefore, \sin \theta = {P \over H} = {{BC} \over {AC}}

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              = {{5k} \over {13k}} = {5 \over {13}}
             \cos \theta = {B \over H} = {{AB} \over {AC}} = {{12k} \over {13k}} = {{12} \over {13}}
             \tan \theta = {P \over B} = {{BC} \over {AB}} = {5 \over {12}}
             \cot \theta = {1 \over {\tan \theta }} = {{12} \over 5}
             \cos ec\theta = {1 \over {\sin \theta }} = {{13} \over 5}


Q.6      If \angle A and \angle B are acute angles such that cos A = cos B, then show that \angle A = \angle B.
Sol.      In rt \Delta ABC, \cos A = {{AC} \over {AB}} = \left[ {{B \over H}} \right]

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            and, \cos B = {{BC} \over {AB}}
            But, cos A = cos B [Given]
             \Rightarrow {{AC} \over {AB}} = {{BC} \over {AB}}
             \Rightarrow AC = BC
            Since, in \Delta ABC, AC = BC
             \Rightarrow \angle A = \angle B [Angles opposite to equal sides are equal]


Q.7     If \cot \theta = {7 \over 8}, evaluate :
           (i)  {{\left( {1 + \sin \theta } \right)\left( {1 - \sin \theta } \right)} \over {\left( {1 + \cos \theta } \right)\left( {1 - \cos \theta } \right)}}
           (ii) {\cot ^2}\theta
Sol.       Consider a \Delta ABC in which \angle A = \theta ^\circ  and \angle B = 90^\circ . Then, Base = AB, Perp = BC and Hyp = AC
5
              Therefore, \cot \theta = {B \over P} = {{AB} \over {BC}} = {7 \over 8}
              Let AB = 7k and BC = 8k.
              Then, AC = \sqrt {B{C^2} + A{B^2}} = \sqrt {{{\left( {8k} \right)}^2} + {{\left( {7k} \right)}^2}}
                                = \sqrt {64{k^2} + 49{k^2}} = \sqrt {113{k^2}} = \sqrt {113k}
              Therefore, \sin \theta = {P \over H} = {{BC} \over {AC}} = {{8k} \over {\sqrt {113k} }} = {8 \over {\sqrt {113} }}
              and \cos \theta = {B \over H} = {{AB} \over {AC}} = {{7k} \over {\sqrt {113k} }} = {7 \over {\sqrt {113} }}

              (i) Therefore, {{\left( {1 + \sin \theta } \right)\left( {1 - \sin \theta } \right)} \over {\left( {1 + \cos \theta } \right)\left( {1 - \cos \theta } \right)}} = {{1 - {{\sin }^2}\theta } \over {1 - {{\cos }^2}\theta }} = {{1 - {{64} \over {113}}} \over {1 - {{49} \over {113}}}}
                                                            = {{113 - 64} \over {113 - 49}} = {{49} \over {64}}

              (ii) {\cot ^2}\theta = {\left( {{7 \over 8}} \right)^2} = {{49} \over {64}}


Q.8     If 3 cot A = 4, check whether {{1 - {{\tan }^2}A} \over {1 + {{\tan }^2}A}} = {\cos ^2} A - {\sin ^2}A or not.
Sol.      Consider a \Delta ABC in which \angle B = 90^\circ .
             For \angle A, we have
             Base = AB, Perp = BC and Hyp = AC

6
             Therefore, \cot A = {B \over P} = {{AB} \over {BC}} = {4 \over 3}       
             Let AB = 4k and BC = 3k [3cot A = 4  \Rightarrow \cot A = {4 \over 3}]

             Then, AC = \sqrt {A{B^2} + B{C^2}} = \sqrt {{{\left( {4k} \right)}^2} + {{\left( {3k} \right)}^2}}
                               = \sqrt {16{k^2} + 9{k^2}} = \sqrt {25{k^2}} = 5k
             Therefore, \sin A = {P \over H} = {{BC} \over {AC}} = {{3k} \over {5k}} = {3 \over 5}
             \cos A = {B \over H} = {{AB} \over {AC}} = {{4k} \over {5k}} = {4 \over 5}
             and, \tan A = {1 \over {\cot A}} = {3 \over 4}
             L.H.S.  = {{1 - {{\tan }^2}A} \over {1 + {{\tan }^2}A}} = {{1 - {9 \over {16}}} \over {1 + {9 \over {16}}}} = {{16 - 9} \over {16 + 9}} = {7 \over {25}}
             R.H.S. = {\cos ^2}A - {\sin ^2}A
              = {\left( {{4 \over 5}} \right)^2} - {\left( {{3 \over 5}} \right)^2} = {{16} \over {25}} - {9 \over {25}} = {7 \over {25}}
               \Rightarrow L.H.S = R.H.S.
              Therefore, {{1 - {{\tan }^2}A} \over {1 + {{\tan }^2}A}} = {\cos ^2}A - {\sin ^2}A


Q.9     In \Delta ABC right angled at B, if tan A = {1 \over {\sqrt 3 }}, find the value of
           (i) sin A cos C + cos A sin C
           (ii) cos A cos C – sin A sin C
Sol.       Consider a \Delta ABC , in which \angle B = 90^\circ .
              For \angle A, we have
              Base = AB, Perp = BC
              and Hyp = AC
              \tan A = {{Perp} \over {Hyp}}
                       = {{BC} \over {AB}} = {1 \over {\sqrt 3 }}
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              Let BC = k and AB = {\sqrt 3 } k .
              Then, AC = \sqrt {A{B^2} + B{C^2}} = \sqrt {3{k^2} + {k^2}}
                                = \sqrt {4{k^2}} = 2k
              Therefore, \sin A = {{Perp} \over {Hyp}} = {{BC} \over {AC}} = {k \over {2k}} = {1 \over 2}
              \cos A = {{Base} \over {Hyp}} = {{AB} \over {AC}} = {{\sqrt {3k} } \over {2k}} = {{\sqrt 3 } \over 2}
              For \angle C, we have
              Base = BC, Perp = AB and Hyp = AC
              Therefore, \sin C = {{Perp} \over {Hyp}} = {{AB} \over {AC}} = {{\sqrt {3k} } \over {2k}} = {{\sqrt 3 } \over 2}
              and, \cos C = {{Base} \over {Hyp}} = {{BC} \over {AC}} = {k \over {2k}} = {1 \over 2}

              (i) sinA cosC + cosA sinC = {1 \over 2} \times {1 \over 2} + {{\sqrt 3 } \over 2} \times {{\sqrt 3 } \over 2}
                                                        = {1 \over 4} + {3 \over 4} = {4 \over 4} = 1

              (ii) cos A cosC – sin A sin C = {{\sqrt 3 } \over 2} \times {1 \over 2} - {1 \over 2} \times {{\sqrt 3 } \over 2} = 0


Q.10     In \Delta PQR, right angled at Q, PR + QR 25 cm and PQ = 5 cm. Determine the values of sin P cos P and tan P.
Sol.         In \Delta PQR, right \angle d at Q,
               PR + QR = 25 cm and PQ = 5cm
               Let QR = x cm

105               Therefore, PR = (25 – x) cm
               By Pythagoras theorem, we have 
               {R{P^2} = R{Q^2} + Q{P^2}}
                \Rightarrow {\left( {25 - x} \right)^2} = {x^2} + {5^2}
                \Rightarrow 625 - 50x + {x^2} = {x^2} + 25
                \Rightarrow - 50x = - 600
                \Rightarrow x = {{ - 600} \over { - 50}} = 12
               Therefore, RQ = 12cm
                \Rightarrow RP = (25 – 12)cm = 13cm
               Now, \sin P = {{RQ} \over {RP}} = {{12} \over {13}}
               \cos P = {{PQ} \over {RP}} = {5 \over {13}}
               and \tan P = {{RQ} \over {PQ}} = {{12} \over 5}


Q.11     State whether the following are true or false. Justify your answer.
             (i) The value of tan A is always less than 1.
             (ii) \sec A = {{12} \over 5} for some value of angle A .
             (iii) cos A is the abbreviation used for the cosecant of angle A.
             (iv) cot A is the product of cot and A.
             (v) \sin \theta = {4 \over 3} for some angle \theta .
Sol.          (i) False because sides of a right triangle may have any length, so tan A may have any value.
                (ii) True as sec A is always greater than 1.
                (iii) False as cos A is the abbreviation used for cosine A.
                (iv) False as cot A is not the product of 'cot' and A. 'cot' separated from A has no meaning.
                (v)  False as sin\theta cannot be > 1.

 



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