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

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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 $\Delta ABC$. 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. 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 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 [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, $\sin \theta = {P \over H} = {{BC} \over {AC}}$ $= {{5k} \over {13k}} = {5 \over {13}}$
$\cos \theta = {B \over H} = {{AB} \over {AC}} = {{12k} \over {13k}} = {{12} \over {13}}$
$\tan \theta = {P \over B} = {{BC} \over {AB}} = {5 \over {12}}$
$\cot \theta = {1 \over {\tan \theta }} = {{12} \over 5}$
$\cos ec\theta = {1 \over {\sin \theta }} = {{13} \over 5}$

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 $\Delta ABC$, $\cos A = {{AC} \over {AB}} = \left[ {{B \over H}} \right]$   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 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 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
$\tan A = {{Perp} \over {Hyp}}$
= ${{BC} \over {AB}} = {1 \over {\sqrt 3 }}$ 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 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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