Mathematics Part ii Solutions Solutions for Class 9 Maths Chapter 8 - Trigonometry

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Page No 104:

Question 1:

In the given Fig, $∠$ R is the right angle of $∆$ PQR. Write the following ratios.
(i) sin P (ii) cos Q (iii) tan P (iv) tan Q

Answer:

(i) sinP = $\frac{Opposite side of ∠ P}{Hypotenuse} = \frac{QR}{PQ}$

(ii) cosQ = $\frac{Adjacent side of ∠ Q}{Hypotenuse} = \frac{QR}{PQ}$

(iii) tanP = $\frac{Opposite side of ∠ P}{Adjacent side of ∠ P} = \frac{QR}{PR}$

(iv) tanQ = $\frac{Opposite side of ∠ Q}{Adjacent side of ∠ Q} = \frac{PR}{QR}$

Page No 104:

Question 2:

In the right angled $∆$ XYZ, $∠$ XYZ = 90^° and a, b, c are the lengths of the sides as shown in the figure. Write the following ratios,
(i) sin X (ii) tan Z (iii) cos X (iv) tan X.

Answer:

(i) sinX = $\frac{Opposite side of ∠ X}{Hypotenuse} = \frac{YZ}{XZ} = \frac{a}{c}$

(ii) tanZ = $\frac{Opposite side of ∠ Z}{Adjacent side of ∠ Z} = \frac{XY}{YZ} = \frac{b}{a}$

(iii) cosX = $\frac{Adjacent side of ∠ X}{Hypotenuse} = \frac{XY}{XZ} = \frac{b}{c}$

(iv) tanX = $\frac{Opposite side of ∠ X}{Adjacent side of ∠ X} = \frac{YZ}{XY} = \frac{a}{b}$

Page No 104:

Question 3:

In right angled

∆

LMN,

∠

LMN = 90^°

∠

L = 50^° and

∠

N = 40^°, write the following ratios.

(i) sin 50^°(ii) cos 50^°(iii) tan 40^° (iv) cos 40^°

Answer:

(i) sin50º = $\frac{Opposite side of ∠ L}{Hypotenuse} = \frac{MN}{LN}$

(ii) cos50º = $\frac{Adjacent side of ∠ L}{Hypotenuse} = \frac{LM}{LN}$

(iii) tan40º = $\frac{Opposite side of ∠ N}{Adjacent side of ∠ N} = \frac{LM}{MN}$

(iv) cos40º = $\frac{Adjacent side of ∠ N}{Hypotenuse} = \frac{MN}{LN}$

Page No 104:

Question 4:

In the given figure, $∠$ PQR = 90^°, $∠$ PQS = 90^°, $∠$ PRQ = $α$ and $∠$ QPS = $θ$ Write the following trigonometric ratios.
(i) sin $α$ , cos $α$ , tan $α$
(ii) sin $θ$ , cos $θ$ , tan $θ$

Answer:

(i)
sin $α$ = $\frac{Opposite side of ∠ PRQ}{Hypotenuse} = \frac{PQ}{PR}$

cos $α$ = $\frac{Adjacent side of ∠ PRQ}{Hypotenuse} = \frac{RQ}{PR}$

tan $α$ = $\frac{Opposite side of ∠ PRQ}{Adjacent side of ∠ PRQ} = \frac{PQ}{RQ}$

(ii)
sinθ = $\frac{Opposite side of ∠ QPS}{Hypotenuse} = \frac{QS}{PS}$

cosθ = $\frac{Adjacent side of ∠ QPS}{Hypotenuse} = \frac{PQ}{PS}$

tanθ = $\frac{Opposite side of ∠ QPS}{Adjacent side of ∠ QPS} = \frac{QS}{PQ}$

Page No 112:

Question 1:

In the following table, a ratio is given in each column. Find the remaining two ratios in the column and complete the table.

sin $θ$		$\frac{11}{61}$		$\frac{1}{2}$				$\frac{3}{5}$
cos $θ$	$\frac{35}{37}$				$\frac{1}{\sqrt{3}}$
tan $θ$			1			$\frac{21}{20}$	$\frac{8}{15}$		$\frac{1}{2 \sqrt{2}}$

Answer:

(i)
$\cos θ = \frac{35}{37}$

Now,

$\sin^{2} θ + \cos^{2} θ = 1 \Rightarrow \sin^{2} θ + {(\frac{35}{37})}^{2} = 1 \Rightarrow \sin^{2} θ = 1 - \frac{1225}{1369} = \frac{1369 - 1225}{1369} = \frac{144}{1369} \Rightarrow \sin^{2} θ = {(\frac{12}{37})}^{2} \Rightarrow \sin θ = \frac{12}{37}$
$∴ \tan θ = \frac{\sin θ}{\cos θ} = \frac{\frac{12}{37}}{\frac{35}{37}} = \frac{12}{35}$

(ii)
$\sin θ = \frac{11}{61}$

Now,

$\sin^{2} θ + \cos^{2} θ = 1 \Rightarrow {(\frac{11}{61})}^{2} + \cos^{2} θ = 1 \Rightarrow \cos^{2} θ = 1 - \frac{121}{3721} = \frac{3721 - 121}{3721} = \frac{3600}{3721} \Rightarrow \cos^{2} θ = {(\frac{60}{61})}^{2} \Rightarrow \cos θ = \frac{60}{61}$
$∴ \tan θ = \frac{\sin θ}{\cos θ} = \frac{\frac{11}{61}}{\frac{60}{61}} = \frac{11}{60}$

(iii)
$\tan θ = 1 \Rightarrow \frac{\sin θ}{\cos θ} = \frac{1}{1} \Rightarrow \sin θ = k, \cos θ = k$
Now,

$\sin^{2} θ + \cos^{2} θ = 1 \Rightarrow k^{2} + k^{2} = 1 \Rightarrow 2 k^{2} = 1 \Rightarrow k^{2} = \frac{1}{2} \Rightarrow k = \frac{1}{\sqrt{2}}$
$∴ \sin θ = k = \frac{1}{\sqrt{2}}$ and $\cos θ = k = \frac{1}{\sqrt{2}}$

(iv)
$\sin θ = \frac{1}{2}$

Now,

$\sin^{2} θ + \cos^{2} θ = 1 \Rightarrow {(\frac{1}{2})}^{2} + \cos^{2} θ = 1 \Rightarrow \cos^{2} θ = 1 - \frac{1}{4} = \frac{4 - 1}{4} = \frac{3}{4} \Rightarrow \cos^{2} θ = {(\frac{\sqrt{3}}{2})}^{2} \Rightarrow \cos θ = \frac{\sqrt{3}}{2}$
$∴ \tan θ = \frac{\sin θ}{\cos θ} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}}$

(v)
$\cos θ = \frac{1}{\sqrt{3}}$

Now,

$\sin^{2} θ + \cos^{2} θ = 1 \Rightarrow \sin^{2} θ + {(\frac{1}{\sqrt{3}})}^{2} = 1 \Rightarrow \sin^{2} θ = 1 - \frac{1}{3} = \frac{3 - 1}{3} = \frac{2}{3} \Rightarrow \sin^{2} θ = {(\sqrt{\frac{2}{3}})}^{2} \Rightarrow \sin θ = \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}}$
$∴ \tan θ = \frac{\sin θ}{\cos θ} = \frac{\frac{\sqrt{2}}{\sqrt{3}}}{\frac{1}{\sqrt{3}}} = \sqrt{2}$

(vi)
$\tan θ = \frac{21}{20} \Rightarrow \frac{\sin θ}{\cos θ} = \frac{21}{20} \Rightarrow \sin θ = 21 k, \cos θ = 20 k$
Now,

$\sin^{2} θ + \cos^{2} θ = 1 \Rightarrow {(21 k)}^{2} + {(20 k)}^{2} = 1 \Rightarrow 441 k^{2} + 400 k^{2} = 1 \Rightarrow 841 k^{2} = 1 \Rightarrow k^{2} = \frac{1}{841} = {(\frac{1}{29})}^{2} \Rightarrow k = \sqrt{{(\frac{1}{29})}^{2}} = \frac{1}{29}$
$∴ \sin θ = 21 k = 21 \times \frac{1}{29} = \frac{21}{29}$ and $\cos θ = 20 k = 20 \times \frac{1}{29} = \frac{20}{29}$

(vii)
$\tan θ = \frac{8}{15} \Rightarrow \frac{\sin θ}{\cos θ} = \frac{8}{15} \Rightarrow \sin θ = 8 k, \cos θ = 15 k$
Now,

$\sin^{2} θ + \cos^{2} θ = 1 \Rightarrow {(8 k)}^{2} + {(15 k)}^{2} = 1 \Rightarrow 64 k^{2} + 225 k^{2} = 1 \Rightarrow 289 k^{2} = 1 \Rightarrow k^{2} = \frac{1}{289} = {(\frac{1}{17})}^{2} \Rightarrow k = \sqrt{{(\frac{1}{17})}^{2}} = \frac{1}{17}$
$∴ \sin θ = 8 k = 8 \times \frac{1}{17} = \frac{8}{17}$ and $\cos θ = 15 k = 15 \times \frac{1}{17} = \frac{15}{17}$

(viii)
$\sin θ = \frac{3}{5}$

Now,

$\sin^{2} θ + \cos^{2} θ = 1 \Rightarrow {(\frac{3}{5})}^{2} + \cos^{2} θ = 1 \Rightarrow \cos^{2} θ = 1 - \frac{9}{25} = \frac{25 - 9}{25} = \frac{16}{25} \Rightarrow \cos^{2} θ = {(\frac{4}{5})}^{2} \Rightarrow \cos θ = \frac{4}{5}$
$∴ \tan θ = \frac{\sin θ}{\cos θ} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4}$

(ix)
$\tan θ = \frac{1}{2 \sqrt{2}} \Rightarrow \frac{\sin θ}{\cos θ} = \frac{1}{2 \sqrt{2}} \Rightarrow \sin θ = k, \cos θ = 2 \sqrt{2} k$
Now,

$\sin^{2} θ + \cos^{2} θ = 1 \Rightarrow k^{2} + {(2 \sqrt{2} k)}^{2} = 1 \Rightarrow k^{2} + 8 k^{2} = 1 \Rightarrow 9 k^{2} = 1 \Rightarrow k^{2} = \frac{1}{9} = {(\frac{1}{3})}^{2} \Rightarrow k = \sqrt{{(\frac{1}{3})}^{2}} = \frac{1}{3}$
$∴ \sin θ = k = \frac{1}{3}$ and $\cos θ = 2 \sqrt{2} k = 2 \sqrt{2} \times \frac{1}{3} = \frac{2 \sqrt{2}}{3}$

The complete table is given below:

sinθ	$\frac{12}{37}$	$\frac{11}{61}$	$\frac{1}{\sqrt{2}}$	$\frac{1}{2}$	$\frac{\sqrt{2}}{\sqrt{3}}$	$\frac{21}{29}$	$\frac{8}{17}$	$\frac{3}{5}$	$\frac{1}{3}$
cosθ	$\frac{35}{37}$	$\frac{60}{61}$	$\frac{1}{\sqrt{2}}$	$\frac{\sqrt{3}}{2}$	$\frac{1}{\sqrt{3}}$	$\frac{20}{29}$	$\frac{15}{17}$	$\frac{4}{5}$	$\frac{2 \sqrt{2}}{3}$
tanθ	$\frac{12}{35}$	$\frac{11}{60}$	1	$\frac{1}{\sqrt{3}}$	$\sqrt{2}$	$\frac{21}{20}$	$\frac{8}{15}$	$\frac{3}{4}$	$\frac{1}{2 \sqrt{2}}$

Page No 112:

Question 2:

Find the values of -

(i) 5 sin 30â€‹^°+ 3 tan 45^° (ii) $\frac{4}{5} \tan^{2} 60^{°} + 3 \sin^{2} 60^{°}$ (iii) 2sin 30^°+ cos 0^°+ 3sin 90°

(iv) $\frac{\tan 60}{\sin 60 + \cos 60}$ (v) $\cos^{2} 45^{°} + \sin^{2} 30^{°}$ (vi) cos 60^°× cos 30^° + sin 60^°× sin 30^°

Answer:

(i)
$5 \sin 30 ° + 3 \tan 45 ° = 5 \times \frac{1}{2} + 3 \times 1 = \frac{5}{2} + 3 = \frac{5 + 6}{2} = \frac{11}{2}$
(ii)
$\frac{4}{5} \tan^{2} 60 ° + 3 \sin^{2} 60 ° = \frac{4}{5} \times {(\sqrt{3})}^{2} + 3 \times {(\frac{\sqrt{3}}{2})}^{2} = \frac{4}{5} \times 3 + 3 \times \frac{3}{4} = \frac{12}{5} + \frac{9}{4} = \frac{48 + 45}{20} = \frac{93}{20}$
(iii)
$2 \sin 30 ° + \cos 0 ° + 3 \sin 90 ° = 2 \times \frac{1}{2} + 1 + 3 \times 1 = 1 + 1 + 3 = 5$
(iv)
$\frac{\tan 60 °}{\sin 60 ° + \cos 60 °} = \frac{\sqrt{3}}{\frac{\sqrt{3}}{2} + \frac{1}{2}} = \frac{\sqrt{3}}{\frac{\sqrt{3} + 1}{2}} = \frac{2 \sqrt{3}}{\sqrt{3} + 1}$
(v)
$\cos^{2} 45 ° + \sin^{2} 30 ° = {(\frac{1}{\sqrt{2}})}^{2} + {(\frac{1}{2})}^{2} = \frac{1}{2} + \frac{1}{4} = \frac{2 + 1}{4} = \frac{3}{4}$
(vi)
$\cos 60 ° \times \cos 30 ° + \sin 60 ° \times \sin 30 ° = \frac{1}{2} \times \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \times \frac{1}{2} = \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} = \frac{2 \sqrt{3}}{4} = \frac{\sqrt{3}}{2}$

Page No 112:

Question 3:

If sin $θ$ = $\frac{4}{5}$ then find cos $θ$

Answer:

$\sin^{2} θ + \cos^{2} θ = 1 \Rightarrow {(\frac{4}{5})}^{2} + \cos^{2} θ = 1 (\sin θ = \frac{4}{5}) \Rightarrow \frac{16}{25} + \cos^{2} θ = 1 \Rightarrow \cos^{2} θ = 1 - \frac{16}{25}$
$\Rightarrow \cos^{2} θ = \frac{25 - 16}{25} = \frac{9}{25} \Rightarrow \cos θ = \sqrt{\frac{9}{25}} = \sqrt{{(\frac{3}{5})}^{2}} \Rightarrow \cos θ = \frac{3}{5}$
Thus, the value of cosθ is $\frac{3}{5}$ .

Page No 112:

Question 4:

If cos $θ$ = $\frac{15}{17}$ then find sin $θ$

Answer:

$\sin^{2} θ + \cos^{2} θ = 1 \Rightarrow \sin^{2} θ + {(\frac{15}{17})}^{2} = 1 (\cos θ = \frac{15}{17}) \Rightarrow \sin^{2} θ + \frac{225}{289} = 1 \Rightarrow \sin^{2} θ = 1 - \frac{225}{289} \Rightarrow \sin^{2} θ = \frac{289 - 225}{289} = \frac{64}{289}$
$\Rightarrow \sin θ = \sqrt{\frac{64}{289}} = \sqrt{{(\frac{8}{17})}^{2}} \Rightarrow \sin θ = \frac{8}{17}$
Thus, the value of sinθ is $\frac{8}{17}$ .

Page No 113:

Question 1:

Choose the correct alternative answer for following multiple choice questions.

(i) Which of the following statements is true ?

(A) sin

θ

= cos (90

-

θ

) (B) cos

θ

= tan (90

-

θ

)
(C) sin

θ

= tan (90

-

θ

) (D) tan

θ

= tan (90

-

â€‹

θ

)

(ii) Which of the following is the value of sin 90^°?

(A)

\frac{\sqrt{3}}{2}

(B) 0

(C)

\frac{1}{2}

(D) 1

(iii) 2 tan 45^°+ cos 45^°

-

sin 45^°= ?

(A) 0 (B) 1

(iv)

\frac{\cos 28^{°}}{\sin 62^{°}}

(A) 2 (B)

-

1 (C) 0 (D) 1

Answer:

(i)
We know, $\sin θ = \cos (90 ° - θ)$ .

Also,

$\cos θ = \sin (90 ° - θ)$ and $\tan θ = \cot (90 ° - θ)$

Hence, the correct answer is option (A).

(ii)
We know, sin90º = 1.

Hence, the correct answer is option (D).

(iii)
$2 \tan 45 ° + \cos 45 ° - \sin 45 ° = 2 \times 1 + \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 2$
Hence, the correct answer is option (C).

(iv)
$\frac{\cos 28 °}{\sin 62 °} = \frac{\sin (90 ° - 28 °)}{\sin 62 °} [\cos θ = \sin (90 ° - θ)] = \frac{\sin 62 °}{\sin 62 °} = 1$
Hence, the correct answer is option (D).

Page No 113:

Question 2:

In right angled $∆$ TSU, TS = 5, $∠$ S = 90^°, SU = 12 then find sin T, cos T, tan T. Similarly find sin U, cos U, tan U.

Answer:

In right âˆ†TSU,

TU² = SU² + TS² (Pythagoras theorem)

⇒ TU² = 12² + 5² = 144 + 25 = 169

⇒ TU² = 13²

⇒ TU = 13

Now,

$sinT = \frac{SU}{TU} = \frac{12}{13} cosT = \frac{TS}{TU} = \frac{5}{13} tanT = \frac{SU}{TS} = \frac{12}{5}$
Also,

$sinU = \frac{TS}{TU} = \frac{5}{13} cosU = \frac{SU}{TU} = \frac{12}{13} tanU = \frac{TS}{SU} = \frac{5}{12}$

Page No 113:

Question 3:

In right angled $∆$ YXZ, $∠$ X = 90^°, XZ = 8 cm, YZ = 17 cm, find sin Y, cos Y, tan Y, sin Z, cos Z, tan Z.

Answer:

In right âˆ†YXZ,

YZ² = XZ² + XY² (Pythagoras theorem)

⇒ XY² = YZ² − XZ²

⇒ XY² = 17² − 8² = 289 − 64 = 225

⇒ XY² = 15²

⇒ XY = 15 cm

Now,

$sinY = \frac{XZ}{YZ} = \frac{8}{17} cosY = \frac{XY}{YZ} = \frac{15}{17} tanY = \frac{XZ}{XY} = \frac{8}{15}$
Also,

$sinZ = \frac{XY}{YZ} = \frac{15}{17} cosZ = \frac{XZ}{YZ} = \frac{8}{17} tanZ = \frac{XY}{XZ} = \frac{15}{8}$

Page No 113:

Question 4:

In right angled $∆$ LMN , if $∠$ N = $θ$ , $∠$ M = $90^{°}$ , $\cos θ = \frac{24}{25}$ , find sin $θ$ and tan $θ$ Similarly, find ( sin² $θ$ ) and ( cos² $θ$ ).

Answer:

In right âˆ†LMN, ∠N = θ.

$\cos θ = \frac{24}{25} \Rightarrow \frac{MN}{LN} = \frac{24}{25}$
Let MN = 24k and LN = 25k.

Using Pythagoras theorem, we have

LN² = LM² + MN²

⇒ (25k)² = LM² + (24k)²

⇒ LM² = 625k² − 576k² = 49k²

⇒ LM² = (7k)²

⇒ LM = 7k

Now,

$\sin θ = \frac{LM}{LN} = \frac{7 k}{25 k} = \frac{7}{25}$

$\tan θ = \frac{LM}{NM} = \frac{7 k}{24 k} = \frac{7}{24}$

Also,

$\sin^{2} θ = {(\frac{7}{25})}^{2} = \frac{49}{625}$

$\cos^{2} θ = {(\frac{24}{25})}^{2} = \frac{576}{625}$

Page No 113:

Question 5:

Fill in the blanks.

(i) sin20⁰ = cos $123^{°}$

(ii) tan30⁰ $\times$ tan $123^{°}$ = 1

(iii) cos40⁰ = sin $123^{°}$

Answer:

(i)
$\sin 20 ° = \cos (90 ° - 20 °) = \cos 70 ° [\sin θ = \cos (90 ° - θ)]$

(ii)
$\tan 30 ° = \frac{1}{\sqrt{3}} \tan 60 ° = \sqrt{3}$
Now,

$\frac{1}{\sqrt{3}} \times \sqrt{3} = 1 \Rightarrow \tan 30 ° \times \tan 60 ° = 1$
(iii)
$\cos 40 ° = \sin (90 ° - 40 °) = \sin 50 ° [\cos θ = \sin (90 ° - θ)]$

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