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#### Question 1: In the given Fig, $\angle$R is the right angle of $∆$PQR. Write the following ratios.
(i) sin P (ii) cos Q (iii) tan P (iv) tan Q (i) sinP =

(ii) cosQ =

(iii) tanP =

(iv) tanQ =

#### Question 2: In the right angled $∆$XYZ, $\angle$XYZ = 90° and abc 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. (i) sinX =

(ii) tanZ =

(iii) cosX =

(iv) tanX =

#### Question 3: In right angled $∆$ LMN, $\angle$LMN = 90° $\angle$L = 50° and $\angle$N = 40°, write the following ratios.
(i) sin 50° (ii) cos 50° (iii) tan 40° (iv) cos 40° (i) sin50º =

(ii) cos50º =

(iii) tan40º =

(iv) cos40º =

#### Question 4: In the given figure, $\angle$PQR = 90° $\angle$PQS = 90° , $\angle$PRQ = $\mathrm{\alpha }$ and$\angle$QPS = $\mathrm{\theta }$  Write the following trigonometric ratios.
(i) sin$\mathrm{\alpha }$, cos$\mathrm{\alpha }$ , tan$\mathrm{\alpha }$
(ii) sin$\mathrm{\theta }$, cos$\mathrm{\theta }$, tan$\mathrm{\theta }$ (i)
sin$\alpha$

cos$\alpha$ =

tan$\alpha$

(ii)
sinθ

cosθ =

tanθ

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

(i)
$\mathrm{cos}\theta =\frac{35}{37}$

Now,

${\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta =1\phantom{\rule{0ex}{0ex}}⇒{\mathrm{sin}}^{2}\theta +{\left(\frac{35}{37}\right)}^{2}=1\phantom{\rule{0ex}{0ex}}⇒{\mathrm{sin}}^{2}\theta =1-\frac{1225}{1369}=\frac{1369-1225}{1369}=\frac{144}{1369}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{sin}}^{2}\theta ={\left(\frac{12}{37}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}\theta =\frac{12}{37}$
$\therefore \mathrm{tan}\theta =\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }=\frac{\frac{12}{37}}{\frac{35}{37}}=\frac{12}{35}$

(ii)
$\mathrm{sin}\theta =\frac{11}{61}$

Now,

${\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta =1\phantom{\rule{0ex}{0ex}}⇒{\left(\frac{11}{61}\right)}^{2}+{\mathrm{cos}}^{2}\theta =1\phantom{\rule{0ex}{0ex}}⇒{\mathrm{cos}}^{2}\theta =1-\frac{121}{3721}=\frac{3721-121}{3721}=\frac{3600}{3721}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{cos}}^{2}\theta ={\left(\frac{60}{61}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}\theta =\frac{60}{61}$
$\therefore \mathrm{tan}\theta =\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }=\frac{\frac{11}{61}}{\frac{60}{61}}=\frac{11}{60}$

(iii)

Now,

${\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta =1\phantom{\rule{0ex}{0ex}}⇒{k}^{2}+{k}^{2}=1\phantom{\rule{0ex}{0ex}}⇒2{k}^{2}=1\phantom{\rule{0ex}{0ex}}⇒{k}^{2}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒k=\frac{1}{\sqrt{2}}$
and $\mathrm{cos}\theta =k=\frac{1}{\sqrt{2}}$

(iv)
$\mathrm{sin}\theta =\frac{1}{2}$

Now,

${\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta =1\phantom{\rule{0ex}{0ex}}⇒{\left(\frac{1}{2}\right)}^{2}+{\mathrm{cos}}^{2}\theta =1\phantom{\rule{0ex}{0ex}}⇒{\mathrm{cos}}^{2}\theta =1-\frac{1}{4}=\frac{4-1}{4}=\frac{3}{4}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{cos}}^{2}\theta ={\left(\frac{\sqrt{3}}{2}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}\theta =\frac{\sqrt{3}}{2}$
$\therefore \mathrm{tan}\theta =\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}$

(v)
$\mathrm{cos}\theta =\frac{1}{\sqrt{3}}$

Now,

${\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta =1\phantom{\rule{0ex}{0ex}}⇒{\mathrm{sin}}^{2}\theta +{\left(\frac{1}{\sqrt{3}}\right)}^{2}=1\phantom{\rule{0ex}{0ex}}⇒{\mathrm{sin}}^{2}\theta =1-\frac{1}{3}=\frac{3-1}{3}=\frac{2}{3}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{sin}}^{2}\theta ={\left(\sqrt{\frac{2}{3}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}\theta =\sqrt{\frac{2}{3}}=\frac{\sqrt{2}}{\sqrt{3}}$
$\therefore \mathrm{tan}\theta =\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }=\frac{\frac{\sqrt{2}}{\sqrt{3}}}{\frac{1}{\sqrt{3}}}=\sqrt{2}$

(vi)

Now,

${\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta =1\phantom{\rule{0ex}{0ex}}⇒{\left(21k\right)}^{2}+{\left(20k\right)}^{2}=1\phantom{\rule{0ex}{0ex}}⇒441{k}^{2}+400{k}^{2}=1\phantom{\rule{0ex}{0ex}}⇒841{k}^{2}=1\phantom{\rule{0ex}{0ex}}⇒{k}^{2}=\frac{1}{841}={\left(\frac{1}{29}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒k=\sqrt{{\left(\frac{1}{29}\right)}^{2}}=\frac{1}{29}$
and $\mathrm{cos}\theta =20k=20×\frac{1}{29}=\frac{20}{29}$

(vii)

Now,

${\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta =1\phantom{\rule{0ex}{0ex}}⇒{\left(8k\right)}^{2}+{\left(15k\right)}^{2}=1\phantom{\rule{0ex}{0ex}}⇒64{k}^{2}+225{k}^{2}=1\phantom{\rule{0ex}{0ex}}⇒289{k}^{2}=1\phantom{\rule{0ex}{0ex}}⇒{k}^{2}=\frac{1}{289}={\left(\frac{1}{17}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒k=\sqrt{{\left(\frac{1}{17}\right)}^{2}}=\frac{1}{17}$
and $\mathrm{cos}\theta =15k=15×\frac{1}{17}=\frac{15}{17}$

(viii)
$\mathrm{sin}\theta =\frac{3}{5}$

Now,

${\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta =1\phantom{\rule{0ex}{0ex}}⇒{\left(\frac{3}{5}\right)}^{2}+{\mathrm{cos}}^{2}\theta =1\phantom{\rule{0ex}{0ex}}⇒{\mathrm{cos}}^{2}\theta =1-\frac{9}{25}=\frac{25-9}{25}=\frac{16}{25}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{cos}}^{2}\theta ={\left(\frac{4}{5}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}\theta =\frac{4}{5}$
$\therefore \mathrm{tan}\theta =\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }=\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{4}$

(ix)

Now,

${\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta =1\phantom{\rule{0ex}{0ex}}⇒{k}^{2}+{\left(2\sqrt{2}k\right)}^{2}=1\phantom{\rule{0ex}{0ex}}⇒{k}^{2}+8{k}^{2}=1\phantom{\rule{0ex}{0ex}}⇒9{k}^{2}=1\phantom{\rule{0ex}{0ex}}⇒{k}^{2}=\frac{1}{9}={\left(\frac{1}{3}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒k=\sqrt{{\left(\frac{1}{3}\right)}^{2}}=\frac{1}{3}$
and $\mathrm{cos}\theta =2\sqrt{2}k=2\sqrt{2}×\frac{1}{3}=\frac{2\sqrt{2}}{3}$

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 $\frac{1}{\sqrt{3}}$ $\sqrt{2}$ $\frac{21}{20}$ $\frac{8}{15}$ $\frac{3}{4}$ $\frac{1}{2\sqrt{2}}$

#### Question 2:

Find the values of -

(i) 5 sin 30​° + 3 tan 45°    (ii)    (iii) 2sin 30° + cos 0° + 3sin 90°

(iv)         (v)           (vi)  cos 60°× cos 30° + sin 60°× sin 30°

(i)
$5\mathrm{sin}30°+3\mathrm{tan}45°\phantom{\rule{0ex}{0ex}}=5×\frac{1}{2}+3×1\phantom{\rule{0ex}{0ex}}=\frac{5}{2}+3\phantom{\rule{0ex}{0ex}}=\frac{5+6}{2}\phantom{\rule{0ex}{0ex}}=\frac{11}{2}$
(ii)

(iii)
$2\mathrm{sin}30°+\mathrm{cos}0°+3\mathrm{sin}90°\phantom{\rule{0ex}{0ex}}=2×\frac{1}{2}+1+3×1\phantom{\rule{0ex}{0ex}}=1+1+3\phantom{\rule{0ex}{0ex}}=5$
(iv)
$\frac{\mathrm{tan}60°}{\mathrm{sin}60°+\mathrm{cos}60°}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{\frac{\sqrt{3}}{2}+\frac{1}{2}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{\frac{\sqrt{3}+1}{2}}\phantom{\rule{0ex}{0ex}}=\frac{2\sqrt{3}}{\sqrt{3}+1}$
(v)
${\mathrm{cos}}^{2}45°+{\mathrm{sin}}^{2}30°\phantom{\rule{0ex}{0ex}}={\left(\frac{1}{\sqrt{2}}\right)}^{2}+{\left(\frac{1}{2}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}+\frac{1}{4}\phantom{\rule{0ex}{0ex}}=\frac{2+1}{4}\phantom{\rule{0ex}{0ex}}=\frac{3}{4}$
(vi)
$\mathrm{cos}60°×\mathrm{cos}30°+\mathrm{sin}60°×\mathrm{sin}30°\phantom{\rule{0ex}{0ex}}=\frac{1}{2}×\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}×\frac{1}{2}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}\phantom{\rule{0ex}{0ex}}=\frac{2\sqrt{3}}{4}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{2}$

#### Question 3:

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

$⇒{\mathrm{cos}}^{2}\theta =\frac{25-16}{25}=\frac{9}{25}\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}\theta =\sqrt{\frac{9}{25}}=\sqrt{{\left(\frac{3}{5}\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}\theta =\frac{3}{5}$
Thus, the value of cosθ is $\frac{3}{5}$.

#### Question 4:

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

$⇒\mathrm{sin}\theta =\sqrt{\frac{64}{289}}=\sqrt{{\left(\frac{8}{17}\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}\theta =\frac{8}{17}$
Thus, the value of sinθ is $\frac{8}{17}$.

#### Question 1:

Choose the correct alternative answer for following multiple choice questions.

(i) Which of the following statements is true ?
(A) sin $\mathrm{\theta }$ = cos (90$-$ $\mathrm{\theta }$) (B) cos $\mathrm{\theta }$ = tan (90$-$$\mathrm{\theta }$ )
(C) sin $\mathrm{\theta }$ = tan (90$-$$\mathrm{\theta }$)    (D) tan $\mathrm{\theta }$ = tan (90$-$$\mathrm{\theta }$)

(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
(C) 2 ( D) 3

(iv)

(A) 2 (B) $-$1 (C) 0 (D) 1

(i)
We know, $\mathrm{sin}\theta =\mathrm{cos}\left(90°-\theta \right)$.

Also,

$\mathrm{cos}\theta =\mathrm{sin}\left(90°-\theta \right)$ and $\mathrm{tan}\theta =\mathrm{cot}\left(90°-\theta \right)$

Hence, the correct answer is option (A).

(ii)
We know, sin90º = 1.

Hence, the correct answer is option (D).

(iii)
$2\mathrm{tan}45°+\mathrm{cos}45°-\mathrm{sin}45°\phantom{\rule{0ex}{0ex}}=2×1+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\phantom{\rule{0ex}{0ex}}=2$
Hence, the correct answer is option (C).

(iv)

Hence, the correct answer is option (D).

#### Question 2:

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

TU2 = SU2 + TS2              (Pythagoras theorem)

⇒ TU2 = 122 + 52 = 144 + 25 = 169

⇒ TU2 = 132

⇒ TU = 13

Now,

$\mathrm{sinT}=\frac{\mathrm{SU}}{\mathrm{TU}}=\frac{12}{13}\phantom{\rule{0ex}{0ex}}\mathrm{cosT}=\frac{\mathrm{TS}}{\mathrm{TU}}=\frac{5}{13}\phantom{\rule{0ex}{0ex}}\mathrm{tanT}=\frac{\mathrm{SU}}{\mathrm{TS}}=\frac{12}{5}$
Also,

$\mathrm{sinU}=\frac{\mathrm{TS}}{\mathrm{TU}}=\frac{5}{13}\phantom{\rule{0ex}{0ex}}\mathrm{cosU}=\frac{\mathrm{SU}}{\mathrm{TU}}=\frac{12}{13}\phantom{\rule{0ex}{0ex}}\mathrm{tanU}=\frac{\mathrm{TS}}{\mathrm{SU}}=\frac{5}{12}$

#### Question 3:

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

YZ2 = XZ2 + XY2              (Pythagoras theorem)

⇒ XY2 = YZ2 − XZ2

⇒ XY2 = 172 − 82 = 289 − 64 = 225

⇒ XY2 = 152

⇒ XY = 15 cm

Now,

$\mathrm{sinY}=\frac{\mathrm{XZ}}{\mathrm{YZ}}=\frac{8}{17}\phantom{\rule{0ex}{0ex}}\mathrm{cosY}=\frac{\mathrm{XY}}{\mathrm{YZ}}=\frac{15}{17}\phantom{\rule{0ex}{0ex}}\mathrm{tanY}=\frac{\mathrm{XZ}}{\mathrm{XY}}=\frac{8}{15}$
Also,

$\mathrm{sinZ}=\frac{\mathrm{XY}}{\mathrm{YZ}}=\frac{15}{17}\phantom{\rule{0ex}{0ex}}\mathrm{cosZ}=\frac{\mathrm{XZ}}{\mathrm{YZ}}=\frac{8}{17}\phantom{\rule{0ex}{0ex}}\mathrm{tanZ}=\frac{\mathrm{XY}}{\mathrm{XZ}}=\frac{15}{8}$

#### Question 4:

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

$\mathrm{cos}\theta =\frac{24}{25}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{MN}}{\mathrm{LN}}=\frac{24}{25}$
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,

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

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

Also,

${\mathrm{sin}}^{2}\theta ={\left(\frac{7}{25}\right)}^{2}=\frac{49}{625}$

${\mathrm{cos}}^{2}\theta ={\left(\frac{24}{25}\right)}^{2}=\frac{576}{625}$

#### Question 5:

Fill in the blanks.

(i) sin20  =  cos

(ii) tan300 $×$ tan   = 1

(iii) cos400  = sin

(i)

(ii)
$\mathrm{tan}30°=\frac{1}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}\mathrm{tan}60°=\sqrt{3}$
Now,

$\frac{1}{\sqrt{3}}×\sqrt{3}=1\phantom{\rule{0ex}{0ex}}⇒\mathrm{tan}30°×\mathrm{tan}\overline{)60}°=1$
(iii)

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