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Standard Values of Trigonometric Ratios

In each of th...

Question

In each of the following, one of the six trigonometric ratios is given. Find the values of the other trigonometric ratios.

(i) $\sin A = \frac{2}{3}$
(ii) $\cos A = \frac{4}{5}$
(iii) tan θ = 11
(iv) $\sin θ = \frac{11}{15}$
(v) $\tan α = \frac{5}{12}$
(vi) $\sin θ = \frac{\sqrt{3}}{2}$
(vii) $\cos θ = \frac{7}{25}$
(viii) $\tan θ = \frac{8}{15}$
(ix) $\cot θ = \frac{12}{5}$
(x) $\sec θ = \frac{13}{5}$
(xi) $cosec θ = \sqrt{10}$
(xii) $\cos θ = \frac{12}{15}$

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Solution

(i) Given: ……(1)

By definition,

…... (2)

By Comparing (1) and (2)

We get,

Perpendicular side = 2 and

Hypotenuse = 3

Therefore, by Pythagoras theorem,

Now we substitute the value of perpendicular side (BC) and hypotenuse (AC) and get the base side (AB)

Therefore,

Hence, Base =

Now,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

(ii) Given: ……(1)

By definition,

…... (2)

By Comparing (1) and (2)

We get,

Base = 4 and

Hypotenuse = 5

Therefore,

By Pythagoras theorem,

Now we substitute the value of base side (AB) and hypotenuse (AC) and get the perpendicular side (BC)

Hence, Perpendicular side = 3

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

(iii) Given: ……(1)

By definition,

…... (2)

By Comparing (1) and (2)

We get,

Base = 1 and

Perpendicular side = 5

Therefore,

By Pythagoras theorem,

Now we substitute the value of base side (AB) and the perpendicular side (BC) and get hypotenuse (AC)

Hence, Hypotenuse =

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

(iv) Given: ……(1)

By definition,

…... (2)

By Comparing (1) and (2)

We get,

Perpendicular side = 11 and

Hypotenuse = 15

Therefore,

By Pythagoras theorem,

Now we substitute the value of perpendicular side (BC) and hypotenuse(AC) and get the base side (AB)

Hence, Base =

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

(v) Given: ……(1)

By definition,

…... (2)

By Comparing (1) and (2)

We get,

Base = 12 and

Perpendicular side = 5

Therefore,

By Pythagoras theorem,

Now we substitute the value of base side (AB) and the perpendicular side (BC) and get hypotenuse (AC)

Hence, Hypotenuse = 13

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

(vi) Given: ……(1)

By definition,

…... (2)

By Comparing (1) and (2)

We get,

Perpendicular side = and

Hypotenuse = 2

Therefore,

By Pythagoras theorem,

Now we substitute the value of perpendicular side (BC) and hypotenuse(AC) and get the base side (AB)

Hence, Base = 1

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

(vii) Given: ……(1)

By definition,

…... (2)

By Comparing (1) and (2)

We get,

Base = 7 and

Hypotenuse = 25

Therefore,

By Pythagoras theorem,

Now we substitute the value of base side (AB) and hypotenuse (AC) and get the perpendicular side (BC)

Hence, Perpendicular side = 24

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

(viii) Given: ……(1)

By definition,

…... (2)

By Comparing (1) and (2)

We get,

Base = 15 and

Perpendicular side = 8

Therefore,

By Pythagoras theorem,

Now we substitute the value of base side (AB) and the perpendicular side (BC) and get hypotenuse (AC)

Hence, Hypotenuse = 17

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

(ix) Given: ……(1)

By definition,

…... (2)

By Comparing (1) and (2)

We get,

Base = 12 and

Perpendicular side = 5

Therefore,

By Pythagoras theorem,

Now we substitute the value of base side (AB) and the perpendicular side (BC) and get hypotenuse (AC)

Hence, Hypotenuse = 13

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

(x) Given: ……(1)

By definition,

…... (2)

By Comparing (1) and (2)

We get,

Base = 5 and

Hypotenuse = 13

Therefore,

By Pythagoras theorem,

Now we substitute the value of base side (AB) and hypotenuse (AC) and get the perpendicular side (BC)

Hence, Perpendicular side = 12

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

(xi) Given:

…… (1)

By definition,

…... (2)

By Comparing (1) and (2)

We get,

Perpendicular side = 1 and

Hypotenuse =

Therefore,

By Pythagoras theorem,

Now we substitute the value of perpendicular side (BC) and hypotenuse (AC) and get the base side (AB)

Hence, Base side = 3

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

(xii) Given: ……(1)

By definition,

…... (2)

By Comparing (1) and (2)

We get,

Base = 12 and

Hypotenuse = 15

Therefore,

By Pythagoras theorem,

Now we substitute the value of base side (AB) and hypotenuse (AC) and get the perpendicular side (BC)

Hence, Perpendicular side = 9

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

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Similar questions

In each of the following, one of the six stigonometric ratios is given. Find the values of the other trigonometric ratios.

(i) $s i n A = \frac{2}{3}$

(ii) $c o s A = \frac{4}{5}$

(iii) $t a n θ = 11$

(iv) $s i n θ = \frac{11}{15}$

(v) $t a n α = \frac{5}{12}$

(vi) $s i n θ = \frac{\sqrt{3}}{2}$

(vii) $c o s θ = \frac{7}{25}$

(viii) $t a n θ = \frac{8}{15}$

(ix) $c o t θ = \frac{12}{5}$

(x) $s e c θ = \frac{13}{5}$

(xi) $c o s e c θ = \sqrt{10}$

(xii) $c o s θ = \frac{12}{15}$

In the given figure,

∠

∠

∠

α

∠

θ

Write the following trigonometric ratios.
(i) sin

α

α

α

θ

θ

θ

Some trigonometric ratios and the interval in which $θ$ lies is given. Match the intervals with the ratios which are positive in those intervals.

$θ$ gives positive values

p. (0, $\frac{π}{2}$ ) 1. Only sin $θ$ , cosec $θ$

q. ( $\frac{π}{2}$ , $π$ ) 2. Only cos $θ$ , sec $θ$

r. $(π, \frac{3 π}{2})$ 3. Only tan $θ$ , cot $θ$

s. $(\frac{3 π}{2}, 2 π)$ 4. All sin $θ$ , cos $θ$ , tan $θ$ , cot $θ$ , sec $θ$ , cosec $θ$

Prove the following trigonometric identities.
(i) $\frac{1 + cos θ + sin θ}{1 + cos θ - sin θ} = \frac{1 + sin θ}{cos θ}$

(ii) $\frac{sin θ - cos θ + 1}{sin θ + cos θ - 1} = \frac{1}{sec θ - tan θ}$

(iii) $\frac{cos θ - sin θ + 1}{cos θ + sin θ - 1} = cosec θ + cot θ$

(iv) $(sin θ + cos θ) (tan θ + cot θ) = sec θ + cosec θ$

In the table, trigonometric ratios and the intervals of $θ$ are given. Match the intervals with the ratios which are positive in those intervals.

\begin{matrix} θ & gives positive values p. (0, \frac{π}{2}) & 1. Only sin θ, cosec θ q. (\frac{π}{2}, π) & 2. Only cos θ, sec θ r. (π, \frac{3 π}{2}) & 3. Only tan θ, cot θ s. (\frac{3 π}{2}, 2 π) & 4. All sin θ, cos θ, tan θ, cot θ, sec θ, cosec θ \end{matrix}

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