Question

Find values of sin 30 ,cos 30,tan30,cosec30,sec30 and cot 30 by using ratio concepts

Answer 1

Step-by-step explanation:

Take an equilateral triangle ∆ABC,Draw a perpendicular which join point C to the midpoint of AB;D

As we know that in Equilateral triangle Altitudes from a vertex,bisects angle as well as side.

To calculate sin 30°

we know that

$\angle ACD= 30°$

$sin \: 30° = \frac{opposite \: side \: of \: angle}{hypotenuse} \\ \\sin \: 30° = \frac{AD}{AC} \\ \\ sin \: 30° = \frac{ \frac{a}{2} }{a} \\ \\ sin \: 30° = \frac{1}{2} \\ \\ cosec \: 30° = \frac{AC}{AD} = 2$

To find

CD,apply Pythagoras Theorem in right triangle CAD

$CD = \sqrt{ {a}^{2} - \frac{ {a}^{2} }{4} } \\ \\ CD = \frac{ \sqrt{3} a}{2}$

$cos\: 30° = \frac{adjacent \: side \: of \: angle}{hypotenuse} \\ \\cos \: 30° = \frac{CD}{AC} \\ \\ cos\: 30° = \frac{ \frac{ \sqrt{3} a}{2} }{a} \\ \\ cos \: 30° = \frac{ \sqrt{3} }{2} \\ \\ sec \: 30° = \frac{AC}{CD} \\ \\ sec30°= \frac{2}{ \sqrt{3} }$

Also

$tan \: 30°= \frac{AD}{CD} \\ \\ = \frac{ \frac{a}{2} }{ \frac{ \sqrt{3} a}{2} } \\ \\ tan \: 30°= \frac{1}{ \sqrt{3} } \\ \\ cot \: 30° = \frac{CD}{AD} \\ \\ cot \: 30° = \sqrt{3}$

Hope it helps you.

Answer 2

Take an equilateral triangle ∆ABC,Draw a perpendicular which join point C to the midpoint of AB;D

As we know that in Equilateral triangle Altitudes from a vertex,bisects angle as well as side.

To calculate sin 30°

we know that

\begin{gathered}\angle ACD= 30° \\\end{gathered}

∠ACD=30°

\begin{gathered}sin \: 30° = \frac{opposite \: side \: of \: angle}{hypotenuse} \\ \\sin \: 30° = \frac{AD}{AC} \\ \\ sin \: 30° = \frac{ \frac{a}{2} }{a} \\ \\ sin \: 30° = \frac{1}{2} \\ \\ cosec \: 30° = \frac{AC}{AD} = 2 \\ \\\end{gathered}

sin30°=

hypotenuse

oppositesideofangle

sin30°=

AC

AD

sin30°=

a

2

a

sin30°=

2

1

cosec30°=

AD

AC

=2

To find

CD,apply Pythagoras Theorem in right triangle CAD

\begin{gathered}CD = \sqrt{ {a}^{2} - \frac{ {a}^{2} }{4} } \\ \\ CD = \frac{ \sqrt{3} a}{2} \\ \\\end{gathered}

CD=

a

2

−

4

a

2

CD=

2

3

a

\begin{gathered}cos\: 30° = \frac{adjacent \: side \: of \: angle}{hypotenuse} \\ \\cos \: 30° = \frac{CD}{AC} \\ \\ cos\: 30° = \frac{ \frac{ \sqrt{3} a}{2} }{a} \\ \\ cos \: 30° = \frac{ \sqrt{3} }{2} \\ \\ sec \: 30° = \frac{AC}{CD} \\ \\ sec30°= \frac{2}{ \sqrt{3} } \\ \\\end{gathered}

cos30°=

hypotenuse

adjacentsideofangle

cos30°=

AC

CD

cos30°=

a

2

3

a

cos30°=

2

3

sec30°=

CD

AC

sec30°=

3

2

Also

\begin{gathered}tan \: 30°= \frac{AD}{CD} \\ \\ = \frac{ \frac{a}{2} }{ \frac{ \sqrt{3} a}{2} } \\ \\ tan \: 30°= \frac{1}{ \sqrt{3} } \\ \\ cot \: 30° = \frac{CD}{AD} \\ \\ cot \: 30° = \sqrt{3} \\ \\\end{gathered}

tan30°=

CD

AD

=

2

3

a

2

a

tan30°=

3

1

cot30°=

AD

CD

cot30°=

3