Question

In right angle triangle ABC 15 cm and 17 cm are the lengths of AB AND AC respectively. Then find out all the six trigonometric ratios of angle A​

Answer 1

Question :-

In a right angled triangle ABC , ∠B = 90° . 15 cm & 17cm are the length of the sides AB and AC . we need to find out all the six trigonometric ratios of angle A ?

Answer :-

Given :-

In a right angled triangle ABC , ∠B = 90°

AB = 15 cm , AC = 17 cm

Required to find :-

• All the six trigonometric ratios of angle A ?

Diagram :-

$\setlength{\unitlength}{20} \begin{picture}(6,6) \put(1,1){\line(1,0){4}} \put(1,1){\line(0,1){4}}\put(5,1){\line( - 1, 1){4}}\put(1.5,1){\line(0,1){0.5}}\put(1,1.5){\line(1,0){0.5}}\qbezier(4.5,1)(4.25,1.25)(4.5,1.5)\put(1,0.5){ \tt B }\put(5,0.5){ \tt A }\put(1,5){ \tt C }\put(3,3){ \tt AC = 17 \: cm }\put(2,0.5){ \tt AB = 15 \: cm }\end{picture}$

Solution :-

Given :-

In right angled triangle ABC ;

∠B = 90° , AB = 15 cm , AC = 17 cm

we need to find the all six trigonometric ratios of angle A .

So,

In order to find all the six trigonometric ratios .

First let's find the unknown side .

Using the Pythagorean theorem ;

$\\ \tt{AB^2 + BC^2 = AC^2 } \\ \\ \rm( \because AB = 15 \: cm , AC = 17 \: cm ) \\ \\ \tt (15 {)}^{2} + (BC {)}^{2} = (17 {)}^{2} \\ \\ \tt 225 + {BC}^{2} = 289 \\ \\ \tt {BC}^{2} = 289 - 225 \\ \\ \tt {BC}^{2} = 64 \\ \\ \tt BC = \sqrt{64} \\ \\ \tt BC = \pm 8 \: cm \: \\ \\ \sf \because Length \: can \: 't be \: in \: negative \\ \\ \rm \implies \: BC = 8 \: cm$

Now,

Let's find the all trigonometric ratios of angle A .

So,

We know that ;

1st trigonometric ratio :-

$\boxed{\rm{\red{ \sin \theta = \dfrac{ Opposite \; \; side }{ Hypotenuse } }}}$

( Note :- Here $\theta$ = A )

So,

Sin A = Opposite side / Hypotenuse

Sin A = BC/AC

Sin A = 8/17

Hence,

Sin A = 8/17

Now,

2nd trigonometric ratio :-

$\boxed{\rm{ \red{ \cos \theta = \dfrac{Adjacent \: \: side}{ Hypotenuse}}}}$

So,

Cos A = Adjacent side / Hypotenuse

cos A = AB/AC

cos A = 15/17

Hence,

cos A = 15/17

Now,

3rd trigonometric ratio :-

$\boxed{ \rm{ \red{ \tan \theta = \dfrac{Opposite \; \: side }{Adjacent \; \: side }}}}$

So,

Tan A = opposite side/ Adjacent side

Tan A = BC/AB

Tan A = 8/15

Hence,

Tan A = 8/15

Now,

4th trigonometric ratio :-

$\boxed{ \rm{ \red{ \cosec \theta = \dfrac{ Hypotenuse}{Opposite \: \: side}}}}$

so,

Cosec A = Hypotenuse/Opposite side

Cosec A = AC/AB

Cosec A = 17/8

Hence,

Cosec A = 17/8

Now,

5th trigonometric ratio :-

$\boxed{ \rm{ \red{ \sec \theta = \dfrac{ Hypotenuse}{Adjacent \: \: side}}}}$

So,

Cosec A = Hypotenuse/ Adjacent side

Cosec A = AC/AB

Cosec A = 17/15

Hence,

Cosec A = 17/15

Now,

6th trigonometric ratio :-

$\: \boxed{ \rm{ \red{ \cot \theta = \dfrac{Adjacent \; \: side }{Opposite \; \: side }}}}$

So,

cot A = Adjacent side/ Opposite side

cot A = AB/BC

cot A = 15/8

Hence,

cot A = 15/8

$\huge{\sf{\pink{\checkmark{ Hence \; solved }}}}$

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{1st  trigonometric ratio :- \: sin \: a = \frac{8}{17} }}}$

$\green{\tt{\therefore{2 \: nd \: trigonometric  \: ratio :- \: cos \: a = \frac{15}{17} }}}$

$\green{\tt{\therefore{3 \: rd \: trigonometric ratio :- \: tan\: a = \frac{8}{15} }}}$

$\green{\tt{\therefore{4th \: trigonometric ratio :- \: cosec \: a = \frac{17}{8} }}}$

$\green{\tt{\therefore{5th \: trigonometric ratio :- \: cosec \: a = \frac{15}{8} }}}$

$\green{\tt{\therefore{6th \: trigonometric ratio :- \: sin \: a = \frac{15}{8} }}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}}$

$\tt: \implies In \: right \: angle \: triangle \: ABC \: 15 \: cm \: and \: 17 \: cm$

$\tt: \implies the \: lengths \: of \: AB \: AND \: AC \: respectively.$

$\red{\underline \bold{To \: Find :}}$

$\tt: \implies find \: out \: all \: the \: six \: trigonometric \: ratios of \: angle \: A$

$\tt Given \: Question$

$\tt: \implies so \: now \: find \: all \: trigonometric \: ratio \: of \: \angle{A}$

$\tt \: \:  \bigg[refer \: to \: the \: attachment\bigg]  \: \:$

$\tt \: \: \: \: \: \: \: \: \: hence \: all \: problem \: solved \: \: \: \: \: \: \: \: \: \:$