In right angle triangle ABC 15 cm and 17 cm are the lengths of AB AND

1.	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» 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 $\huge{\sf{\pink{\checkmark{ Hence \; solved }}}}$

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»

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

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