This can be a method to keep in mind how values the trigonometric ratios sin, cosine and tangent of an angle will be computed.

Allow us to see, how this shortcut works to recollect the above talked about trigonometric ratios.

Earlier than we talk about this shortcut, tell us the title of every aspect of a proper triangle from the determine given under.

To grasp the shortcut, first we now have to divide SOHCAHTOA into three components as given under.

What do SOH, CAH and TOA stand for?

Right here is the reply

From the above figures, we are able to derive formulation for the three trigonometric ratios sin, cos and tan as given under.

## Reciprocal Relations

The trigonometric ratios cscÎ¸, secÎ¸ and cotÎ¸ are the reciprocals of sinÎ¸, cosÎ¸ and tanÎ¸ respectively.

sinÎ¸ = 1/cscÎ¸ cosÎ¸ = 1/secÎ¸ tanÎ¸ = 1/cotÎ¸ tanÎ¸ = sinÎ¸/cosÎ¸ |
cscÎ¸ = 1/sinÎ¸ secÎ¸ = 1/cosÎ¸ cotÎ¸ = 1/tanÎ¸ cotÎ¸ = cosÎ¸/sinÎ¸ |

## Solved Issues

Downside 1 :

In the appropriate triangle PQR proven under, discover the six trigonometric ratios of the angleÂ Î¸.

Resolution :

Within the above proper angled triangle, notice that for the given angle Î¸, PQ is the â€˜reverseâ€™ aspect and PR is the â€˜adjoiningâ€™ aspect.

Then,

sinÎ¸ = reverse aspect/hypotenuse = PQ/QR = 5/13

cosÎ¸ = adjoining aspect/hypotenuse = PR/QR = 12/13

tanÎ¸ = reverse aspect/adjoining aspect = PQ/PR = 5/12

cscÎ¸ = 1/sinÎ¸ = 13/5

secÎ¸ = 1/cosÎ¸ = 13/12

cotÎ¸ = 1/tanÎ¸ = 12/5

Downside 2 :

Within the determine proven under, discover the six trigonometric ratios of the angleÂ Î¸.

Resolution :

In the appropriate angled triangle ABC proven above,

AC = 24

BC = 7

By Pythagorean theorem,

AB^{2}Â = BC^{2}Â + CA^{2}

AB^{2}Â = 7^{2}Â + 24^{2}

AB^{2}Â = 49 + 576

AB^{2}Â = 625

AB^{2}Â = 25^{2}

AB = 25Â

Now, we are able to use the three sides to search out the six trigonometric ratios of angleÂ Î¸.

sinÎ¸ = reverse aspect/hypotenuse = BC/AB = 7/25

cosÎ¸ = adjoining aspect/hypotenuse = AC/AB = 24/25

tanÎ¸ = reverse aspect/adjoining aspect = BC/AC = 7/24

cscÎ¸ = 1/sinÎ¸ = 25/7

secÎ¸ = 1/cosÎ¸ = 25/24

cotÎ¸ = 1/tanÎ¸ = 24/7

Downside 3 :

In triangle ABC, proper angled at B, 15sin A = 12. Discover the opposite 5 trigonometric ratios of the angle A.

Resolution :

15sinA = 12

sinA = 12/15

sinA = reverse aspect/hypotenuse

sinA = 12/15

By Pythagorean theorem,

AC^{2}Â = AB^{2} + BC^{2}

15^{2}Â = AB^{2}Â + 12^{2}

225 = AB^{2}Â + 144

Subtract 144 from either side.

81 = AB^{2}

9^{2}Â = AB^{2}

9 = ABÂ

Now, we are able to use the three sides to search out the 5 trigonometric ratios of angle A and 6 trigonometric ratios of angle C.

cosA = adjoining aspect/hypotenuse

= AB/AC

= 9/15

= 3/5

tanA = reverse aspect/adjoining aspect

= BC/AB

= 12/9

= 4/3

cscAÂ = 1/sinA

= 15/12

= 5/4

secAÂ = 1/cosA

= 5/3

cotAÂ = 1/tanA

= 3/4

Downside 4 :

Within the determine proven under, discover the values of

sinB, secB, cotB, cosC, tanC and cscC

Resolution :

In the appropriateÂ Î”ABD, by Pythagorean Theorem,

AB^{2}Â = AD^{2}Â + BD^{2}

13^{2}Â = AD^{2}Â + 5^{2}

169 = AD^{2}Â + 25

Subtract 25 from either side.

144 = AD^{2}

12^{2}Â = AD^{2}

12 = AD

In the appropriateÂ Î”ACD, by Pythagorean Theorem,

AC^{2}Â = AD^{2}Â + CD^{2}

AC^{2}Â = 12^{2}Â + 16^{2}

AC^{2}Â = 144 + 256

AC^{2}Â = 400

AC^{2}Â = 20^{2}

AC = 20

Then,

sinBÂ = reverse aspect/hypotenuse = AD/AB = 12/13

secBÂ = hypotenuse/adjoining aspect = AB/BD = 13/5

cotBÂ = adjoining aspect/reverse aspect = BD/AD = 5/12

cosC =Â adjoining aspect/hypotenuseÂ

= CD/AC

= 16/20

= 4/5

tanC = reverse aspect/adjoining aspect

= AD/CD

= 12/16

= 3/4

cscC = hypotenuse/reverse aspect

= AC/AD

= 20/12

= 5/3

Downside 5 :

A tower stands vertically on the bottom. From a degree on the bottom, which is 48 m away from the foot of the tower, the angle of elevation of the highest of the tower is 30Â°. Discover the peak of the tower.

Resolution :

Let PQ be the peak of the tower.

Take PQ = h and QR is the space between the tower and the purpose R.Â

In proper triangle PQR above, contemplating âˆ PRQ = 30Â°, PQ being the alternative aspect and QR being the adjoining aspect.

We all know the size of the adjoining aspect (= 48 m) and we now have to search out the size of the alternative aspect (top of the tower).Â Â

We all know that

tanÎ¸ = reverse aspect/adjoining aspect

In proper triangle PQR,

tanâˆ PRQ = PQ/QR

tan30Â° = h/48

From trigonometric ratio desk, we now have tan30Â° = âˆš3/3.

âˆš3/3 = h/48

Multiply either side by 48.

48âˆš3/3 = h

16âˆš3 = h

The peak of the tower is 16âˆš3 m.

Downside 6 :

A kite is flying at a top of 75 m above the bottom. The string connected to the kite is briefly tied to a degree on the bottom. The inclination of the string with the bottom is 60Â°. Discover the size of the string, assuming that there isn’t a slack within the string.

Resolution :

Let AB be the peak of the kite above the bottom. Then, AB = 75. Let AC be the size of the string.

In proper triangle ABCÂ above, contemplating âˆ ACB = 60Â°, AB being the alternative aspect and AC being the hypotenuse.

We all know the size of the alternative aspect (= 75 m) and we now have to search out the size of the hypotenuse (size of the string).

We all know that

sinÎ¸ = reverse aspect/hypotenuse

In proper triangle PQR,

sinâˆ ACB = AB/AC

sin60Â° = 75/AC

From trigonometric ratio desk, we now have sin60Â° = âˆš3/2.

âˆš3/2 = 75/AC

Take reciprocal on either side.

2/âˆš3 = AC/75

Multiply either side by 75.

150/âˆš3 = AC

150âˆš3/3 = AC

50âˆš3 = AC

The size of the string is 50âˆš3.

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