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# SOHCAHTOA

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 cscθ, secθ and cotθ are the reciprocals of sinθ, cosθ and tanθ respectively.

 sinθ = 1/cscθ cosθ = 1/secθ tanθ = 1/cotθ tanθ = sinθ/cosθ cscθ = 1/sinθ secθ = 1/cosθ cotθ = 1/tanθ cotθ = cosθ/sinθ

## Solved Issues

Downside 1 :

In the appropriate triangle PQR proven under, discover the six trigonometric ratios of the angle θ. Resolution :

Within the above proper angled triangle, notice that for the given angle θ, PQ is the ‘reverse’ aspect and PR is the ‘adjoining’ aspect.

Then,

sinθ = reverse aspect/hypotenuse = PQ/QR = 5/13

cosθ = adjoining aspect/hypotenuse = PR/QR = 12/13

tanθ = reverse aspect/adjoining aspect = PQ/PR = 5/12

cscθ = 1/sinθ = 13/5

secθ = 1/cosθ = 13/12

cotθ = 1/tanθ = 12/5

Downside 2 :

Within the determine proven under, discover the six trigonometric ratios of the angle θ. Resolution :

In the appropriate angled triangle ABC proven above,

AC = 24

BC = 7

By Pythagorean theorem,

AB2 = BC2 + CA2

AB2 = 72 + 242

AB2 = 49 + 576

AB2 = 625

AB2 = 252

AB = 25

Now, we are able to use the three sides to search out the six trigonometric ratios of angle θ.

sinθ = reverse aspect/hypotenuse = BC/AB = 7/25

cosθ = adjoining aspect/hypotenuse = AC/AB = 24/25

tanθ = reverse aspect/adjoining aspect = BC/AC = 7/24

cscθ = 1/sinθ = 25/7

secθ = 1/cosθ = 25/24

cotθ = 1/tanθ = 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,

AC2 = AB2 + BC2

152 = AB2 + 122

225 = AB2 + 144

Subtract 144 from either side.

81 = AB2

92 = AB2

9 = AB

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

cscA = 1/sinA

= 15/12

= 5/4

secA = 1/cosA

= 5/3

cotA = 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 appropriate ΔABD, by Pythagorean Theorem,

AB2 = AD2 + BD2

132 = AD2 + 52

169 = AD2 + 25

Subtract 25 from either side.

In the appropriate ΔACD, by Pythagorean Theorem,

AC2 = AD2 + CD2

AC2 = 122 + 162

AC2 = 144 + 256

AC2 = 400

AC2 = 202

AC = 20

Then,

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

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

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

cosC =  adjoining aspect/hypotenuse

= CD/AC

= 16/20

= 4/5

tanC = reverse aspect/adjoining aspect

= 12/16

= 3/4

cscC = hypotenuse/reverse aspect

= 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 30°. 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. In proper triangle PQR above, contemplating ∠PRQ = 30°, 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).

We all know that

tanθ = reverse aspect/adjoining aspect

In proper triangle PQR,

tan∠PRQ = PQ/QR

tan30° = h/48

From trigonometric ratio desk, we now have tan30° = √3/3.

√3/3 = h/48

Multiply either side by 48.

48√3/3 = h

16√3 = h

The peak of the tower is 16√3 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 60°. 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 ABC above, contemplating ∠ACB = 60°, 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

sinθ = reverse aspect/hypotenuse

In proper triangle PQR,

sin∠ACB = AB/AC

sin60° = 75/AC

From trigonometric ratio desk, we now have sin60° = √3/2.

√3/2 = 75/AC

Take reciprocal on either side.

2/√3 = AC/75

Multiply either side by 75.

150/√3 = AC

150√3/3 = AC

50√3 = AC

The size of the string is 50√3. Kindly mail your suggestions to v4formath@gmail.com

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