# If Sin A = 3/4, Calculate cos A and tan A

Trigonometry is one of the branches of Mathematics. It is the study of the property of triangles like it is the relation between lengths of its sides and angles between them. The ratios involved in the trigonometric problems are called trigonometric ratios. Trigonometry helps in understanding any topic that involves distances, angles, waves problems, etc. The trigonometric functions (sin, cos, and tan) are key to study science, technology, and engineering.

The above-given problem is simple and is solved using trigonometric ratios and the Pythagoras theorem.

### Trigonometric ratios

Trigonometric ratios are the very beginning thing to learn in trigonometry. These ratios from the basic foundation required to understand this branch of mathematics. Here are all the Trigonometric ratios that we are talking about,

sin(θ) = Perpendicular/Hypotenuse

cos(θ) = Base/Hypotenuse

tan(θ) = Perpendicular/Base

cot (θ) = 1/tan (θ) = Base/Perpendicular

sec (θ) = 1/cos (θ) = Hypotenuse/Base

cosec (θ) = 1/sin (θ) = Hypotenuse/Perpendicular

### Pythagoras Theorem

The Pythagoras Theorem states that if a triangle is a right angled triangle then the sum of squares of perpendicular and base of the triangle is equal to the square of a hypotenuse of a triangle i.e.

If in a Right-angled triangle, AB =Perpendicular, BC = Base and AC = Hypotenuse

Then, (Perpendicular)^{2} + (Base)^{2} = (Hypotenuse)^{2}

Therefore, AB^{2} + BC^{2} = AC^{2}

This is known as Pythagoras Theorem.

### If Sin A = 3/4, Calculate cos A and tan A.

**Solution: **

Given that, Sin A = 3/4 i.e.

For a given problem, let Base = X

Given: AB = Perpendicular = 3 , AC = Hypotenuse = 4

Sin(A) = Perpendicular/Hypotenuse = 3/4

Using above Trigonometric ratios:

Cos(A) = Base/Hypotenuse= X/4 ⇢ (equation 1)

Tan(A) = Perpendicular/Base = 3/X ⇢ (equation 2)

Now to find X we need to apply Pythagoras theorem

Therefore, AB

^{2}+ BC^{2}= AC^{2}3

^{2}+ X^{2}= 4^{2 }9 + X

^{2}= 16X

^{2}= 16 – 9X

^{2}= 7X = √ 7.

Now substituting X in equation 1 and 2 we get:

Cos(A) = Base/Hypotenuse= √7/4

Tan(A) = Perpendicular/Base = 3/√7

### Similar Problems

**Question 1: If Sin A = 4/5, Calculate Cot A and Cosec A.**

**Solution: **

Given that, Sin A = 4/5 i.e.

AB = Perpendicular = 4, AC = Hypotenuse = 5

For a given problem, let Base = X

Sin(A) = Perpendicular/Hypotenuse = 4/5

Using Trigonometric ratios:

(1) cot (A) = Base/Perpendicular = X/4

Now to find X we need to apply Pythagoras theorem

Therefore, AB

^{2}+ BC^{2}= AC^{2}4

^{2}+ BC^{2}= 5^{2}BC

^{2}= 25 – 16BC

^{2}= 9BC = √9

BC = 3

Therefore, cot (A) = Base/Perpendicular = 3/4

(2)

cosec (A) = 1/Sin (A) = Hypotenuse/Perpendicular = 5/4

**Question 2: If Tan = √7/3, Calculate Sec A and Cos A. **

**Solution:**

Given that Tan A = √7/3

Tan A = Perpendicular/Base

So, Perpendicular = √7, Base = 3

Let Hypotenuse be X

Using Pythagoras theorem

√7

^{2}+ 3^{2}= X^{2}7

^{2}+ 9 = X^{2}16 = X

^{2}Therefore, X = 4

Hypotenuse = X = 4

(1)

Sec A = Hypotenuse/Base = 4/3(2) Cos A = 1/Sec A = 3/4

**Question 3: If Cot A = √3, Prove that Tan A + Sin A = (2+√3) / 2√3.**

**Solution: **

Given Cot A = √3

Cot A = Base/Perpendicular

So, Base = √3, Perpendicular = 1

Let Hypotenuse be X,

Using Pythagoras theorem

1

^{2}+ (√3)^{2}= X^{2}1 + 3 = X

^{2}4 = X

^{2}Therefore, X= 2

Hypotenuse = X = 2

Tan A + Sin A = ?

Tan A =1/Cot A = 1/√ 3

Sin A = Perpendicular/Hypotenuse = 1/2

Tan A + Sin A = 1/√3 + 1/2

= (2 + √ 3) / 2√3

Therefore, Tan A + Sin A = (2 + √3) / 2√3.

**Question 4: If cot ^{2} A = 3 (sec^{2} A – tan^{2} A) , then find 3(sin A)?**

**Solution: **

Given: cot

^{2}A = 3 (sec^{2}A -tan^{2}A)sec

^{2}A – tan^{2}A = 1Therefore, cot

^{2}A = 3(1)So, Cot

^{ }A = √3Cot A = Base/Perpendicular = √3/1

Base = √3, Perpendicular = 1

Let Hypotenuse be X,

Using Pythagoras theorem

1

^{2}+ (√3)^{2}= X^{2}1 + 3 = X

^{2}4 = X

^{2}Therefore, X = 2

Hypotenuse = X = 2

3(Sin A) = ?

Sin A = Perpendicular/Hypotenuse = 3(Perpendicular/Hypotenuse)

Therefore, Sin A = 3(1/2) = 3/2

**Question 5: If sin ^{2}A sec A + cos A = 5/4 ,then find tan A?**

**Solution: **

Given: sin

^{2}A sec A + cos A = 5/4Multiply and Divide L.H.S by Cos A ,

Sin

^{2}A Sec A + cos A × (Cos A/Cos A) = 5/4(Sin

^{2}A × Sec A × Cos A + Cos^{2}A)/Cos A = 5/4Sec A = 1/Cos A

So, (Sin

^{2}A × 1/ Cos A × Cos A + Cos^{2}A)/Cos A = 5/4(Sin

^{2}A + Cos^{2}A)/Cos A = 5/41/Cos A = 5/4

Therefore, Cos A = 4/5

Cos A = Base/Hypotenuse = 4/5

Let Perpendicular be X

Using Pythagoras theorem,

4

^{2}+ X^{2}= 5^{2}16 + X

^{2 }= 25X

^{2}= 25 – 16X

^{2}= 9X = √9

X = 3

Tan A = ?

Hence, Tan A = Perpendicular/Base = 3/4

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