Question

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

Answer 1

Given :-

• ${ \rm{sin \:A = \frac{3}{4} }}$

Need to find :-

• cos A and tan A ?

Solution :-

$\leadsto{ \rm{sin \: A = \frac{3}{4} - - - (1) }}$

• We know that :

$\leadsto{ \rm{sin \: A = \frac{P}{H} - - - (2) }}$

• We can say that :

$\leadsto{ \rm{ \frac{P}{H} = \frac{3}{4} }}$

Here,

• Perpendicular (P) = 3
• Hypotenuse (H) = 4

Finding Base :-

• By Pythagoras Theorem :

→ (H)² = (B)² + (P)²

→ 4² = B² + 3²

→ 16 = B² + 9

→ 16 - 9 = B²

→ 7 = B²

→ √7 = B

Hence,

• Base is √7

• Now, we've :

• Perpendicular (P) = 3
• Hypotenuse (H) = 4
• Base (B) = √7

So, we know that,

$\leadsto{ \boxed{ \rm{cos \: A = \frac{B}{H} }}}$

• Now, Substitute the values :

$\leadsto{ \rm{cos \: A = \frac{ \sqrt{7} }{4} }}$

Or,

$\leadsto{ \boxed{ \rm{tan \: A = \frac{P}{B} }}}$

• Now, Substitute the values :

$\leadsto{ \rm{tan\: A = \frac{3 }{√7} }}$

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Answer 2

We will use the basic formula of sine, cosine, and tangent functions to solve the question.

Let's draw a figure according to the given question.

If sin A = 3/4, calculate cos A and tan A.

Let ∆ABC be a right-angled triangle, right-angled at point B.

Given that:

sin A = 3/4

⇒ BC/AC = 3/4

Let BC be 3k. Therefore, hypotenuse AC will be 4k where k is a positive integer.

Applying Pythagoras theorem on ∆ABC, we obtain:

AC2 = AB2 + BC2

AB2 = AC2 - BC2

AB2 = (4k)2 - (3k)2

AB2 = 16k2 - 9k2

AB2 = 7 k2

AB = √7 k

cos A = side adjacent to ∠A / hypotenuse = AB/AC = √7 k / 4k = √7/4

tan A = side opposite to ∠A / side adjacent to ∠A = BC/AB = 3k / √7 k = 3/√7

Thus, cos A= √7/4 and tan A = 3/√7