Question

If sin A = 3/4 , then find value of sec A.

Answer 1

Solution :

Here, it is given that sinA = 3/4

Formula used :

• sinA = P/H
• secA = H/B

Where

• P = Perpendicular of the triangle
• H = Hypotenuse of the triangle
• B = Base of the triangle

Here

Perpendicular and hypotenuse of the triangle are 3units and 4units respectively.

Let the ratio be 3k and 4k respectively.

It's a right angled triangle. So, sum of the squares of perpendicular and base will be equal to the square of hypotenuse of the triangle.

Now

=> H² = P² + B²

=> (4k)² = (3k)² + B²

=> 16k² = 9k² + B²

=> B² = 16k² - 9k²

=> B² = 7k²

=> B = √7k

Now

=> secA = H/B

=> 4k/√7k

=> 4/√7 units

More to know :

• sinA = P/H
• cosA = B/H
• tanA = P/B
• cosecA = H/P
• secA = H/B
• cotA = B/P

Answer 2

Correct Question:-

If sin A = 3/4 , then find value of sec A.

To find:-

• Sec A

Solution:-

Let,

• A = Adjacent
• H = Hypotenuse
• O = opposite

Given,

$\: \: \: \: \: \: \: \: \bullet{ \large{ \: \: \: \: \: \: \rm{sin \: A = \frac{O}{H} = \frac{3}{5} }}}$

Now,

• O = 3
• H = 5

By Pythagoras theorem,

$\underline{ \boxed{ \large{ {A}^{2} = \sqrt{ {H}^{2} - {O}^{2} } }}}$

According to question,

$\large { \longmapsto\sf \: \: \: \: \: \: \: \: \: A = \sqrt{ {5}^{2} - {3}^{2} } }$

$\large { \longmapsto\sf \: \: \: \: \: \: \: \: \: A = \sqrt{16} }$

$\large { \longmapsto\sf \: \: \: \: \: \: \: \:{ \boxed { \boxed{ \purple{ \tt{ A =4}}}}}} \star$

Hence:-

$\: \: \: \: \: \: \: \: \bullet{ \large{ \: \: \: \: \: \: \rm{sec \: A = \frac{H}{A} = \boxed{ \large{ \green{\frac{5}{4} }}}}}}$