Question

IF sinA =3/4 ,then calculate cosA and tanA

Answer 1

Answer:

cos A=(√7)/4 and tan A=3/(√7)

Step-by-step explanation:

We have,

sin A=3/4=p/b

so by Pythagoras theorem

h²=p²+b²

4²=3²+b²

16-9=b²=>b=√7

so h=4,p=3 and b=√7

by this we know

cosA=b/h=(√7)/4 and tanA=p/b=3/(√7)

Is your answer

Answer 2

Question:

If sinA = 3/4, Calculate the value of cosA and tanA.

 

Step-by-step explanation:

 

Given:

sinA = 3/4

 

To find:

The values of tanA and cosA.

 

Solution:

Let us draw a triangle ABC. We know that;

 

$\dashrightarrow \sf \ cosA = \dfrac{Side \ adjacent \ to \ A}{Hypotenuse}$

 

$\dashrightarrow \sf \ tanA = \dfrac{Side \ opposite \ to \ A}{Side \ adjacent \ to \ A}$

 

According to the question;

 

$\dashrightarrow \sf \ sinA = \dfrac{3}{4}$

 

Where "3x" is the measure of the side opposite to A and "4x" is the measure of the hypotenuse.

 

In ΔABC, ∠B = 90°

Using Pythagoras' Theorem;

 

➝ Hypotenuse² = Base² + Altitude²

➝ AC² = BC² + AB²

➝ (4x)² = (3x)² + AB²

➝ 16x² = 9x² + AB²

➝ 16x² - 9x² = AB²

➝ AB² = 7x²

➝ AB = √(7x²)

➝ AB = √7x cm.

 

Now, value of cosA is;

 

$\dashrightarrow \sf \ cosA = \dfrac{Side \ adjacent \ to \ A}{Hypotenuse}$

 

$\dashrightarrow \sf \ cosA = \dfrac{AB}{AC}$

 

$\dashrightarrow \sf \ cosA = \dfrac{\sqrt{7}x}{4x}$

 

$\dashrightarrow \sf \ cosA = \bold{\dfrac{\sqrt{7}}{4}}$

 

Now, value of tanA is;

 

$\dashrightarrow \sf \ tanA = \dfrac{Side \ opposite \ to \ A}{Side \ adjacent \ to \ A}$

 

$\dashrightarrow \sf \ tanA = \dfrac{BC}{AB}$

 

$\dashrightarrow \sf \ tanA = \dfrac{3x}{\sqrt{7}x}$

 

$\dashrightarrow \sf \ tanA = \bold{\dfrac{3}{\sqrt{7}}}$

 

Hence, solved.