Question

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

Answer 1

Answer:

Height =3 base=4

Hypotenuse = 2x(3) and (4)

=9+16

=25

25 square is 5

So cos A =base/Hypotenuse

=4/5

And tan A =base/height

=4/3

Answer 2

Answeram:-

Givenam:-

• $\rm{sin \ A = \frac{3}{4}}$

To Findam:-

• $\rm{ cos \ A}$ &
• $\rm{tan \ A}$.

Let ABC be the required triangle where angle C = 90°.

We know,

$\rm{sin∅ = \frac{Opposite \ side \ of \ ∅}{Hypotenuse}}$

Ratio of opposite side of A and Hypotenuse = 3:4

Let common be = $k$

So, Hypotenuse = 4k (AC)

Opposite side of A = 3k (BC)

Finding $AB$:-

${AB}^{2} = {AC}^{2} - {BC}^{2}$

$\implies AB = \sqrt{{AC}^{2} - {BC}^{2}}$

$\implies AB = \sqrt{{(4k)}^{2} - {(3k)}^{2}}$

$\implies AB = \sqrt{{16k}^{2} - {9k}^{2}}$

$\implies AB = \sqrt{{7k}^2}$

$\implies AB = k \sqrt{7}$

NOW,

Finding:-

$\underline{cos \ A :-}$

We know,

$\rm{ cos \ ∅ = \frac{Adjacent \ side \ of \ ∅}{Hypotenuse}}$

So,

$\rm{\implies cos \ A = \frac{Adjacent \ side \ of \ A}{Hypotenuse}}$

$\rm{\implies cos \ A = \frac{AB}{AC}}$

$\rm{\implies cos \ A = \frac{k \sqrt{7}}{4k}}$

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

$\underline{tan \ A:-}$

We know,

$\rm{tan \ ∅ = \frac{Opposite \ side \ of \ ∅}{Adjacent \ side \ of \ ∅}}$

So,

$\rm{ tan \ A = \frac{Opposite \ side \ of \ A}{Adjacent \ side \ of \ A}}$

$\rm{ \implies tan \ A = \frac{BC}{AB}}$

$\rm{\implies tan \ A = \frac{3k}{k \sqrt{7}}}$

$\rm{\implies tan \ A = \frac{3}{\sqrt{7}}}$

Thuz,

• $tan \ A = \frac{3}{\sqrt{7}}$
• $cos \ A = \frac{\sqrt{7}}{4}$.

Edi okay na?