Question

$\huge{\mathcal\ \textless \ br /\ \textgreater \ {\underline{\green{QuEsTiOn}}}}$

If Sin A =
$\frac{3}{4}$
, Calculate cos A and tan A.

​

Answer 1

Answer:

$\huge{\mathcal{\underline{\green{AnSWER}}}}$

Let us say, ABC is a right-angled triangle, right-angled at B.

Sin A = $\frac{3}{4}$

As we know,

Sin A = Opposite Side/Hypotenuse Side = $\frac{3}{4}$

Now, let BC be 3k and AC will be 4k.

where k is the positive real number.

As per the Pythagoras theorem, we know;

Hypotenuse2 = Perpendicular2+ Base2

AC2 = AB2 + BC2

Substitute the value of AC and BC in the above expression to get;

(4k)2 = (AB)2 + (3k)2

16k2 – 9k2 = AB2

AB2 = 7k2

Hence, AB = √7 k

Now, as per the question, we need to find the value of cos A and tan A.

cos A = Adjacent Side/Hypotenuse side = AB/AC

cos A = √7 k/4k = √7/4

And,

tan A = Opposite side/Adjacent side = BC/AB

tan A = 3k/√7 k = 3/√7

Answer 2

Answer:

AnSWER

Let us say, ABC is a right-angled triangle, right-angled at B.

Sin A = \frac{3}{4}

4

3

As we know,

Sin A = Opposite Side/Hypotenuse Side = \frac{3}{4}

4

3

Now, let BC be 3k and AC will be 4k.

where k is the positive real number.

As per the Pythagoras theorem, we know;

Hypotenuse2 = Perpendicular2+ Base2

AC2 = AB2 + BC2

Substitute the value of AC and BC in the above expression to get;

(4k)2 = (AB)2 + (3k)2

16k2 – 9k2 = AB2

AB2 = 7k2

Hence, AB = √7 k

Now, as per the question, we need to find the value of cos A and tan A.

cos A = Adjacent Side/Hypotenuse side = AB/AC

cos A = √7 k/4k = √7/4

And,

tan A = Opposite side/Adjacent side = BC/AB

tan A = 3k/√7 k = 3/√7