Question

In ∆ ABC, right-angled at B, AB = 24 cm, BC = 7 cm.



Determine:
(i) sin A, cos A
(ii) sin C, cos C

Answer 1

Given :-

Triangle ABC, right angled at B = ∠B = 90°

AB = 24 cm

BC = 7 cm

To Find :-

Determine:

(i) sin A, cos A

(ii) sin C, cos C

Solution :-

In a given triangle ABC, right angled at B = ∠B = 90°

Given that,

AB = 24 cm

BC = 7 cm

According to the Pythagoras theorem,

In a right angled triangle, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides.

By applying Pythagoras theorem,

$\underline{\boxed{\sf AC^{2}=AB^{2}+BC^{2} }}$

Substituting their values, we get

$\sf AC^{2} = (24)^{2}+7^{2}$

$\sf AC^{2} = (576+49)$

$\sf AC^{2} = 625 \ cm^{2}$

$\sf AC=\sqrt{625}=25$

Therefore, AC = 25 cm

Solution I :

To find Sin (A), Cos (A)

We know that,

Sin function is the equal to the ratio of length of the opposite side to the hypotenuse side.

According to the question,

$\underline{\boxed{\sf sin \ A=\dfrac{Opposite \ side}{Hypotenuse} }}$

Substituting their values,

$\sf sin \ A=\dfrac{AB}{AC} =\dfrac{24}{25}$

Cos function is equal to the ratio of the length of the adjacent side to the hypotenuse side.

Hence, it becomes

$\sf sin \ A=\dfrac{Opposite \ side}{Hypotenuse} =\dfrac{AB}{AC} =\dfrac{24}{25}$

Solution II :-

To find sin (C), cos (C)

$\sf Sin \ (C) =\dfrac{AB}{AC} =\dfrac{24}{25}$

$\sf Cos \ (C) =\dfrac{BC}{AC} =\dfrac{7}{25}$