# In a ∆ABC, right angled at B, AB = 24 cm, BC = 7 cm. Determine

Question:

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

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

Solution:

(i) The given triangle is below:-

$B C=7 \mathrm{~cm}$

$\angle A B C=90^{\circ}$

To Find: $\sin A, \cos A$

In this problem, Hypotenuse side is unknown

Hence we first find Hypotenuse side by Pythagoras theorem

By Pythagoras theorem,

We get,

$A C^{2}=A B^{2}+B C^{2}$

$A C^{2}=24^{2}+7^{2}$

$A C^{2}=576+49$

$A C^{2}=625$

$A C=\sqrt{625}$

$A C=25$

Hypotenuse $=25$

By definition,

$\sin A=\frac{\text { Perpendicular side opposite to } \angle A}{\text { Hypotenuse }}$

$\sin A=\frac{B C}{A C}$

$\sin A=\frac{7}{25}$

By definition,

$\cos A=\frac{\text { Base side adjacent to } \angle A}{\text { Hypotenuse }}$

$\cos A=\frac{A B}{A C}$

$\cos A=\frac{24}{25}$

$\sin A=\frac{7}{25} \cos A=\frac{24}{25}$

(ii) The given triangle is below:

Given: In $\triangle \mathrm{ABC}, A B=24 \mathrm{~cm}$

$B C=7 \mathrm{~cm}$

$\angle A B C=90^{\circ}$

To Find: $\sin C, \cos C$

In this problem, Hypotenuse side is unknown

Hence we first find Hypotenuse side by Pythagoras theorem

By Pythagoras theorem,

We get,

$A C^{2}=A B^{2}+B C^{2}$

$A C^{2}=24^{2}+7^{2}$

$A C^{2}=576+49$

$A C^{2}=625$

$A C=\sqrt{625}$

$A C=25$

Hypotenuse $=25$

By definition,

$\sin C=\frac{\text { Perpendicular side opposite to } \angle \mathrm{C}}{\text { Hypotenuse }}$

$\sin C=\frac{A B}{A C}$

$\sin C=\frac{24}{25}$

By definition,

$\cos C=\frac{\text { Base side adjacent to } \angle C}{\text { Hypotenuse }}$

$\cos C=\frac{B C}{A C}$

$\cos C=\frac{7}{25}$

$\sin C=\frac{24}{25} \cos C=\frac{7}{25}$