In a right-angled triangle ABC with a right angle at B, given AB = 24 cm and BC = 7 cm, we find sin A = 7/24, cos A = 24/25, sin C = 24/25, and cos C = 7/25.
In a right-angled triangle ABC, where B is the right angle, AB is the base, and BC is the perpendicular side, we can use trigonometric ratios to determine sin A, cos A, sin C, and cos C.
(i) To find sin A and cos A:
sin A = Opposite/Hypotenuse = BC/AB = 7/24
cos A = Adjacent/Hypotenuse = AB/BC = 24/25
(ii) To find sin C and cos C:
sin C = Opposite/Hypotenuse = AB/AC
Using the Pythagorean theorem (AC^2 = AB^2 + BC^2):
AC^2 = 24^2 + 7^2 = 625
AC = √625 = 25
sin C = AB/AC = 24/25
cos C = Adjacent/Hypotenuse = BC/AC = 7/25
The question probable may be:
In ∆ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine :
(i) sin A, cos A (ii) sin C, cos C AB is base and BC is perpendicular