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

Answer 1

(i) Given: In △ABC, AB = 24 cm, BC = 7 cm and ∠ABC = 90° 

To find: sin A, cos A 

By using Pythagoras theorem in △ABC we have 

AC² = AB² + BC² 

AC² = 24² + 7² 

AC² = 576 + 49 

AC² = 625 

AC = √625 

AC= 25 

Hence, Hypotenuse = 25 

By definition,

sin A = \(\frac{Perpendicular\, side\, opposite\, to\, angle\, A}{ Hypotenuse\, }\)

sin A = \(\frac{BC}{AC}\) 

sin A = \(\frac{7}{25}\) 

And, 

cos A = \(\frac{Base\, side\, adjacent\, to\, angle\, A}{Hypotenuse }\)

cos A = \(\frac{AB}{AC}\) 

cos A = \(\frac{24}{25}\) 

(ii) Given: In △ABC , AB = 24 cm and BC = 7 cm and ∠ABC = 90°

To find: sin C, cos C 

By using Pythagoras theorem in △ABC we have 

AC² = AB² + BC² 

AC² = 24² + 7² 

AC² = 576 + 49 

AC² = 625 

AC = √625 

AC= 25 

Hence, Hypotenuse = 25 

By definition, 

sin C =\(\frac{ Perpendicular\, side\, opposite\, to\, angle\, C}{Hypotenuse }\)

sin C = \(\frac{AB}{AC}\) 

sin C = \(\frac{24}{25 }\)

And, 

cos C = \(\frac{Base\, side\, adjacent\, to\, angle\, C}{Hypotenuse }\)

cos A = \(\frac{BC}{AC}\) 

cos A =\(\frac{ 7}{25}\)