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Q1. In \( ∆ \ ABC \) , right angled at B, AB = 24 cm, BC = 7 cm. Determine :

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


Answer :


Given, in \( ∆ \ ABC \),
AB = 24 cm, BC = 7 cm and right angled at B



By using the Pythagoras theorem, we have

\( AC^2 \ = \ AB^2 \ + \ BC^2 \)
=> \( AC^2 \ = \ (24)^2 \ + \ (7)^2 \)
=> \( AC^2 \ = \ 576 \ + \ 49 \ = \ 625 \)

=>   \( AC \ = \ \sqrt{625} \ = \ 25 \) cm

(i) \( sinA \ = \ \frac{BC}{AC} \ = \ \frac{7}{25} \)      [∵ \( sin\theta \ = \ \frac{P}{H} \) ]

\( cosA \ = \ \frac{AB}{AC} \ = \ \frac{24}{25} \)      [∵ \( cos\theta \ = \ \frac{B}{H} \) ]

(ii) \( sinC \ = \ \frac{AB}{AC} \ = \ \frac{24}{25} \)

\( cosC \ = \ \frac{BC}{AC} \ = \ \frac{7}{25} \)