Premium Online Home Tutors
3 Tutor System
Starting just at 265/hour
(i) sin A, cos A
(ii) sin C, cos C
Answer :
Given, in \( \triangle ABC \),
AB = 24 cm, BC = 7 cm and right angled at B
By using the Pythagoras theorem, we have
\(\Rightarrow AC^2 \ = \ AB^2 \ + \ BC^2 \)
\(\Rightarrow AC^2 \ = \ (24)^2 \ + \ (7)^2 \)
\( \Rightarrow AC^2 \ = \ 576 \ + \ 49 \ = \ 625 \)
\(\Rightarrow AC \ = \ \sqrt{625} \ = \ 25 \) cm
(i)
\( sinA \ = \ \frac{BC}{AC} \ = \ \frac{7}{25} \) [\(\because \) \( sin\theta \ = \ \frac{P}{H} \) ]
\( cosA \ = \ \frac{AB}{AC} \ = \ \frac{24}{25} \) [\(\because \)\( cos\theta \ = \ \frac{B}{H} \) ]
(ii) \( sinC \ = \ \frac{AB}{AC} \ = \ \frac{24}{25} \)
\( cosC \ = \ \frac{BC}{AC} \ = \ \frac{7}{25} \)