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# In $$\triangle ABC$$ , right angled at B, AB = 24 cm, BC = 7 cm. Determine : (i) sin A, cos A (ii) sin C, cos C

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) $$sinC \ = \ \frac{AB}{AC} \ = \ \frac{24}{25}$$

$$cosC \ = \ \frac{BC}{AC} \ = \ \frac{7}{25}$$