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Introduction To Trigonometry
EXERCISE 8.1
CBSE_NCERT_TEXTBOOK
Grade 10
Question:
In ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine : (i) sin A, cos A (ii) sin C, cos C
Step-by-Step Solution
Key Concept: Use the definitions of sine and cosine in a right‑angled triangle: for an acute angle, \(\sin\theta = \frac{\text{opposite side}}{\text{hypotenuse}}\) and \(\cos\theta = \frac{\text{adjacent side}}{\text{hypotenuse}}\). First find the hypotenuse \(AC\) using Pythagoras theorem.
1. Find the hypotenuse \(AC\)\ Since \(\triangle ABC\) is right‑angled at \(B\),\ $$AC^2 = AB^2 + BC^2$$\ $$AC^2 = 24^2 + 7^2 = 576 + 49 = 625$$\ $$AC = \sqrt{625} = 25\text{ cm}$$
2. For angle \(A\)\ - Opposite side to \(A\) is \(BC = 7\) cm.\ - Adjacent side to \(A\) is \(AB = 24\) cm.\ $$\sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{BC}{AC} = \frac{7}{25}$$\ $$\cos A = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{AB}{AC} = \frac{24}{25}$$
3. For angle \(C\)\ - In a right triangle, \(\angle C = 90^{\circ} - \angle A\); therefore the opposite side to \(C\) is \(AB\) and the adjacent side is \(BC\).\ $$\sin C = \frac{AB}{AC} = \frac{24}{25}$$\ $$\cos C = \frac{BC}{AC} = \frac{7}{25}$$
Thus the required trigonometric ratios are obtained.
Correct Answer:(i) \(\sin A = \frac{7}{25},\; \cos A = \frac{24}{25}\)\
(ii) \(\sin C = \frac{24}{25},\; \cos C = \frac{7}{25}\)
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