1. In A ABC, right-angled at B, AB= 24 cm, BC = 7 cm. Determine:(1) si

1.	1. In A ABC, right-angled at B, AB= 24 cm, BC = 7 cm. Determine:(1) sin A, cos A(ü) sin C, cos C
Answer» In ∆ ABC, AB = 24cm, BC = 7 cm. AC = ? ↪ By USING Pythagoras Theorem, we get the VALUE of AC. (HYPOTENUSE)² = (SIDE)² + (SIDE)² ➡ (AC)² = (BC)² + (AB)² ➡ (AC)² = (7)² + (24)² ➡ (AC)² = 49 + 576 ➡ (AC)² = 625 ➡ AC = √625 ➡ AC = 25. ∴ AC = 25 cm. (1) sin A, cos A ↪ sin A : ➡ sin A = opposite/hypotenuse ➡ sin A = BC/AC ➡ sin A = 7/25 ↪ cos A : ➡ cos A = adjacent/hypotenuse ➡ cos A = AB/AC ➡ cos A = 24/25 (2) sin C, cos C ↪ sin C : ➡ sin C = opposite/hypotenuse ➡ sin C = AB/AC ➡ sin C = 24/25 ↪ cos C : ➡ cos C = adjacent/hypotenuse ➡ cos C = BC/AC ➡ cos C = 7/25 Hence, it is solved... Step-by-step explanation: þLÈÄ§È MÄRK MÈ Ä§ ßRÄÌñLÌÈ§†✌✌✌

1. In A ABC, right-angled at B, AB= 24 cm, BC = 7 cm. Determine:(1) sin A, cos A(ü) sin C, cos C

Answer»

In ∆ ABC, AB = 24cm, BC = 7 cm. AC = ?

↪ By USING Pythagoras Theorem, we get the VALUE of AC.

(HYPOTENUSE)² = (SIDE)² + (SIDE)²

➡ (AC)² = (BC)² + (AB)²

➡ (AC)² = (7)² + (24)²

➡ (AC)² = 49 + 576

➡ (AC)² = 625

➡ AC = √625

➡ AC = 25.

∴ AC = 25 cm.

(1) sin A, cos A

↪ sin A :

➡ sin A = opposite/hypotenuse

➡ sin A = BC/AC

➡ sin A = 7/25

↪ cos A :

➡ cos A = adjacent/hypotenuse

➡ cos A = AB/AC

➡ cos A = 24/25

(2) sin C, cos C

↪ sin C :

➡ sin C = opposite/hypotenuse

➡ sin C = AB/AC

➡ sin C = 24/25

↪ cos C :

➡ cos C = adjacent/hypotenuse

➡ cos C = BC/AC

➡ cos C = 7/25

Hence, it is solved...

Step-by-step explanation:

þLÈÄ§È MÄRK MÈ Ä§ ßRÄÌñLÌÈ§†✌✌✌