In ∆ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine: (i) sin A, cos A (ii) sin C, cos C
Answers
Answer:
Brainliest please
Step-by-step explanation:
Let AB be the vertical pole and AC be the rope.
GiveN :
Length of rope (AC) = 20m
Angel made by the rope with the ground level (∠ACB=30) = 30°
To finD :
Height of the pole (AB)
Solution :
We know that sin θ = \small{\sf{\frac{side\: opposite\:to\: θ}{hypotenuse} }}
hypotenuse
sideoppositetoθ
or \small{\sf{\frac{AB}{AC} }}
AC
AB
Hence,
Sin30° = \large{\sf{ \frac{AB }{AC} }}
AC
AB
Answer:
(i) sin A = 7/25
cos A = 24/25
(ii) sin C = 24/25
cos C = 7/25
Step-by-step explanation:
Given: ∆ ABC is right-angled at B.
AB = 24 cm
BC = 7 cm
Pythagoras formula:
(Hypotenuse)² = (Perpendicular)² + (Base)²
(AC)² = (AB)² + (BC)²
(AC)² = (24)² + (7)²
AC = √(576 + 49)
AC = √625 = 25
NOW:
In ∆CBA,
Perpendicular = BC
Base = AB
Hypotenuse = AC
(i) sin A = p/h = BC/AC = 7/25
cos A = b/h = AB/AC = 24/25
In ∆ ABC,
Perpendicular = AB
Base = BC
Hypotenuse = AC
(ii) sin C = p/h = AB/AC = 24/25
cos C = b/h = BC/AC = 7/25