Math, asked by ankitansu659, 9 months ago

3. In triangle ABC right angled at B, BC = 3 cm and AC – AB = 1 cm.

Determines the value of sin A, sin C and cosec A and cot C.​

Answers

Answered by SohamAgarwal
0

Answer:

sinA = 0.6

sin C = 0.8

cosec A = 5/3

cot C = 0.75

Step-by-step explanation:

As the triangle is right angled at B, therefore, AC is the hypotenuse, and AB and BC are its other 2 sides.

AC - AB = 1

AB = AC - 1

By Pythagoras theorem,

(Base)^2 + (Height)^2 = (Hypotenuse)^2

(3)^2 + (AC - 1)^2 = AC^2

9 + AC^2 + 1 - 2AC = AC^2

    (We eliminate AC^2 on both sides)

9 + 1 = 2AC

AC = 5 cm

AB = AC - 1 = 4 cm

sin A = Opposite/Hypotenuse = BC/AC = 3/5

sinA = 0.6

sin C = Opposite/Hypotenuse = AB/AC = 4/5

sin C = 0.8

cosec A = 1 / sin A

cosec A = 5/3

cot C = Adjacent/Opposite = BC/AB = 3/4

cot C = 0.75

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