Question

In the given figure, ABC is a right angled triangle .Find the value of
(i) Sec A
(ii) Tan A
(iii) Sec A

Answer 1

$\huge\bigstar\underbrace{\mathtt\red{A} \mathtt\red{N} \mathtt\purple{S} \mathtt\blue{W}\mathtt\purple{E} \mathtt\green{R}} \bigstar\:$

Given ; AB = 3 units , BC = 4 units AC = 5 units

(i)Sin A

Opposite Side/Hypotenuse = BC/AC = 4/5

(ii) Tan A

Opposite Side/Adjacent Side = AB/BC = 3/4

(iii) Sec A

= 1/Cos A = 1/[Adjacent/Hypotenuse]

= Hypotenuse/Adjacent Side = 1/[AB/AC] = 5/3

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Trigonometric Ratios

• Cos = Adjacent side/Hypotenuse

• Tan = Opposite Side/Adjacent Side

• Sin = Opposite Side/Hypotenuse

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Answer 2

Answer:

Given ; AB = 3 units , BC = 4 units AC = 5 units

(i)Sin A

Opposite Side/Hypotenuse = BC/AC = 4/5

(ii) Tan A

Opposite Side/Adjacent Side = AB/BC = 3/4

(iii) Sec A

= 1/Cos A = 1/[Adjacent/Hypotenuse]

= Hypotenuse/Adjacent Side = 1/[AB/AC] = 5/3

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Trigonometric Ratios

• Cos = Adjacent side/Hypotenuse

• Tan = Opposite Side/Adjacent Side

• Sin = Opposite Side/Hypotenuse

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