Question

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Answer 1

Given:-

• AB of ∆ABC is 2 cm.
• AC of ∆ABC is 3 cm.
• Angle B of ∆ABC is 90°.

To find:-

• Value of sin A

Solution:

We know that,

◆ sinθ = $\frac{Perpendicular}{Hypotenuse}$

Sin A = $\frac{BC}{AC}$

• BC is not given.
• ∆ABC is right angled triangle. We can use Pythagoras theorem.

Pythagoras theorem:-

(Base)²+(Perpendicular)² = (Hypotenuse)²

Base = BC = ?

Perpendicular = AB = 2 cm

Hypotenuse = AC = 3 cm

Put the values in above formula

⇒(BC)²+(AB)² = (AC)²

⇒(BC)² + (2)² = (3)²

⇒(BC)² = (3)²-(2)²

⇒(BC)² = 9 - 4

⇒(BC)² = 5

⇒BC = √5

BC = √5

Sin A = $\frac{BC}{AC}$

Sin A = $\frac{</strong><strong>√</strong><strong>5</strong><strong>}{</strong><strong>3</strong><strong>}$

Therefore,

◆ Sin A = $\frac{√5}{3}$

______________________

More ratio's:-

◆ cos θ = $\frac{Base}{Hypotenuse}$

◆ tan θ = $\frac{Perpendicular}{Base}$

◆ cot θ = $\frac{Base}{Perpendicular}$

◆ sec θ = $\frac{Hypotenuse}{Base}$

◆ cosec θ = $\frac{Hypotenuse}{Perpendicular}$

Answer 2

Answer:

Given:-

AB of ∆ABC is 2 cm.

AC of ∆ABC is 3 cm.

Angle B of ∆ABC is 90°.

To find:-

Value of sin A

Solution:

We know that,

◆ sinθ = \frac{Perpendicular}{Hypotenuse}

Hypotenuse

Perpendicular

Sin A = \frac{BC}{AC}

AC

BC

BC is not given.

∆ABC is right angled triangle. We can use Pythagoras theorem.

Pythagoras theorem:-

(Base)²+(Perpendicular)² = (Hypotenuse)²

Base = BC = ?

Perpendicular = AB = 2 cm

Hypotenuse = AC = 3 cm

Put the values in above formula

⇒(BC)²+(AB)² = (AC)²

⇒(BC)² + (2)² = (3)²

⇒(BC)² = (3)²-(2)²

⇒(BC)² = 9 - 4

⇒(BC)² = 5

⇒BC = √5

BC = √5

Sin A = \frac{BC}{AC}

AC

BC

Sin A = \frac{√5}{3}

3

√5

Therefore,

◆ Sin A = \frac{√5}{3}

3

√5

______________________

More ratio's:-

◆ cos θ = \frac{Base}{Hypotenuse}

Hypotenuse

Base

◆ tan θ = \frac{Perpendicular}{Base}

Base

Perpendicular

◆ cot θ = \frac{Base}{Perpendicular}

Perpendicular

Base

◆ sec θ = \frac{Hypotenuse}{Base}

Base

Hypotenuse

◆ cosec θ = \frac{Hypotenuse}{Perpendicular}

Perpendicular

Hypotenuse