Question

Find AB in the given diagram ? ​

Answer 1

$\large\underline{\sf{Given- }}$

A right angle triangle ABC right-angled at C and BC = 2 units and ∠BAC = 25°.

$\blue{\large\underline{\sf{To\:Find - }}}$

Length of AB

$\red{\large\underline{\sf{Solution-}}}$

It is given that

A right angle triangle ABC right-angled at C and BC = 2 units and ∠BAC = 25°.

In right-angled triangle ABC,

We know from Trigonometric ratios

$\red{\rm :\longmapsto\:sin \angle \: BAC \: = \: \dfrac{opposite \: side}{hypotenuse} \: }$

$\red{\rm :\longmapsto\:sin 25 \degree \: = \: \dfrac{BC}{BA} \: }$

So, [ using Sine Angles Table, attached file ]

$\rm :\longmapsto\:0.4226 = \dfrac{2}{BA}$

$\rm :\longmapsto\:BA = \dfrac{2}{0.4226}$

$\bf\implies \:BA = 4.7326 \: units$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1

Answer 2

$\large\underline{\sf{Given- }}$

A right angle triangle ABC right-angled at C and BC = 2 units and ∠BAC = 25°.

$\blue{\large\underline{\sf{To\:Find - }}}$

Length of AB

$\red{\large\underline{\sf{Solution-}}}$

It is given that

A right angle triangle ABC right-angled at C and BC = 2 units and ∠BAC = 25°.

In right-angled triangle ABC,

We know from Trigonometric ratios

$\red{\rm :\longmapsto\:sin \angle \: BAC \: = \: \dfrac{opposite \: side}{hypotenuse} \: }$

$\red{\rm :\longmapsto\:sin 25 \degree \: = \: \dfrac{BC}{BA} \: }$

So, [ using Sine Angles Table, attached file ]

$\rm :\longmapsto\:0.4226 = \dfrac{2}{BA}$

$\rm :\longmapsto\:BA = \dfrac{2}{0.4226}$

$\bf\implies \:BA = 4.7326 \: units$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1

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