Question

if Ø and Ø+36 are acute angles and sin(ø+36)= cos ø then the value of ø is​

Answer 1

GIVEN :-

• θ and (θ + 36°) are acute angles.
• sin(θ + 36°) = cosθ.

TO FIND :-

• The value of θ .

SOLUTION :-

In the question it is given that "θ" is an acute angle , so we can say that , θ < 90°. [ Acute angles = less than 90° ]

Also , it is given that "θ + 36°" is an acute angle , so we can say that , (θ + 36°) < 90°.

⇒ sin(θ + 36°) = cosθ.

⇒ sin(θ + 36°) = sin(90° - θ)

⇒ θ + 36° = [{sin(90° - θ)}/{sin}]

On cancelling sin with sin we get,

⇒ θ + 36° = 90° - θ

⇒ θ + θ = 90° - 36°

⇒ 2θ = 54

⇒ θ = 54/2

⇒ θ = 27°

Hence , θ < 90° = 27° < 90°.

Also , (θ + 36°) < 90° = 27° + 36° < 90° = 63° < 90°

Hence the required value of θ is 27°.

Answer 2

Correct question :

If θ and θ + 36 are acute angles and sin ( θ + 36 ) = cos θ , then the value of θ is

Answer :

Given :

• θ and θ + 36° are acute angles

• sin ( θ + 36° ) = cos θ

To find :

• Value of θ

According to the question :

↦sin ( θ + 36° ) = cos θ = sin ( 90° - θ )

We know that,

⇛Acute angle = less than 90°

⇛[ sin ( 90° - θ ) = cos θ ]

On comparing, we get :

⟹ θ + 36° = 90° - θ

⟹ θ + θ = 90° - 36°

⟹ 2θ = 54°

⟹ θ = 54° / 2

⟹ ∴ θ = 27°

So, It's Done !!