Answer: The value of B is 60°.

Step-by-step explanation:

Required trigonmetric identities:

cos(2\theta)=cos^2\theta-sin^2\thetacos(2θ)=cos

2

θ−sin

2

θ

sin^2\theta+cos^2\theta=1sin

2

θ+cos

2

θ=1

We have, \begin{gathered}sinA=1-cosB \\\implies sinA=(sin^2(B/2)+cos^2(B/2))-(cos^2(B/2)-sin^2(B/2))\\\implies sinA=2sin^2(B/2)\\\implies 1/2=2sin^2(B/2) \\\implies sin^2(B/2)=1/4 \\\implies sin(B/2)= \pm 1/2\end{gathered}

sinA=1−cosB

⟹sinA=(sin

2

(B/2)+cos

2

(B/2))−(cos

2

(B/2)−sin

2

(B/2))

⟹sinA=2sin

2

(B/2)

⟹1/2=2sin

2

(B/2)

⟹sin

2

(B/2)=1/4

⟹sin(B/2)=±1/2

But,B is given to be acute,so is B/2 .Thus sin(B/2) is non-negative,that is,sin(B/2)=1/2.

Therefore,B/2=30°,that is,B=60°

Step-by-step explanation:

Required trigonmetric identities:

cos(2\theta)=cos^2\theta-sin^2\thetacos(2θ)=cos

2

θ−sin

2

θ

sin^2\theta+cos^2\theta=1sin

2

θ+cos

2

θ=1

We have, \begin{gathered}sinA=1-cosB \\\implies sinA=(sin^2(B/2)+cos^2(B/2))-(cos^2(B/2)-sin^2(B/2))\\\implies sinA=2sin^2(B/2)\\\implies 1/2=2sin^2(B/2) \\\implies sin^2(B/2)=1/4 \\\implies sin(B/2)= \pm 1/2\end{gathered}

sinA=1−cosB

⟹sinA=(sin

2

(B/2)+cos

2

(B/2))−(cos

2

(B/2)−sin

2

(B/2))

⟹sinA=2sin

2

(B/2)

⟹1/2=2sin

2

(B/2)

⟹sin

2

(B/2)=1/4

⟹sin(B/2)=±1/2

But,B is given to be acute,so is B/2 .Thus sin(B/2) is non-negative,that is,sin(B/2)=1/2.

Therefore,B/2=30°,that is,B=60°

Verified answer

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