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Use the complementary angle identity x = (90^ - x).

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6 parts
a

α=50∘\alpha = 50^\circ

b

β=10∘\beta = 10^\circ

c

θ=18∘\theta = 18^\circ

d

γ=35∘\gamma = 35^\circ

e

ϕ=45∘\phi = 45^\circ

f

δ=50∘\delta = 50^\circ

Step-by-step solution

22 steps

Here are the solutions to the equations:

Part a· 3 steps

Solve cos⁡α=sin⁡40∘\cos \alpha = \sin 40^\circ

  1. Use the complementary angle identity sin⁡x=cos⁡(90∘−x)\sin x = \cos(90^\circ - x).

    sin⁡40∘=cos⁡(90∘−40∘)=cos⁡50∘\sin 40^\circ = \cos(90^\circ - 40^\circ) = \cos 50^\circ

  2. Substitute this back into the equation.

    cos⁡α=cos⁡50∘\cos \alpha = \cos 50^\circ

  3. Solve for α\alpha.

    α=50∘\alpha = 50^\circ

    The solution is α=50∘\boxed{\alpha = 50^\circ}.

Part b

β=10∘\beta = 10^\circ

Solve sin⁡β=cos⁡80∘\sin \beta = \cos 80^\circ

  1. Use the complementary angle identity cos⁡x=sin⁡(90∘−x)\cos x = \sin(90^\circ - x).

    cos⁡80∘=sin⁡(90∘−80∘)=sin⁡10∘\cos 80^\circ = \sin(90^\circ - 80^\circ) = \sin 10^\circ

  2. Substitute this back into the equation.

    sin⁡β=sin⁡10∘\sin \beta = \sin 10^\circ

  3. Solve for β\beta.

    β=10∘\beta = 10^\circ

    The solution is β=10∘\boxed{\beta = 10^\circ}.

Part c

θ=18∘\theta = 18^\circ

Solve sin⁡2θ=cos⁡3θ\sin 2\theta = \cos 3\theta

  1. Use the complementary angle identity cos⁡x=sin⁡(90∘−x)\cos x = \sin(90^\circ - x).

    cos⁡3θ=sin⁡(90∘−3θ)\cos 3\theta = \sin(90^\circ - 3\theta)

  2. Substitute this back into the equation.

    sin⁡2θ=sin⁡(90∘−3θ)\sin 2\theta = \sin(90^\circ - 3\theta)

  3. Equate the angles.

    2θ=90∘−3θ2\theta = 90^\circ - 3\theta

  4. Solve for θ\theta.

    2θ+3θ=90∘2\theta + 3\theta = 90^\circ 5θ=90∘5\theta = 90^\circ θ=90∘5\theta = \frac{90^\circ}{5} θ=18∘\theta = 18^\circ

    The solution is θ=18∘\boxed{\theta = 18^\circ}.

Part d

γ=35∘\gamma = 35^\circ

Solve sin⁡γ=cos⁡(γ+20∘)\sin \gamma = \cos(\gamma + 20^\circ)

  1. Use the complementary angle identity cos⁡x=sin⁡(90∘−x)\cos x = \sin(90^\circ - x).

    cos⁡(γ+20∘)=sin⁡(90∘−(γ+20∘))\cos(\gamma + 20^\circ) = \sin(90^\circ - (\gamma + 20^\circ)) cos⁡(γ+20∘)=sin⁡(90∘−γ−20∘)\cos(\gamma + 20^\circ) = \sin(90^\circ - \gamma - 20^\circ) cos⁡(γ+20∘)=sin⁡(70∘−γ)\cos(\gamma + 20^\circ) = \sin(70^\circ - \gamma)

  2. Substitute this back into the equation.

    sin⁡γ=sin⁡(70∘−γ)\sin \gamma = \sin(70^\circ - \gamma)

  3. Equate the angles.

    γ=70∘−γ\gamma = 70^\circ - \gamma

  4. Solve for γ\gamma.

    γ+γ=70∘\gamma + \gamma = 70^\circ 2γ=70∘2\gamma = 70^\circ γ=70∘2\gamma = \frac{70^\circ}{2} γ=35∘\gamma = 35^\circ

    The solution is γ=35∘\boxed{\gamma = 35^\circ}.

Part e

ϕ=45∘\phi = 45^\circ

Solve sin⁡ϕ−cos⁡ϕ=0\sin \phi - \cos \phi = 0

  1. Rearrange the equation.

    sin⁡ϕ=cos⁡ϕ\sin \phi = \cos \phi

  2. Divide both sides by cos⁡ϕ\cos \phi (assuming cos⁡ϕ≠0\cos \phi \neq 0).

    sin⁡ϕcos⁡ϕ=1\frac{\sin \phi}{\cos \phi} = 1

  3. Use the identity tan⁡ϕ=sin⁡ϕcos⁡ϕ\tan \phi = \frac{\sin \phi}{\cos \phi}.

    tan⁡ϕ=1\tan \phi = 1

  4. Solve for ϕ\phi.

    ϕ=arctan⁡(1)\phi = \arctan(1) ϕ=45∘\phi = 45^\circ

    The solution is ϕ=45∘\boxed{\phi = 45^\circ}.

Part f

δ=50∘\delta = 50^\circ

Solve cos⁡δ=sin⁡(δ−10∘)\cos \delta = \sin(\delta - 10^\circ)

  1. Use the complementary angle identity cos⁡x=sin⁡(90∘−x)\cos x = \sin(90^\circ - x).

    cos⁡δ=sin⁡(90∘−δ)\cos \delta = \sin(90^\circ - \delta)

  2. Substitute this back into the equation.

    sin⁡(90∘−δ)=sin⁡(δ−10∘)\sin(90^\circ - \delta) = \sin(\delta - 10^\circ)

  3. Equate the angles.

    90∘−δ=δ−10∘90^\circ - \delta = \delta - 10^\circ

  4. Solve for δ\delta.

    90∘+10∘=δ+δ90^\circ + 10^\circ = \delta + \delta 100∘=2δ100^\circ = 2\delta δ=100∘2\delta = \frac{100^\circ}{2} δ=50∘\delta = 50^\circ

    The solution is δ=50∘\boxed{\delta = 50^\circ}.

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