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Find the angle theta for each equation, rounded to one decimal place.

Mathematics

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5 parts
1.1

78.5∘78.5^\circ

1.2

4.4∘4.4^\circ

1.3

49.9∘49.9^\circ

1.4

26.6∘26.6^\circ

1.5

14.5∘14.5^\circ

Step-by-step solution

17 steps

Here are the solutions for finding the angle θ\theta for each equation, rounded to one decimal place.

Part 1.1· 2 steps

  1. Isolate θ\theta.

    cos⁡θ=0.2\cos \theta = 0.2 θ=arccos⁡(0.2)\theta = \arccos(0.2)

  2. Calculate the value and round to one decimal place.

    θ≈78.463∘\theta \approx 78.463^\circ θ≈78.5∘\theta \approx \boxed{78.5^\circ}

Part 1.2

4.4∘4.4^\circ

  1. Isolate sin⁡4θ\sin 4\theta.

    sin⁡4θ−0.3=0\sin 4\theta - 0.3 = 0 sin⁡4θ=0.3\sin 4\theta = 0.3

  2. Find the principal value for 4θ4\theta.

    4θ=arcsin⁡(0.3)4\theta = \arcsin(0.3) 4θ≈17.457∘4\theta \approx 17.457^\circ

  3. Solve for θ\theta.

    θ=17.457∘4\theta = \frac{17.457^\circ}{4} θ≈4.364∘\theta \approx 4.364^\circ

  4. Round to one decimal place.

    θ≈4.4∘\theta \approx \boxed{4.4^\circ}

Part 1.3

49.9∘49.9^\circ

  1. Find the principal value for (θ−25∘)(\theta - 25^\circ).

    tan⁡(θ−25∘)=0.465\tan(\theta - 25^\circ) = 0.465 θ−25∘=arctan⁡(0.465)\theta - 25^\circ = \arctan(0.465) θ−25∘≈24.928∘\theta - 25^\circ \approx 24.928^\circ

  2. Solve for θ\theta.

    θ=24.928∘+25∘\theta = 24.928^\circ + 25^\circ θ≈49.928∘\theta \approx 49.928^\circ

  3. Round to one decimal place.

    θ≈49.9∘\theta \approx \boxed{49.9^\circ}

Part 1.4

26.6∘26.6^\circ

  1. Isolate tan⁡θ\tan \theta.

    2tan⁡θ=12 \tan \theta = 1 tan⁡θ=12\tan \theta = \frac{1}{2} tan⁡θ=0.5\tan \theta = 0.5

  2. Find the principal value for θ\theta.

    θ=arctan⁡(0.5)\theta = \arctan(0.5) θ≈26.565∘\theta \approx 26.565^\circ

  3. Round to one decimal place.

    θ≈26.6∘\theta \approx \boxed{26.6^\circ}

Part 1.5

14.5∘14.5^\circ

  1. Isolate cos⁡2θ\cos 2\theta.

    4cos⁡2θ=3.54 \cos 2\theta = 3.5 cos⁡2θ=3.54\cos 2\theta = \frac{3.5}{4} cos⁡2θ=0.875\cos 2\theta = 0.875

  2. Find the principal value for 2θ2\theta.

    2θ=arccos⁡(0.875)2\theta = \arccos(0.875) 2θ≈28.955∘2\theta \approx 28.955^\circ

  3. Solve for θ\theta.

    θ=28.955∘2\theta = \frac{28.955^\circ}{2} θ≈14.477∘\theta \approx 14.477^\circ

  4. Round to one decimal place.

    θ≈14.5∘\theta \approx \boxed{14.5^\circ}

Part 1.6· 1 step

  1. Multiply both sides by 3.

    2sin⁡(θ−10∘)3=0.102\frac{2 \sin(\theta - 10^\circ)}{3} = 0.102 2sin⁡(θ−10∘)=0.102×32 \sin(\theta - 10^\circ) = 0.102 \times 3
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