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Solve for θ with restriction 0° < θ < 90° correct to one decimal place. a) sin θ = 1.9/2.6 b) 2 tan θ = 1 c) tan 20°/2 = 5.102 d) cos θ = sin 20° e) sin θ = tan 40°

Mathematics

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Answers

4 parts
a

47.0∘47.0^\circ

b

26.6∘26.6^\circ

68.6∘68.6^\circ

d

70.0∘70.0^\circ

e

57.0∘57.0^\circ

Step-by-step solution

15 steps

We need to solve for θ\theta in the range 0∘<θ<90∘0^\circ < \theta < 90^\circ and round the answers to one decimal place.

Part a· 3 steps

Solve for θ\theta: sin⁡θ=1.92.6\sin \theta = \frac{1.9}{2.6}

  1. Calculate the value of the ratio.

    sin⁡θ=1.92.6≈0.730769\sin \theta = \frac{1.9}{2.6} \approx 0.730769

  2. Use the inverse sine function to find θ\theta.

    θ=sin⁡−1(0.730769)\theta = \sin^{-1}(0.730769) θ≈46.959∘\theta \approx 46.959^\circ

  3. Round to one decimal place.

    θ≈47.0∘\theta \approx \boxed{47.0^\circ}

Part b

26.6∘26.6^\circ

Solve for θ\theta: 2tan⁡θ=12 \tan \theta = 1

  1. Isolate tan⁡θ\tan \theta.

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

  2. Use the inverse tangent function to find θ\theta.

    θ=tan⁡−1(0.5)\theta = \tan^{-1}(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}

    c) Solve for θ\theta: The question as written is tan⁡20∘=5.1022\tan 20^\circ = \frac{5.102}{2}, which does not contain θ\theta. Assuming there was a typo and the question intended to be tan⁡θ=5.1022\tan \theta = \frac{5.102}{2}.

  4. Calculate the value of the ratio.

    tan⁡θ=5.1022\tan \theta = \frac{5.102}{2} tan⁡θ=2.551\tan \theta = 2.551

  5. Use the inverse tangent function to find θ\theta.

    θ=tan⁡−1(2.551)\theta = \tan^{-1}(2.551) θ≈68.599∘\theta \approx 68.599^\circ

  6. Round to one decimal place.

    θ≈68.6∘\theta \approx \boxed{68.6^\circ}

Part d

70.0∘70.0^\circ

Solve for θ\theta: cos⁡θ=sin⁡20∘\cos \theta = \sin 20^\circ

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

    cos⁡θ=cos⁡(90∘−20∘)\cos \theta = \cos (90^\circ - 20^\circ) cos⁡θ=cos⁡70∘\cos \theta = \cos 70^\circ

  2. Since 0∘<θ<90∘0^\circ < \theta < 90^\circ, we can equate the angles.

    θ=70∘\theta = 70^\circ

  3. Round to one decimal place.

    θ=70.0∘\theta = \boxed{70.0^\circ}

Part e

57.0∘57.0^\circ

Solve for θ\theta: sin⁡θ=tan⁡40∘\sin \theta = \tan 40^\circ

  1. Calculate the value of tan⁡40∘\tan 40^\circ.

    tan⁡40∘≈0.8390996\tan 40^\circ \approx 0.8390996

  2. Use the inverse sine function to find θ\theta.

    sin⁡θ=0.8390996\sin \theta = 0.8390996 θ=sin⁡−1(0.8390996)\theta = \sin^{-1}(0.8390996) θ≈57.019∘\theta \approx 57.019^\circ

  3. Round to one decimal place.

    θ≈57.0∘\theta \approx \boxed{57.0^\circ}
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