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Math  /  Trigonometry

Question2.2.1 sin⁡2θ+1,283=tan⁡62∘\sin 2 \theta+1,283=\tan 62^{\circ} 2.2.2 2cot⁡(θ−25∘)=52 \cot \left(\theta-25^{\circ}\right)=5 3∣Page3 \mid \mathrm{Page}

Studdy Solution

STEP 1

1. The problem consists of two separate trigonometric equations to solve for θ\theta.
2. We will use trigonometric identities and algebraic manipulation to solve each equation.

STEP 2

1. Solve equation 2.2.1: sin⁡2θ+1.283=tan⁡62∘\sin 2 \theta + 1.283 = \tan 62^{\circ}.
2. Solve equation 2.2.2: 2cot⁡(θ−25∘)=52 \cot (\theta - 25^{\circ}) = 5.

STEP 3

First, solve equation 2.2.1 by isolating sin⁡2θ\sin 2 \theta.
sin⁡2θ+1.283=tan⁡62∘ \sin 2 \theta + 1.283 = \tan 62^{\circ}
Subtract 1.283 from both sides:
sin⁡2θ=tan⁡62∘−1.283 \sin 2 \theta = \tan 62^{\circ} - 1.283

STEP 4

Calculate tan⁡62∘\tan 62^{\circ} using a calculator.
tan⁡62∘≈1.8807 \tan 62^{\circ} \approx 1.8807
Substitute this value back into the equation:
sin⁡2θ=1.8807−1.283 \sin 2 \theta = 1.8807 - 1.283
sin⁡2θ≈0.5977 \sin 2 \theta \approx 0.5977

STEP 5

Solve for 2θ2\theta using the inverse sine function:
2θ=sin⁡−1(0.5977) 2\theta = \sin^{-1}(0.5977)
Calculate the principal value:
2θ≈36.77∘ 2\theta \approx 36.77^{\circ}
Divide by 2 to find θ\theta:
θ≈18.39∘ \theta \approx 18.39^{\circ}

STEP 6

Now, solve equation 2.2.2 by isolating cot⁡(θ−25∘)\cot (\theta - 25^{\circ}).
2cot⁡(θ−25∘)=5 2 \cot (\theta - 25^{\circ}) = 5
Divide both sides by 2:
cot⁡(θ−25∘)=52 \cot (\theta - 25^{\circ}) = \frac{5}{2}

STEP 7

Use the definition of cotangent to find θ−25∘\theta - 25^{\circ}.
θ−25∘=cot⁡−1(52) \theta - 25^{\circ} = \cot^{-1}\left(\frac{5}{2}\right)
Calculate the principal value:
θ−25∘≈21.8∘ \theta - 25^{\circ} \approx 21.8^{\circ}
Add 25 to both sides to solve for θ\theta:
θ≈21.8∘+25∘ \theta \approx 21.8^{\circ} + 25^{\circ}
θ≈46.8∘ \theta \approx 46.8^{\circ}
The solutions for θ\theta are approximately:
For 2.2.1: θ≈18.39∘\theta \approx 18.39^{\circ}
For 2.2.2: θ≈46.8∘\theta \approx 46.8^{\circ}

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