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Math  /  Trigonometry

Question*1) Solve for the two angle susing SOH, CAH, TOH: a) 25θ21σ1=0\frac{25 \theta_{2}}{1 \sigma_{1}}=0 b) c)

Studdy Solution

STEP 1

1. We are using trigonometric ratios: sine (SOH), cosine (CAH), and tangent (TOH).
2. The triangles are right triangles, allowing the use of these trigonometric ratios.
3. We need to solve for angles θ1\theta_1 and θ2\theta_2 in each triangle.

STEP 2

1. Solve for angles in triangle a) using tangent.
2. Solve for angles in triangle b) using sine and cosine.
3. Solve for angles in triangle c) using tangent.

STEP 3

For triangle a), use the tangent ratio:
tan⁡(θ2)=oppositeadjacent=2025 \tan(\theta_2) = \frac{\text{opposite}}{\text{adjacent}} = \frac{20}{25}
Calculate θ2\theta_2:
θ2=tan⁡−1(2025) \theta_2 = \tan^{-1}\left(\frac{20}{25}\right)
STEP_1.1: Calculate θ1\theta_1 using the complementary angle property of right triangles:
θ1=90∘−θ2 \theta_1 = 90^\circ - \theta_2

STEP 4

For triangle b), use the sine ratio for θ2\theta_2:
sin⁡(θ2)=oppositehypotenuse=1525 \sin(\theta_2) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{15}{25}
Calculate θ2\theta_2:
θ2=sin⁡−1(1525) \theta_2 = \sin^{-1}\left(\frac{15}{25}\right)
STEP_2.1: Use the cosine ratio for θ1\theta_1:
cos⁡(θ1)=adjacenthypotenuse=1525 \cos(\theta_1) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{15}{25}
Calculate θ1\theta_1:
θ1=cos⁡−1(1525) \theta_1 = \cos^{-1}\left(\frac{15}{25}\right)

STEP 5

For triangle c), use the tangent ratio for θ1\theta_1:
tan⁡(θ1)=oppositeadjacent=57 \tan(\theta_1) = \frac{\text{opposite}}{\text{adjacent}} = \frac{5}{7}
Calculate θ1\theta_1:
θ1=tan⁡−1(57) \theta_1 = \tan^{-1}\left(\frac{5}{7}\right)
STEP_3.1: Calculate θ2\theta_2 using the complementary angle property of right triangles:
θ2=90∘−θ1 \theta_2 = 90^\circ - \theta_1
The angles for each triangle are calculated as follows:
Triangle a): θ2=tan⁡−1(2025) \theta_2 = \tan^{-1}\left(\frac{20}{25}\right) θ1=90∘−θ2 \theta_1 = 90^\circ - \theta_2
Triangle b): θ2=sin⁡−1(1525) \theta_2 = \sin^{-1}\left(\frac{15}{25}\right) θ1=cos⁡−1(1525) \theta_1 = \cos^{-1}\left(\frac{15}{25}\right)
Triangle c): θ1=tan⁡−1(57) \theta_1 = \tan^{-1}\left(\frac{5}{7}\right) θ2=90∘−θ1 \theta_2 = 90^\circ - \theta_1

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