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Math  /  Trigonometry

QuestionSOH⇒sin⁡x=0hCAH⇒cos⁡x=ahTOA⇒tan⁡x=0/aS O H \Rightarrow \sin x=\frac{0}{h} \quad C A H \Rightarrow \cos x=\frac{a}{h} \quad T O A \Rightarrow \tan x=0 / a (1)) Solve for the missing angle θ\theta. a) b)

Studdy Solution

STEP 1

1. We are dealing with right triangles.
2. We need to find the missing angle θ\theta for each triangle using trigonometric ratios.
3. For triangle (a), use the sine ratio.
4. For triangle (b), use the tangent ratio.

STEP 2

1. Solve for θ\theta in triangle (a) using the sine function.
2. Solve for θ\theta in triangle (b) using the tangent function.

STEP 3

For triangle (a), use the sine function:
sin⁡θ=oppositehypotenuse \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}
Substitute the given values:
sin⁡θ=415 \sin \theta = \frac{4}{15}

STEP 4

Solve for θ\theta using the inverse sine function:
θ=sin⁡−1(415) \theta = \sin^{-1} \left(\frac{4}{15}\right)
Calculate θ\theta:
θ≈15.47∘ \theta \approx 15.47^\circ

STEP 5

For triangle (b), use the tangent function:
tan⁡θ=oppositeadjacent \tan \theta = \frac{\text{opposite}}{\text{adjacent}}
Substitute the given values:
tan⁡θ=1721 \tan \theta = \frac{17}{21}

STEP 6

Solve for θ\theta using the inverse tangent function:
θ=tan⁡−1(1721) \theta = \tan^{-1} \left(\frac{17}{21}\right)
Calculate θ\theta:
θ≈39.29∘ \theta \approx 39.29^\circ
The missing angles are:
For triangle (a): θ≈15.47∘\theta \approx 15.47^\circ
For triangle (b): θ≈39.29∘\theta \approx 39.29^\circ

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