Math Problem Statement

Identify the trigonometric ratio needed to find the unknown side length for each triangle, then solve for the unknown side length.

Solution

Let's go through each triangle one by one, identify the relevant trigonometric ratio, and solve for the unknown side.

a) Triangle ECD\triangle ECD

  • Given: Angle E=25\angle E = 25^\circ, adjacent side ED=3.4mED = 3.4 \, \text{m}, and we are solving for the opposite side e=CDe = CD.
  • Trigonometric Ratio: Since we are dealing with the opposite and adjacent sides relative to the given angle, we use tangent: tan(25)=oppositeadjacent=e3.4\tan(25^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{e}{3.4}
  • Solving for ee: e=3.4×tan(25)e = 3.4 \times \tan(25^\circ) Using a calculator: e3.4×0.4663=1.5854me \approx 3.4 \times 0.4663 = 1.5854 \, \text{m} So, e1.59me \approx 1.59 \, \text{m}.

b) Triangle PMQ\triangle PMQ

  • Given: Angle P=13\angle P = 13^\circ, hypotenuse PQ=103ftPQ = 103 \, \text{ft}, and we are solving for the opposite side q=MQq = MQ.
  • Trigonometric Ratio: Since we have the opposite and hypotenuse, we use sine: sin(13)=oppositehypotenuse=q103\sin(13^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{q}{103}
  • Solving for qq: q=103×sin(13)q = 103 \times \sin(13^\circ) Using a calculator: q103×0.2249=23.1667ftq \approx 103 \times 0.2249 = 23.1667 \, \text{ft} So, q23.17ftq \approx 23.17 \, \text{ft}.

c) Triangle HGF\triangle HGF

  • Given: Angle H=43\angle H = 43^\circ, adjacent side HG=43inHG = 43 \, \text{in}, and we are solving for the hypotenuse g=HFg = HF.
  • Trigonometric Ratio: Since we are working with the adjacent and hypotenuse sides, we use cosine: cos(43)=adjacenthypotenuse=43g\cos(43^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{43}{g}
  • Solving for gg: g=43cos(43)g = \frac{43}{\cos(43^\circ)} Using a calculator: g430.7314=58.78ing \approx \frac{43}{0.7314} = 58.78 \, \text{in} So, g58.78ing \approx 58.78 \, \text{in}.

Let me know if you need further explanation or details on any step.

Follow-up questions:

  1. How do you decide which trigonometric function to use for each triangle?
  2. Can we solve for all sides of a right triangle given just one side and an angle?
  3. What is the difference between using sine, cosine, and tangent for right triangles?
  4. How would you solve for an angle if given two sides of a triangle?
  5. How can we apply these trigonometric ratios to solve real-world problems?

Tip: Always remember that for right triangles, the sine function relates the opposite side and hypotenuse, the cosine function relates the adjacent side and hypotenuse, and the tangent function relates the opposite and adjacent sides.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Right Triangles
Trigonometric Ratios

Formulas

Tangent: tan(θ) = opposite / adjacent
Sine: sin(θ) = opposite / hypotenuse
Cosine: cos(θ) = adjacent / hypotenuse

Theorems

Trigonometric functions (sine, cosine, tangent) for right triangles

Suitable Grade Level

Grades 9-12