Math Problem Statement

Solve the following math word problems using the method called Solving a Right Triangle. And here are the word problems provided in the photo.

Solve the following math word problems using the method called Solving a Right Triangle.

Solution

To solve the given problems, we'll apply trigonometric functions (sine, cosine, and tangent) as well as the Pythagorean theorem, which are the standard methods for solving right triangles.

Problem 6:

You have a right triangle where:

  • The side opposite to the 66° angle is 5.5m5.5 \, m (this is side aa).
  • The hypotenuse is hh.
  • The adjacent side is qq (this is side bb).

a) Finding the hypotenuse hh:

We use the sine function since we know the opposite side and the angle: sin(66)=oppositehypotenuse=ah=5.5h\sin(66^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{a}{h} = \frac{5.5}{h} Solving for hh: h=5.5sin(66)h = \frac{5.5}{\sin(66^\circ)} h5.50.91356.0mh \approx \frac{5.5}{0.9135} \approx 6.0 \, m

b) Finding side bb (adjacent side):

We use the cosine function for this: cos(66)=adjacenthypotenuse=bh\cos(66^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{b}{h} b=hcos(66)=6.00.40672.4mb = h \cdot \cos(66^\circ) = 6.0 \cdot 0.4067 \approx 2.4 \, m


Problem 7:

You have a right triangle where:

  • The hypotenuse hh is 35.5cm35.5 \, cm.
  • The angle is 7070^\circ.
  • Side bb is opposite to the 70° angle.
  • Side cc is adjacent to the 70° angle.

a) Finding side bb (opposite side):

We use the sine function here: sin(70)=bh=b35.5\sin(70^\circ) = \frac{b}{h} = \frac{b}{35.5} b=35.5sin(70)=35.50.939733.4cmb = 35.5 \cdot \sin(70^\circ) = 35.5 \cdot 0.9397 \approx 33.4 \, cm

b) Finding side cc (adjacent side):

We use the cosine function: cos(70)=ch=c35.5\cos(70^\circ) = \frac{c}{h} = \frac{c}{35.5} c=35.5cos(70)=35.50.342012.1cmc = 35.5 \cdot \cos(70^\circ) = 35.5 \cdot 0.3420 \approx 12.1 \, cm


Problem 8:

You have a right triangle where:

  • The adjacent side cc is 100m100 \, m.
  • The angle is 2727^\circ.
  • Side aa is opposite to the 27° angle.
  • The hypotenuse is hh.

a) Finding side aa (opposite side):

We use the tangent function here: tan(27)=ac=a100\tan(27^\circ) = \frac{a}{c} = \frac{a}{100} a=100tan(27)=1000.509550.95ma = 100 \cdot \tan(27^\circ) = 100 \cdot 0.5095 \approx 50.95 \, m

b) Finding the hypotenuse hh:

We use the cosine function: cos(27)=ch\cos(27^\circ) = \frac{c}{h} h=ccos(27)=1000.8910112.3mh = \frac{c}{\cos(27^\circ)} = \frac{100}{0.8910} \approx 112.3 \, m


These are the solutions to the problems. Would you like more details or have any further questions?

5 Related Questions:

  1. What is the general formula for solving right triangles?
  2. How do you know when to use sine, cosine, or tangent?
  3. Can you explain the Pythagorean theorem and how it relates to right triangles?
  4. How do angle and side length changes affect the trigonometric ratios in a triangle?
  5. Can you use trigonometry to solve non-right triangles?

Tip:

When solving for side lengths in right triangles, always make sure to choose the correct trigonometric function based on the known and unknown sides (opposite, adjacent, or hypotenuse).

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Right Triangle
Sine
Cosine
Tangent
Pythagorean Theorem

Formulas

sin(θ) = opposite / hypotenuse
cos(θ) = adjacent / hypotenuse
tan(θ) = opposite / adjacent
h = sqrt(a^2 + b^2) (Pythagorean theorem)

Theorems

Pythagorean Theorem

Suitable Grade Level

Grades 9-11