Math Problem Statement

Solve the right triangles. Find all missing sides and angles for triangles in Problems 7 and 8. Also, identify the correct equations to solve for missing sides or angles in Problem 9.

Solution

Let's break down each of the triangles and solve for the missing sides and angles step by step.

Problem 7:

You have a right triangle with:

  • Hypotenuse RS=18\overline{RS} = 18
  • Side RQ=6\overline{RQ} = 6
  • You need to find QS\overline{QS}, mQm\angle Q, and mSm\angle S.

Since this is a right triangle, we can apply the Pythagorean theorem and trigonometric ratios to find the missing values.

Step 1: Find QS\overline{QS}

Use the Pythagorean theorem: QS2=RS2RQ2QS^2 = RS^2 - RQ^2 QS2=18262=32436=288QS^2 = 18^2 - 6^2 = 324 - 36 = 288 QS=28816.97QS = \sqrt{288} \approx 16.97

Step 2: Find mQm\angle Q

Using trigonometry (since we know the opposite side and adjacent side relative to Q\angle Q): tan(Q)=RQQS=616.970.354\tan(Q) = \frac{RQ}{QS} = \frac{6}{16.97} \approx 0.354 mQ=tan1(0.354)19.5m\angle Q = \tan^{-1}(0.354) \approx 19.5^\circ

Step 3: Find mSm\angle S

Since the sum of the angles in a triangle is 180°, and Q\angle Q and S\angle S are complementary (because it's a right triangle): mS=90mQ=9019.5=70.5m\angle S = 90^\circ - m\angle Q = 90^\circ - 19.5^\circ = 70.5^\circ

Problem 8:

This is another right triangle, where we are given:

  • Hypotenuse XY=26.3\overline{XY} = 26.3
  • Angle mX=46m\angle X = 46^\circ
  • We need to find the side WX\overline{WX}, side XY\overline{XY}, and mXm\angle X.

Step 1: Find WXWX

We can use the sine function: sin(46)=WX26.3\sin(46^\circ) = \frac{WX}{26.3} WX=26.3×sin(46)26.3×0.719318.9WX = 26.3 \times \sin(46^\circ) \approx 26.3 \times 0.7193 \approx 18.9

Step 2: Find mYm\angle Y

Since it's a right triangle: mY=9046=44m\angle Y = 90^\circ - 46^\circ = 44^\circ

Problem 9:

For triangle ABC\triangle ABC, we are tasked with finding the correct equations to solve for the missing sides or angles.

From the options provided:

  • Option A: BC=16×tan(29)BC = 16 \times \tan(29^\circ) is correct.
    • This would help solve for BCBC, the side opposite to B\angle B.

Other options are incorrect based on the standard trigonometric relationships for solving right triangles.


Would you like further details or explanations on any part of the solution?

Here are some related questions you might find interesting:

  1. How does the Pythagorean theorem relate to trigonometric functions?
  2. When do we use sine, cosine, or tangent in solving right triangles?
  3. How do you solve for an unknown angle if you are given two sides of a right triangle?
  4. What happens if you’re given all sides of a right triangle but no angles?
  5. Can you calculate missing side lengths without using trigonometric ratios?

Tip: Always check whether you are working with a right triangle when solving for angles or sides. The relationships between the sides and angles are simpler and more direct in right triangles!

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Pythagorean Theorem
Angles in a Triangle

Formulas

Pythagorean theorem: a^2 + b^2 = c^2
Trigonometric ratios: sin(θ) = opposite/hypotenuse, cos(θ) = adjacent/hypotenuse, tan(θ) = opposite/adjacent
Angle sum property: Sum of angles in a triangle = 180°

Theorems

Pythagorean Theorem
Angle Sum Theorem
Basic Trigonometric Relationships

Suitable Grade Level

Grades 9-10