Math Problem Statement

uppose a triangle has vertices A, B and C and has the following measurements: m∠C=91∘¯¯¯¯¯¯¯¯AC=40.1 cm¯¯¯¯¯¯¯¯BC=32.5 cm

Start by drawing a diagram of this triangle and labeling the known values.

What is the length of AB

in cm?

¯¯¯¯¯¯¯¯AB=

cm

What is the degree measure of ∠A

?

m∠A=

°

What is the degree measure of ∠B

?

m∠B=

°

Solution

Let's start by understanding the given problem:

  1. Vertices: The triangle has vertices AA, BB, and CC.
  2. Angle at CC: mC=91m\angle C = 91^\circ (an obtuse angle).
  3. Side lengths:
    • AC=40.1AC = 40.1 cm
    • BC=32.5BC = 32.5 cm
  4. Unknowns:
    • The length of side ABAB (which we will call cc).
    • The measures of angles AA and BB.

Step 1: Drawing the Diagram

We have a triangle with vertices AA, BB, and CC where:

  • AC=40.1AC = 40.1 cm
  • BC=32.5BC = 32.5 cm
  • mC=91m\angle C = 91^\circ
   A
   /\
  /  \
c/____\b
B       C

Step 2: Applying the Law of Cosines

To find the length of side ABAB (denoted as cc), we use the Law of Cosines:

c2=a2+b22abcos(C)c^2 = a^2 + b^2 - 2ab \cdot \cos(C)

Substituting the values: c2=(40.1)2+(32.5)22×40.1×32.5×cos(91)c^2 = (40.1)^2 + (32.5)^2 - 2 \times 40.1 \times 32.5 \times \cos(91^\circ)

Note that cos(91)\cos(91^\circ) is slightly negative because 91° is slightly greater than 90°.

Step 3: Calculating cc

Let's calculate cc:

c2=1608.01+1056.252×40.1×32.5×(0.017452)c^2 = 1608.01 + 1056.25 - 2 \times 40.1 \times 32.5 \times (-0.017452) c2=1608.01+1056.25+45.45c^2 = 1608.01 + 1056.25 + 45.45 c2=2709.71c^2 = 2709.71 c2709.7152.06 cmc \approx \sqrt{2709.71} \approx 52.06 \text{ cm}

So, AB52.06 cm\overline{AB} \approx 52.06 \text{ cm}.

Step 4: Finding Angles AA and BB

Now, let's use the Law of Sines to find the remaining angles.

The Law of Sines states: asinA=bsinB=csinC\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

  1. Find mAm\angle A: sinA=asinCc\sin A = \frac{a \cdot \sin C}{c} sinA=40.1×sin(91)52.0640.1×152.060.7704\sin A = \frac{40.1 \times \sin(91^\circ)}{52.06} \approx \frac{40.1 \times 1}{52.06} \approx 0.7704 mAarcsin(0.7704)50.3m\angle A \approx \arcsin(0.7704) \approx 50.3^\circ

  2. Find mBm\angle B: Since the sum of angles in a triangle is 180180^\circ: mB=180mCmAm\angle B = 180^\circ - m\angle C - m\angle A mB=1809150.338.7m\angle B = 180^\circ - 91^\circ - 50.3^\circ \approx 38.7^\circ

Final Results:

  • AB52.06\overline{AB} \approx 52.06 cm
  • mA50.3m\angle A \approx 50.3^\circ
  • mB38.7m\angle B \approx 38.7^\circ

Would you like a more detailed explanation on any of the steps? Here are some related questions to explore:

  1. How is the Law of Cosines derived?
  2. What is the significance of the Law of Sines in solving triangles?
  3. How can you determine if a triangle is acute, obtuse, or right using side lengths?
  4. Why does the sine of angles greater than 9090^\circ result in a negative value?
  5. How do you calculate the area of a triangle using the given sides and angle?
  6. Can the Law of Cosines be used to solve any type of triangle?
  7. What would happen if one of the angles was given as exactly 9090^\circ?
  8. How does changing one side of the triangle affect the angles and the other sides?

Tip: When dealing with angles slightly above 9090^\circ, always check the sign of trigonometric functions, as it affects calculations significantly.

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Math Problem Analysis

Mathematical Concepts

Geometry
Trigonometry
Law of Cosines
Law of Sines

Formulas

Law of Cosines
Law of Sines

Theorems

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Suitable Grade Level

High School