Math Problem Statement

Solve the following triangle problems: 1. In triangle ABC: b = 120, m(∠A) = 32°, and m(∠C) = 48°. Find sides AB and BC. 2. In triangle ABC: b = √6 cm, m(∠A) = 75°, and m(∠C) = 45°. Find angles A, B, and C, and then find the perimeter of the triangle.

Solution

The image contains two geometry problems involving triangles. Let's analyze and solve both.

Problem 1:

In triangle ΔABC\Delta ABC:

  • b=120b = 120
  • m(A)=32m(\angle A) = 32^\circ
  • m(C)=48m(\angle C) = 48^\circ

You are asked to find sides ABAB and BCBC.

Solution to Problem 1:

  1. Using the angle sum property of triangles: m(B)=180(m(A)+m(C))=180(32+48)=100m(\angle B) = 180^\circ - (m(\angle A) + m(\angle C)) = 180^\circ - (32^\circ + 48^\circ) = 100^\circ
  2. Use the Law of Sines to find the sides: asinA=bsinB=csinC\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} Here, b=120b = 120, so we calculate aa and cc as follows:
    • For side aa (opposite angle A\angle A): asin32=120sin100\frac{a}{\sin 32^\circ} = \frac{120}{\sin 100^\circ} Solve for aa: a=120×sin32sin100a = 120 \times \frac{\sin 32^\circ}{\sin 100^\circ}

    • For side cc (opposite angle C\angle C): csin48=120sin100\frac{c}{\sin 48^\circ} = \frac{120}{\sin 100^\circ} Solve for cc: c=120×sin48sin100c = 120 \times \frac{\sin 48^\circ}{\sin 100^\circ}

You can compute these values to get the exact lengths of sides ABAB and BCBC.


Problem 2:

In triangle ΔABC\Delta ABC:

  • b=6b = \sqrt{6} cm
  • m(A)=75m(\angle A) = 75^\circ
  • m(C)=45m(\angle C) = 45^\circ

You are asked to find the angles AA, BB, and CC, as well as the perimeter.

Solution to Problem 2:

  1. Use the angle sum property to find m(B)m(\angle B): m(B)=180(m(A)+m(C))=180(75+45)=60m(\angle B) = 180^\circ - (m(\angle A) + m(\angle C)) = 180^\circ - (75^\circ + 45^\circ) = 60^\circ

  2. Again, apply the Law of Sines to find sides: asinA=bsinB=csinC\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} Since b=6b = \sqrt{6}, calculate aa and cc.

    • For side aa (opposite angle A\angle A): asin75=6sin60\frac{a}{\sin 75^\circ} = \frac{\sqrt{6}}{\sin 60^\circ} Solve for aa: a=6×sin75sin60a = \sqrt{6} \times \frac{\sin 75^\circ}{\sin 60^\circ}

    • For side cc (opposite angle C\angle C): csin45=6sin60\frac{c}{\sin 45^\circ} = \frac{\sqrt{6}}{\sin 60^\circ} Solve for cc: c=6×sin45sin60c = \sqrt{6} \times \frac{\sin 45^\circ}{\sin 60^\circ}

Finally, calculate the perimeter by adding a+b+ca + b + c.


Would you like me to calculate the exact values for sides and perimeter?

Here are some follow-up questions:

  1. What are the other methods to solve triangles apart from the Law of Sines?
  2. How does the Law of Cosines differ from the Law of Sines?
  3. Can you calculate the angles of a triangle if two sides are known?
  4. How can trigonometric identities help in solving non-right triangles?
  5. What are the real-world applications of solving triangles using trigonometry?

Tip: Always check if the given triangle is valid using triangle inequality rules before solving!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Trigonometry
Geometry
Law of Sines
Triangle Properties

Formulas

Law of Sines: a/sin(A) = b/sin(B) = c/sin(C)
Angle Sum Property of a Triangle: A + B + C = 180°

Theorems

Law of Sines
Angle Sum Property of Triangles

Suitable Grade Level

Grades 9-12