In triangle ABC, angle A=45°, and C=70°, the side opposite angle C is 40 m long.
a) find the length of the opposite to angle A
b) find the angle B
c) find the length opposite to angle B
d) find the area if triangle ABC using heron's formula
e) check the area using (A=1/2 a b sinθ)

In triangle \( ABC \), we are given:

- \( \angle A = 45^\circ \)
- \( \angle C = 70^\circ \)
- The side opposite to \( \angle C \) (denoted as \( c \)) is 40 m.

### a) Find the length of the side opposite to angle \( A \) (denoted as \( a \)).

To find this side, we can use the **Law of Sines**:
\[
\frac{a}{\sin A} = \frac{c}{\sin C}
\]
Given:
- \( \sin A = \sin 45^\circ \)
- \( \sin C = \sin 70^\circ \)
- \( c = 40 \) m.

Using this equation, we can solve for \( a \).

### b) Find the angle \( B \).

Using the **angle sum property** of a triangle:
\[
\angle A + \angle B + \angle C = 180^\circ
\]
We can calculate \( \angle B \).

### c) Find the length of the side opposite to angle \( B \) (denoted as \( b \)).

Using the **Law of Sines** again:
\[
\frac{b}{\sin B} = \frac{c}{\sin C}
\]
We can solve for \( b \) using the calculated \( \sin B \).

### d) Find the area of triangle ABC using **Heron’s formula**.

First, calculate the semi-perimeter \( s \):
\[
s = \frac{a + b + c}{2}
\]
Then, the area \( A \) can be found using:
\[
A = \sqrt{s(s - a)(s - b)(s - c)}
\]

### e) Check the area using \( A = \frac{1}{2} a b \sin \theta \), where \( \theta \) is the included angle \( \angle C \).

To cross-check the area, we use:
\[
A = \frac{1}{2} a b \sin C
\]

Let's calculate the required values step by step.Here are the results for each part of the problem:

### a) Length of the side opposite to angle \( A \):
\[
a \approx 30.10 \, \text{m}
\]

### b) Angle \( B \):
\[
B = 65^\circ
\]

### c) Length of the side opposite to angle \( B \):
\[
b \approx 38.58 \, \text{m}
\]

### d) Area of triangle ABC using **Heron's formula**:
\[
\text{Area} \approx 545.59 \, \text{m}^2
\]

### e) Area using \( A = \frac{1}{2} a b \sin C \):
\[
\text{Area} \approx 545.59 \, \text{m}^2
\]

The areas from both methods are consistent.

Do you have any further questions or need more details?

Here are 5 related questions to explore:
1. How does the Law of Cosines compare to the Law of Sines in triangle problems?
2. Can we use Heron’s formula for any type of triangle?
3. What is the significance of the semi-perimeter in Heron’s formula?
4. How do the sine, cosine, and tangent functions relate to angles in triangles?
5. How would the calculations change if we knew only two sides and an included angle?

**Tip:** Always double-check angle conversions (degrees to radians) when solving trigonometric equations in calculators.

In triangle ABC, angle A=45°, and C=70°, the side opposite angle C is 40 m long. a) find the length of the opposite to angle A b) find the angle B c) find the length opposite to angle B d) find the area of triangle ABC using heron's formula e) check the area using (A=1/2 a b sinθ)

Learn how to solve triangle ABC with given angles and sides. Find the side lengths using the Law of Sines, calculate the unknown angle, and compute the area using Heron's Formula. Verify the result with the formula A = 1/2 a b sin θ. Ideal for high school students (Grades 9-12).

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation