Solving the Right-Angle Triangle Problem

The problem asks us to find the value of \(\sin A + \sin B + \sin C\) for a right-angle triangle \(\triangle\)ABC, where the right angle is at B, and we are given that \(\tan A = \frac{3}{4}\).

Understanding the Given Information

We have a right-angled triangle at vertex B. This means the angle at B, \(\angle B\), is \(90^{\circ}\). We are also given the value of \(\tan A\), which is the ratio of the side opposite to angle A to the side adjacent to angle A.

• \(\angle B = 90^{\circ}\)
• \(\tan A = \frac{\text{Opposite Side to A}}{\text{Adjacent Side to A}} = \frac{3}{4}\)

In \(\triangle\)ABC, with the right angle at B:

• The side opposite to A is BC.
• The side adjacent to A is AB.
• The hypotenuse is AC.

So, \(\frac{BC}{AB} = \frac{3}{4}\). We can assume BC = \(3k\) and AB = \(4k\) for some positive constant \(k\).

Finding the Hypotenuse using the Pythagorean Theorem

For a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides (\(a^2 + b^2 = c^2\)). In \(\triangle\)ABC:

\(AC^2 = AB^2 + BC^2\)

Substitute the values \(AB = 4k\) and \(BC = 3k\):

\(AC^2 = (4k)^2 + (3k)^2\)

\(AC^2 = 16k^2 + 9k^2\)

\(AC^2 = 25k^2\)

Taking the square root of both sides:

\(AC = \sqrt{25k^2} = 5k\)

So, the lengths of the sides of the triangle can be considered as AB = \(4k\), BC = \(3k\), and AC = \(5k\).

Calculating the Sine Values

Now we can find the sine of each angle:

For angle A:

\(\sin A = \frac{\text{Opposite Side to A}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{3k}{5k} = \frac{3}{5}\)

For angle B:

Since \(\angle B = 90^{\circ}\), we know the value of \(\sin B\):

\(\sin B = \sin 90^{\circ} = 1\)

For angle C:

In a right-angled triangle at B, angles A and C are complementary, meaning \(A + C = 90^{\circ}\). Therefore, \(C = 90^{\circ} - A\).

Using trigonometric identities, \(\sin C = \sin (90^{\circ} - A) = \cos A\).

We can find \(\cos A\) from the side lengths:

\(\cos A = \frac{\text{Adjacent Side to A}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{4k}{5k} = \frac{4}{5}\)

So, \(\sin C = \frac{4}{5}\).

Alternatively, we can find \(\sin C\) directly using the sides relative to angle C:

\(\sin C = \frac{\text{Opposite Side to C}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{4k}{5k} = \frac{4}{5}\)

Both methods give the same result.

Calculating the Sum sin A + sin B + sin C

Now we add the values we found:

• \(\sin A = \frac{3}{5}\)
• \(\sin B = 1\)
• \(\sin C = \frac{4}{5}\)

Sum = \(\sin A + \sin B + \sin C = \frac{3}{5} + 1 + \frac{4}{5}\)

Combine the fractions first:

Sum = \(\left(\frac{3}{5} + \frac{4}{5}\right) + 1 = \frac{3+4}{5} + 1 = \frac{7}{5} + 1\)

To add \(\frac{7}{5}\) and 1, convert 1 to a fraction with a denominator of 5:

\(1 = \frac{5}{5}\)

Sum = \(\frac{7}{5} + \frac{5}{5} = \frac{7+5}{5} = \frac{12}{5}\)

Converting the Result to a Mixed Number

The sum is \(\frac{12}{5}\). To convert this improper fraction to a mixed number, divide 12 by 5:

\(12 \div 5\)

12 divided by 5 is 2 with a remainder of 2.

So, \(\frac{12}{5} = 2\) and \(\frac{2}{5}\), which is written as \(2\frac{2}{5}\).

Therefore, \(\sin A + \sin B + \sin C = 2\frac{2}{5}\).

Revision Table: Right Triangle Trigonometry

Additional Information: Trigonometric Ratios

Trigonometric ratios relate the angles of a right-angled triangle to the ratios of the lengths of its sides. The primary ratios are sine, cosine, and tangent.

• Sine (sin): Ratio of the length of the side opposite the angle to the length of the hypotenuse. \(\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}}\)
• Cosine (cos): Ratio of the length of the side adjacent to the angle to the length of the hypotenuse. \(\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}}\)
• Tangent (tan): Ratio of the length of the side opposite the angle to the length of the side adjacent to the angle. \(\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}}\)

These ratios are constant for a given angle, regardless of the size of the right triangle.

Understanding these fundamental trigonometric ratios and the Pythagorean theorem is crucial for solving problems involving right-angled triangles.