ΔABC is a right triangle. If ∠B = 90° and tan A = \(\frac{1}{\sqrt2}\), then the value of sin A cos C + cos A sin C is :
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We are given a right triangle \(\Delta ABC\) where \(\angle B = 90^\circ\). We are also given that \(\tan A = \frac{1}{\sqrt{2}}\). We need to find the value of the expression \(\sin A \cos C + \cos A \sin C\).
Let's look at the expression we need to evaluate: \(\sin A \cos C + \cos A \sin C\).
This expression is the standard expansion for the sine of the sum of two angles, i.e., \(\sin(A+C)\).
So, \(\sin A \cos C + \cos A \sin C = \sin(A+C)\).
Now, let's consider the angles in triangle \(\Delta ABC\). The sum of angles in any triangle is \(180^\circ\). So, for \(\Delta ABC\), we have:
\(A + B + C = 180^\circ\)
We are given that \(\angle B = 90^\circ\). Substituting this into the equation:
\(A + 90^\circ + C = 180^\circ\)
Subtracting \(90^\circ\) from both sides, we get:
\(A + C = 180^\circ - 90^\circ\)
\(A + C = 90^\circ\)
This means that angles A and C are complementary angles in this right triangle.
Now, we can substitute this value of \((A+C)\) back into the expression \(\sin(A+C)\):
Value = \(\sin(90^\circ)\)
The sine of \(90^\circ\) is a standard trigonometric value:
\(\sin(90^\circ) = 1\)
Therefore, the value of \(\sin A \cos C + \cos A \sin C\) is 1.
Notice that in this method, the information about \(\tan A = \frac{1}{\sqrt{2}}\) was not strictly needed to find the value of the expression, as it simplified based on the properties of a right triangle and trigonometric identities.
We are given \(\tan A = \frac{1}{\sqrt{2}}\) in the right triangle \(\Delta ABC\) where \(\angle B = 90^\circ\). We know that \(\tan A = \frac{\text{Opposite side to A}}{\text{Adjacent side to A}}\).
Let the side opposite to angle A be BC, and the side adjacent to angle A (excluding the hypotenuse) be AB. Let the hypotenuse be AC.
So, \(\tan A = \frac{BC}{AB} = \frac{1}{\sqrt{2}}\).
We can let \(BC = 1k\) and \(AB = \sqrt{2}k\) for some positive constant \(k\). For simplicity, we can just use \(BC = 1\) and \(AB = \sqrt{2}\).
Now, we can find the hypotenuse AC using the Pythagorean theorem:
\(AC^2 = AB^2 + BC^2\)
\(AC^2 = (\sqrt{2})^2 + (1)^2\)
\(AC^2 = 2 + 1\)
\(AC^2 = 3\)
\(AC = \sqrt{3}\) (Since length must be positive)
Now we can find the sine and cosine of angle A:
Next, let's find the sine and cosine of angle C. In a right triangle, angles A and C are complementary, meaning \(A + C = 90^\circ\). This implies:
Alternatively, from angle C's perspective:
These values match what we got using the complementary angle relationships.
Now, substitute these values into the expression \(\sin A \cos C + \cos A \sin C\):
\(\sin A \cos C + \cos A \sin C = \left(\frac{1}{\sqrt{3}}\right) \left(\frac{1}{\sqrt{3}}\right) + \left(\frac{\sqrt{2}}{\sqrt{3}}\right) \left(\frac{\sqrt{2}}{\sqrt{3}}\right)\)
\(= \frac{1}{\sqrt{3} \times \sqrt{3}} + \frac{\sqrt{2} \times \sqrt{2}}{\sqrt{3} \times \sqrt{3}}\)
\(= \frac{1}{3} + \frac{2}{3}\)
\(= \frac{1+2}{3}\)
\(= \frac{3}{3}\)
\(= 1\)
Both methods give the same result, confirming the answer.
The value of \(\sin A \cos C + \cos A \sin C\) in the given right triangle \(\Delta ABC\) with \(\angle B = 90^\circ\) is 1. This is because the expression simplifies to \(\sin(A+C)\), and for a right triangle with angle B = \(90^\circ\), the sum of the other two angles A and C must be \(90^\circ\).
| Given Information | Value |
|---|---|
| Triangle Type | Right Triangle \(\Delta ABC\) |
| Angle B | \(90^\circ\) |
| \(\tan A\) | \(\frac{1}{\sqrt{2}}\) |
| Expression to find | \(\sin A \cos C + \cos A \sin C\) |
| Concept | Description | Formula/Identity |
|---|---|---|
| Sum of Angles in a Triangle | The sum of interior angles in any triangle is \(180^\circ\). | \(A + B + C = 180^\circ\) |
| Angles in a Right Triangle | If one angle is \(90^\circ\), the other two angles are complementary. | If \(\angle B = 90^\circ\), then \(A + C = 90^\circ\). |
| Sine Addition Formula | The sine of the sum of two angles. | \(\sin(x+y) = \sin x \cos y + \cos x \sin y\) |
| Complementary Angle Identities | Relationships between trigonometric functions of complementary angles. | \(\sin(90^\circ - x) = \cos x\) \(\cos(90^\circ - x) = \sin x\) |
Trigonometry is the study of the relationships between the angles and sides of triangles. Right triangles are fundamental in trigonometry because the trigonometric ratios (sine, cosine, tangent) are defined based on the ratios of the sides relative to the acute angles in a right triangle.
Understanding these basic concepts is key to solving problems involving right triangles and trigonometry, like the one discussed here. The problem beautifully illustrates how combining the angle properties of a right triangle with trigonometric identities can lead to a quick solution, even when specific side ratios are given.
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