Answer :
To solve for [tex]\( x \)[/tex] in right triangle [tex]\( ABC \)[/tex], where angle [tex]\( A \)[/tex] is the right angle (90 degrees), we need to utilize the relationship of the sides of the triangle with the sine and cosine functions.
Given:
[tex]\[ \cos B = 2x - 0.15 \][/tex]
[tex]\[ \sin C = x + 0.24 \][/tex]
We know from trigonometry that for any right triangle, the following Pythagorean trigonometric identity holds:
[tex]\[ \cos^2(\theta) + \sin^2(\theta) = 1 \][/tex]
Since angle [tex]\( B \)[/tex] and angle [tex]\( C \)[/tex] are the non-right angles in our right triangle, and they are complementary ([tex]\( B + C = 90^\circ \)[/tex]), then:
[tex]\[ \sin C = \sin(90^\circ - B) = \cos B \][/tex]
We can now set up the equality using the Pythagorean trigonometric identity for angle [tex]\( B \)[/tex]:
[tex]\[ (2x - 0.15)^2 + (x + 0.24)^2 = 1 \][/tex]
This simplifies to a quadratic equation in [tex]\( x \)[/tex]:
[tex]\[ (4x^2 - 0.6x + 0.0225) + (x^2 + 0.48x + 0.0576) = 1 \][/tex]
Combining like terms, we get:
[tex]\[ 5x^2 - 0.12x + 0.0801 = 1 \][/tex]
Move 1 to the left side of the equation:
[tex]\[ 5x^2 - 0.12x + 0.0801 - 1 = 0 \][/tex]
Simplify the constants:
[tex]\[ 5x^2 - 0.12x - 0.9199 = 0 \][/tex]
Now we have a quadratic equation in the standard form [tex]\( ax^2 + bx + c = 0 \)[/tex]. To solve for [tex]\( x \)[/tex], we can use the quadratic formula where [tex]\( a = 5 \)[/tex], [tex]\( b = -0.12 \)[/tex], and [tex]\( c = -0.9199 \)[/tex]:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
[tex]\[ x = \frac{0.12 \pm \sqrt{(-0.12)^2 - 4(5)(-0.9199)}}{2(5)} \][/tex]
[tex]\[ x = \frac{0.12 \pm \sqrt{0.0144 + 18.398}}{10} \][/tex]
[tex]\[ x = \frac{0.12 \pm \sqrt{18.4124}}{10} \][/tex]
[tex]\[ x = \frac{0.12 \pm 4.2912}{10} \][/tex]
We have two possible solutions:
[tex]\[ x = \frac{0.12 + 4.2912}{10} \][/tex]
[tex]\[ x = \frac{4.4112}{10} \][/tex]
[tex]\[ x = 0.44112 \][/tex]
or
[tex]\[ x = \frac{0.12 - 4.2912}{10} \][/tex]
[tex]\[ x = \frac{-4.1712}{10} \][/tex]
[tex]\[ x = -0.41712 \][/tex]
However, since [tex]\( x \)[/tex] is a measure of the sine and cosine of an angle in a right triangle, it must be positive and cannot be greater than 1 because the range of sine and cosine functions for angles in a triangle is [tex]\( [0, 1] \)[/tex]. Therefore, the negative value is not an acceptable solution in this context.
Thus, the only feasible solution is:
[tex]\[ x = 0.44112 \][/tex]
This is the value of [tex]\( x \)[/tex] that satisfies the trigonometric conditions for right triangle [tex]\( ABC \)[/tex].
Given:
[tex]\[ \cos B = 2x - 0.15 \][/tex]
[tex]\[ \sin C = x + 0.24 \][/tex]
We know from trigonometry that for any right triangle, the following Pythagorean trigonometric identity holds:
[tex]\[ \cos^2(\theta) + \sin^2(\theta) = 1 \][/tex]
Since angle [tex]\( B \)[/tex] and angle [tex]\( C \)[/tex] are the non-right angles in our right triangle, and they are complementary ([tex]\( B + C = 90^\circ \)[/tex]), then:
[tex]\[ \sin C = \sin(90^\circ - B) = \cos B \][/tex]
We can now set up the equality using the Pythagorean trigonometric identity for angle [tex]\( B \)[/tex]:
[tex]\[ (2x - 0.15)^2 + (x + 0.24)^2 = 1 \][/tex]
This simplifies to a quadratic equation in [tex]\( x \)[/tex]:
[tex]\[ (4x^2 - 0.6x + 0.0225) + (x^2 + 0.48x + 0.0576) = 1 \][/tex]
Combining like terms, we get:
[tex]\[ 5x^2 - 0.12x + 0.0801 = 1 \][/tex]
Move 1 to the left side of the equation:
[tex]\[ 5x^2 - 0.12x + 0.0801 - 1 = 0 \][/tex]
Simplify the constants:
[tex]\[ 5x^2 - 0.12x - 0.9199 = 0 \][/tex]
Now we have a quadratic equation in the standard form [tex]\( ax^2 + bx + c = 0 \)[/tex]. To solve for [tex]\( x \)[/tex], we can use the quadratic formula where [tex]\( a = 5 \)[/tex], [tex]\( b = -0.12 \)[/tex], and [tex]\( c = -0.9199 \)[/tex]:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
[tex]\[ x = \frac{0.12 \pm \sqrt{(-0.12)^2 - 4(5)(-0.9199)}}{2(5)} \][/tex]
[tex]\[ x = \frac{0.12 \pm \sqrt{0.0144 + 18.398}}{10} \][/tex]
[tex]\[ x = \frac{0.12 \pm \sqrt{18.4124}}{10} \][/tex]
[tex]\[ x = \frac{0.12 \pm 4.2912}{10} \][/tex]
We have two possible solutions:
[tex]\[ x = \frac{0.12 + 4.2912}{10} \][/tex]
[tex]\[ x = \frac{4.4112}{10} \][/tex]
[tex]\[ x = 0.44112 \][/tex]
or
[tex]\[ x = \frac{0.12 - 4.2912}{10} \][/tex]
[tex]\[ x = \frac{-4.1712}{10} \][/tex]
[tex]\[ x = -0.41712 \][/tex]
However, since [tex]\( x \)[/tex] is a measure of the sine and cosine of an angle in a right triangle, it must be positive and cannot be greater than 1 because the range of sine and cosine functions for angles in a triangle is [tex]\( [0, 1] \)[/tex]. Therefore, the negative value is not an acceptable solution in this context.
Thus, the only feasible solution is:
[tex]\[ x = 0.44112 \][/tex]
This is the value of [tex]\( x \)[/tex] that satisfies the trigonometric conditions for right triangle [tex]\( ABC \)[/tex].
