In a right triangle, the hypotenuse is always the longest side. So, in this case, the hypotenuse (denoted as x) is longer than the adjacent side (9) and the opposite side (15).
We can use the trigonometric function cosine (cos) to solve for x. The cosine of an angle in a right triangle is calculated by dividing the length of the adjacent side by the length of the hypotenuse. Therefore, we have:
cos(angle) = adjacent/hypotenuse
cos(angle) = 9/x
To isolate x, we can rearrange the equation:
x = 9/cos(angle)
Now we need to find the angle. We can use the trigonometric function sine (sin) to solve for the angle. The sine of an angle in a right triangle is calculated by dividing the length of the opposite side by the length of the hypotenuse. Therefore, we have:
sin(angle) = opposite/hypotenuse
sin(angle) = 15/x
To find the angle, we can use the inverse sine function (sin^(-1)):
angle = sin^(-1)(15/x)
Now we substitute the value of the angle into the equation for x:
x = 9/cos(sin^(-1)(15/x))
To solve for x, we apply the trigonometric identity:
cos(sin^(-1)(θ)) = sqrt(1 - θ^2)
x = 9/sqrt(1 - (15/x)^2)
Now we can simplify the equation and solve for x.
Let's multiply both sides of the equation by sqrt(1 - (15/x)^2):
x * sqrt(1 - (15/x)^2) = 9
Now we square both sides of the equation to remove the square root:
x^2 * (1 - (15/x)^2) = 9^2
Expand and simplify:
x^2 - 15^2 = 81
x^2 - 225 = 81
x^2 = 306
Take the square root of both sides:
x = sqrt(306)
Now we approximate the square root:
x ≈ 17.5
Therefore, the hypotenuse of the right triangle (x) is approximately 17.5 when the adjacent side is 9 and the opposite side is 15.
A right triangle has a hypotneuse of x an adjacent of 9 and an oppisite of 15
Solve for x. Round to the nearest tenth. Please show all work including the trig function used in order to receive full credit.
Please use trig wit sin cos or tan
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1 answer