Answer:
299.99 miles
Step-by-step explanation:
Since the plane traveled due west,
The total angle is 49.17 + 90
Represent that with θ
θ = 49.17 + 90
θ = 139.17.
Represent the sides as
A = 170
B = 150
C = unknown
Since, θ is opposite side C, side C can be calculated using cosine formula as;
C² = A² + B² - 2ABCosθ
Substitute values for A, B and θ
C² = 150² + 170² - 2 * 150 * 170 * Cos 139.17
C² = 22500 + 28900 - 51000 * Cos 139.17
C² = 51400 - 51000 (−0.7567)
C² = 51400 + 38,591.7
C² = 89,991.7
Take Square Root of both sides
C = 299.9861663477167
C = 299.99 miles (Approximated)
Hence, the distance between the plane and the airport is 299.99 miles
Check the picture below.
we know that the arcST is 30°, meaning the inscribed angle intercepting it will be half that or 15°.
in the triangle RWT, two sides of it are RT and RW, both of which are radius segments and thus equal, meaning that triangle RWT is an isosceles, and in an isosceles the twin sides also make twin angles, meaning that ∡RTW is a twin of the inscribed angle ∡RWT.
Answer:
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Step-by-step explanation:
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Answer:
Step-by-step explanation:
Th correct one is the third box
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