A plane flying a straight course observes a mountain at a bearing of 32.6° to the right of its course. At that time the plane is 6 km from the mountain. A short time later, the bearing to the mountain becomes 42.6°. How far is the plane from the mountain when the second bearing is taken (to the nearest tenth of a km)?
6 sin 32.6 / sin 42.6 = 4.77
To solve this problem, we can use trigonometry, specifically the Law of Sines. Let's go step by step:
Step 1: Draw a diagram:
- Draw a line to represent the course of the plane.
- Mark a point on the line where the plane is located.
- Draw a line segment from that point to represent the mountain.
- Label the angle between the course and the first bearing as A, and the angle between the course and the second bearing as B.
Step 2: Write down the known values:
- Bearing A = 32.6°
- Bearing B = 42.6°
- Distance from the plane to the mountain when bearing A is taken = 6 km
Step 3: Use the Law of Sines to set up a proportion:
The Law of Sines states that the ratio of the length of a side of a triangle to the sine of the opposite angle is constant. In this case, let's call the length of the side opposite angle A as a and the length of the side opposite angle B as b. The proportion can be written as follows:
a / sin(B) = b / sin(A)
Step 4: Substitute the known values into the proportion:
Using the information from step 2, we have:
6 / sin(42.6°) = b / sin(32.6°)
Step 5: Solve for b:
To find b, cross-multiply and solve for b:
b = (sin(32.6°) / sin(42.6°)) * 6
Step 6: Calculate the value of b:
Using a calculator, the value of b is approximately equal to 6.5 km (rounded to one decimal place).
Therefore, the plane is approximately 6.5 km from the mountain when the second bearing is taken.
To solve this problem, we can use trigonometry and the concept of bearing angles.
First, let's draw a diagram to visualize the situation. We have a plane flying a straight course and observing a mountain. Let's label the necessary information:
- The initial bearing of the mountain is 32.6° to the right of the plane's course.
- The distance from the plane to the mountain at that time is 6 km.
- The later bearing of the mountain is 42.6°.
Now, we need to find the distance between the plane and the mountain when the second bearing is taken.
To determine the distance, we can use the concept of a right-angled triangle. Let's consider the triangle formed by the plane, the mountain, and the perpendicular line dropped from the mountain to the plane's course.
The given bearing angle of 32.6° to the right of the course tells us that the triangle is an isosceles triangle, where the two equal sides are the distances from the plane to the mountain.
Now, we can find the angles of the triangle:
- The angle between the perpendicular line and the bearing line is 90°.
- The angle between the perpendicular line and the second bearing line (42.6°) is (90 - 42.6) = 47.4°.
Since we have the included angle (47.4°) and the opposite side (6 km), we can use the Law of Sines to find the length of the other two sides of the triangle.
Using the Law of Sines:
a/sinA = b/sinB = c/sinC,
where 'a', 'b', and 'c' are the lengths of the three sides of the triangle, and 'A', 'B', and 'C' are the opposite angles:
Let's denote the unknown equal sides as 'x'.
x/sin(32.6°) = 6/sin(47.4°).
We can rearrange the equation to find 'x':
x = 6 * sin(32.6°) / sin(47.4°).
Calculating this using a calculator, we get:
x ≈ 3.87 km.
Therefore, the plane is approximately 3.9 km from the mountain when the second bearing is taken.