To calculate the height of a tree, Sophie measures the angle of elevation from a point “A” to be 34°. She then walks 10 m directly toward the tree, and finds the angle of elevation from the new point “B” to be 41°. What is the height of the tree, to the nearest tenth of a metre?

30.1 Metres

Step-by-step explanation:

For this problem, we will use the law of sines twice to find the height of the tree.

First, you need to draw the image of the tree with two triangles, one with an angle of 34 degrees, and the second with an angle of 41 degrees, both being a right triangle toward the tree.  The distance between point A and point B should be 10 meters.

From here, your bottom line of the triangle should be represented as 10 + x for the unknown distance from where Sophie stood to the tree.  The x represents the distance from point B to the tree, whereas 10 + x represents the distance from point A to the tree.

Now, let's use the scalene triangle created by the distance between point A and point B to the top of the tree, to calculate the diagonal distance from point B to the top of the tree.

Using this scalene, we have the angles, 34, 7, and 139.  Now, use the law of sines to find the diagonal of point B to the top of the tree represented by r.

sin(7) / 10 == sin (34) / r

r == 10 * sin (34) / sin(7)

r == 10 * .55919 / .12187

r == 45.88

Now that we know this diagonal distance, we can use the law of sines again with the second triangle, 41, 49, 90, to find the height of the tree represented by y.

sin(41) / y == sin(90) / 45.88

y == 45.88 * sin(41) / sin(90)

y == 45.88 * .65606 / 1

y == 30.1

So the height of the tree to the nearest tenth of a metre would be 30.1 metres.

Cheers.

Step-by-step explanation:

tan(34°)≈0.6745         tg(41°)≈0.8693