Question

the screen of a surveying device is 0.0026 m wide and is 0.1 m away from the lens. if the surveyor wants the image of the 2 - m target to fit on the screen, what distance d should the lens be from the target? explain.
the triangles on either side of the lens are similar because the angles are
angles, and both triangles are isosceles, so that the corresponding sides adjacent to these angles must have lengths in the same ratio. consider the right triangles that have the black, dashed lines as hypotenuses. a proportion for the side lengths is
=
so, the distance d is
m. (round to the nearest integer as needed.)

Explanation:

Step1: Identify similar triangle angles

The triangles share vertical angles, and are isosceles, so corresponding angles are equal, confirming similarity.

Step2: Set up proportion for similar triangles

Use ratios of target/screen width and target/screen distance from lens:
$$\frac{2}{0.0026} = \frac{d}{0.1}$$

Step3: Solve for distance $d$

Rearrange to isolate $d$:
$$d = \frac{2 \times 0.1}{0.0026}$$
Calculate the value:
$$d = \frac{0.2}{0.0026} \approx 76.92$$
Round to nearest integer.

Answer:

The triangles on either side of the lens are similar because the angles are vertical angles, and both triangles are isosceles, so that the corresponding sides adjacent to these angles must have lengths in the same ratio. Consider the right triangles that have the black, dashed lines as hypotenuses. A proportion for the side lengths is $\boldsymbol{\frac{2}{0.0026}} = \boldsymbol{\frac{d}{0.1}}$. So, the distance $d$ is $\boldsymbol{77}$ m.