Question

distance the player traveled?
module 13 • lesson 13.1

1. use repeated reasoning suppose $\triangle lmn$ and $\triangle opq$ are similar right triangles. for $\triangle lmn$, the side opposite $\angle m$ is 10 m, the side opposite $\angle n$ is 24 m, and $nm = 26$ m.

a. what is $\tan p$ and $\tan q$?
b. what is a possible length of each side of $\triangle opq$?

1. attend to precision the tangent ratios of the acute angles from three different right triangles are mixed together. determine which angles belong to the same right triangle.

$\tan\angle3 = 1.25$
$\tan\angle4 = 0.6$
$\tan\angle1 = 0.8$
$\tan\angle2 = 1.67$
$\tan\angle5 = 0.25$
$\tan\angle6 = 4$

1. find the measure of $\angle c$. round to the nearest tenth.

spiral review • assessment readiness

1. $\triangle abc$ and $\triangle xyz$ are similar, and $\triangle abc$ is a scaled drawing of $\triangle xyz$. the length of $\overline{ab}$ is 7.2 cm, the length of $\overline{bc}$ is 6 cm,

1. in right triangle $\triangle xyz$, the side opposite $\angle x$ is 6 m long, the side adjacent to $\angle x$ is 8 m long, and the hypotenuse is 10 m long. what is

Explanation:

Problem 23

Step1: Identify triangle sides

In $\triangle LMN$, opposite $\angle M = 10$ m, opposite $\angle N = 24$ m, hypotenuse $NM=26$ m.

Step2: Match similar angles

$\triangle LMN \sim \triangle OPQ$, so $\angle P \cong \angle M$, $\angle Q \cong \angle N$.

Step3: Calculate $\tan P$

$\tan P = \tan M = \frac{\text{opposite}}{\text{adjacent}} = \frac{24}{10} = 2.4$

Step4: Calculate $\tan Q$

$\tan Q = \tan N = \frac{\text{opposite}}{\text{adjacent}} = \frac{10}{24} = \frac{5}{12} \approx 0.42$

Step5: Find scaled sides (scale factor $k$)

Choose any positive scale factor, e.g., $k=2$.
New side opposite $\angle P$: $10 \times 2 = 20$ m
New side opposite $\angle Q$: $24 \times 2 = 48$ m
New hypotenuse: $26 \times 2 = 52$ m

Step1: Recall tangent property

In a right triangle, tangents of acute angles are reciprocals: $\tan\theta \times \tan(90^\circ-\theta) = 1$.

Step2: Check reciprocal pairs

• $1.25 \times 0.8 = 1$, so $\angle1$ and $\angle3$ are a pair.
• $1.67 \times 0.6 \approx 1$, so $\angle2$ and $\angle4$ are a pair.
• $4 \times 0.25 = 1$, so $\angle5$ and $\angle6$ are a pair.

Step1: Find length of $BD$

In $\triangle ABD$, $\tan40^\circ = \frac{BD}{AB}$, so $BD = AB \times \tan40^\circ = 8 \times \tan40^\circ \approx 8 \times 0.8391 = 6.7128$ mm

Step2: Calculate $\tan\angle C$

$\tan\angle C = \frac{BD}{BC} = \frac{6.7128}{18} \approx 0.3729$

Step3: Find $\angle C$

$\angle C = \arctan(0.3729) \approx 20.4^\circ$

Answer:

A. $\tan P = 2.4$, $\tan Q = \frac{5}{12} \approx 0.42$
B. One possible set: 20 m, 48 m, 52 m (any positive scale factor works)

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Problem 24