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.

in the right triangle diagram: $\angle a=40^\circ$, $ab=8$ mm, $bc=18$ mm, $db\perp ac$

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 right triangle angles

$\triangle LMN$ is right-angled, so right angle is $\angle L$ (since $NM=26$ is hypotenuse, longest side). $\angle M$ opposite side $LN=24$, $\angle N$ opposite side $LM=10$.

Step2: Calculate $\tan M, \tan N$

$\tan M = \frac{\text{opposite}}{\text{adjacent}} = \frac{24}{10} = 2.4$
$\tan N = \frac{\text{opposite}}{\text{adjacent}} = \frac{10}{24} = \frac{5}{12} \approx 0.4167$

Step3: Apply similarity for $\tan P, \tan Q$

Similar triangles have equal corresponding tangent ratios. $\angle P \cong \angle M$, $\angle Q \cong \angle N$, so $\tan P = \tan M$, $\tan Q = \tan N$.

Step4: Choose scale factor for $\triangle OPQ$

Pick scale factor $k=2$ (any positive $k$ works). Multiply $\triangle LMN$ sides by $k$:
$LM' = 10 \times 2 = 20$ m, $LN' = 24 \times 2 = 48$ m, $NM' = 26 \times 2 = 52$ m

Step1: Pair complementary tangent ratios

For right triangles, acute angles are complementary, so $\tan \theta = \frac{1}{\tan(90^\circ-\theta)}$.
Check pairs:

• $\tan\angle3=1.25 = \frac{5}{4}$, $\tan\angle4=0.8 = \frac{4}{5}$ (reciprocal pair)
• $\tan\angle1=0.8 = \frac{4}{5}$, $\tan\angle2=1.25 = \frac{5}{4}$ (reciprocal pair)
• $\tan\angle5=0.25 = \frac{1}{4}$, $\tan\angle6=4$ (reciprocal pair)

Step1: Find length of $DB$

In $\triangle ABD$, $\tan 40^\circ = \frac{DB}{AB}$, so $DB = AB \times \tan40^\circ = 8 \times \tan40^\circ \approx 8 \times 0.8391 = 6.7128$ mm

Step2: Calculate $\tan C$

In $\triangle CBD$, $\tan C = \frac{DB}{BC} = \frac{6.7128}{18} \approx 0.3729$

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 scaled version of 10 m, 24 m, 26 m is valid)

---

Problem 24