Matemática Elementar

Problema 1

In right triangle $ABC$ , $\angle C = 90^\circ$ . Altitude $\overline{CD}$ is drawn to hypotenuse $\overline{AB}$ , with $D$ on $\overline{AB}$ . Given $AD = 9$ and $DB = 16$ :

a) Find $AB$ .

b) Find $CD$ .

c) Find $AC$ and $BC$ .

Problemas

1. In right triangle $ABC$ , $\angle C = 90^\circ$ . Altitude $\overline{CD}$ is drawn to hypotenuse $\overline{AB}$ , with $D$ on $\overline{AB}$ . Given $AD = 9$ and $DB = 16$ :

a) Find $AB$ .

b) Find $CD$ .

c) Find $AC$ and $BC$ .

2. In right triangle $RST$ , $\angle T = 90^\circ$ , $RS = 50$ , $RT = 30$ , and $ST = 40$ . Altitude $\overline{TU}$ is drawn to hypotenuse $\overline{RS}$ .

a) Find $RU$ .

b) Find $US$ .

c) Find $TU$ .

3. Triangle $ABC$ has vertices $A(0,0)$ , $B(12,0)$ , and $C(12,5)$ .

a) Find the side lengths of $\triangle ABC$ .

b) For angle $A$ , find $\sin A$ , $\cos A$ , and $\tan A$ .

c) For angle $C$ , find $\sin C$ , $\cos C$ , and $\tan C$ .

4. An observer stands $60$ feet from the base of a building. The angle of elevation from the observer's eye to the top of the building is $35^\circ$ . The observer's eye is $5$ feet above the ground.

a) Draw or describe the right triangle used for the calculation.

b) Find the height of the building to the nearest tenth of a foot.

5. From the top of a cliff, the angle of depression to a boat is $28^\circ$ . A second boat is on the same straight line from the cliff and is $180$ meters farther from the base of the cliff; its angle of depression is $18^\circ$ .

a) Let $x$ be the horizontal distance from the base of the cliff to the nearer boat. Write two equations involving the cliff height $h$ .

b) Find the height of the cliff to the nearest tenth of a meter.

Gabarito — Right Triangle Similarity and Trigonometry

1. a) The hypotenuse is split into two parts:

$AB = AD + DB = 9 + 16 = 25$

b) In a right triangle, the altitude to the hypotenuse is the geometric mean of the two hypotenuse segments:

$CD^2 = AD \cdot DB$

$CD^2 = 9 \cdot 16 = 144$

$CD = 12$

c) Each leg is the geometric mean of the full hypotenuse and its adjacent hypotenuse segment:

$AC^2 = AB \cdot AD = 25 \cdot 9 = 225$

$AC = 15$

$BC^2 = AB \cdot DB = 25 \cdot 16 = 400$

$BC = 20$

Verification: $15^2 + 20^2 = 225 + 400 = 625 = 25^2$ , so the side lengths form a right triangle. $\checkmark$

2. Use the right-triangle altitude relationships. The leg $RT$ is adjacent to segment $RU$ , and the leg $ST$ is adjacent to segment $US$ .

$RT^2 = RS \cdot RU$

$30^2 = 50 \cdot RU$

$900 = 50RU$

$RU = 18$

$US = RS - RU = 50 - 18 = 32$

$TU^2 = RU \cdot US$

$TU^2 = 18 \cdot 32 = 576$

$TU = 24$

Verification: $30^2 + 40^2 = 900 + 1600 = 2500 = 50^2$ . Also, the altitude $24$ is consistent with the area:

$\frac{1}{2}(30)(40) = 600$

$\frac{1}{2}(50)(24) = 600$

3. a) The horizontal distance from $A$ to $B$ is $12$ , and the vertical distance from $B$ to $C$ is $5$ , so:

$AB = 12, \quad BC = 5$

The hypotenuse is:

$AC = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13$

b) For angle $A$ , the opposite side is $BC = 5$ , the adjacent side is $AB = 12$ , and the hypotenuse is $AC = 13$ :

$\sin A = \frac{5}{13}$

$\cos A = \frac{12}{13}$

$\tan A = \frac{5}{12}$

c) For angle $C$ , the opposite side is $AB = 12$ , the adjacent side is $BC = 5$ , and the hypotenuse is $AC = 13$ :

$\sin C = \frac{12}{13}$

$\cos C = \frac{5}{13}$

$\tan C = \frac{12}{5}$

The ratios for $A$ and $C$ are complementary, which is expected because the two acute angles in a right triangle add to $90^\circ$ .

4. a) The right triangle has horizontal leg $60$ feet and vertical leg equal to the building height above the observer's eye. The angle at the observer is $35^\circ$ .

b) Let $y$ be the height from the observer's eye to the top of the building. Then:

$\tan(35^\circ) = \frac{y}{60}$

$y = 60\tan(35^\circ)$

Using $\tan(35^\circ) \approx 0.7002$ :

$y \approx 60(0.7002) = 42.012$

Add the observer's eye height:

$42.012 + 5 = 47.012$

The building is approximately:

$\boxed{47.0 \text{ feet tall}}$

5. a) Angles of depression are congruent to the angles of elevation from the boats. For the nearer boat:

$\tan(28^\circ) = \frac{h}{x}$

So:

$h = x\tan(28^\circ)$

For the farther boat, the horizontal distance is $x + 180$ :

$\tan(18^\circ) = \frac{h}{x + 180}$

So:

$h = (x + 180)\tan(18^\circ)$

b) Set the two expressions for $h$ equal:

$x\tan(28^\circ) = (x + 180)\tan(18^\circ)$

Use $\tan(28^\circ) \approx 0.5317$ and $\tan(18^\circ) \approx 0.3249$ :

$0.5317x = 0.3249(x + 180)$

$0.5317x = 0.3249x + 58.482$

$0.2068x = 58.482$

$x \approx 282.8$

Now find $h$ :

$h = x\tan(28^\circ)$

$h \approx 282.8(0.5317) \approx 150.4$

The cliff is approximately:

$\boxed{150.4 \text{ meters tall}}$