6.8 Chapter 6 Exam Đề kiểm tra Chương 6

The altitude $x$ cuts the hypotenuse into $c_1 = 3$ and $c_2 = 12,$ so $c = 15.$ By Corollary 1 and Corollary 2 of Theorem 6-1:

Answer: $x = 6,$ $y = 3\sqrt{5},$ $z = 6\sqrt{5}.$

The leg $12$ and the piece next to it, $c_1 = 8,$ are given; the hypotenuse is $x.$ By Corollary 2 of Theorem 6-1, a leg squared is the hypotenuse times the piece next to it:

Answer: $x = 18,$ $y = 10,$ $z = 6\sqrt{5}.$

The altitude $h$ is a leg of both right triangles. In the right one, the hypotenuse is $13$ and the other leg is $5$; by Theorem 6-2:

In the left one, the hypotenuse is $15$:

The base of the large triangle is $x + 5 = 14,$ so its sides are $13,$ $14$ and $15.$ The longest side is $15$:

Since $225 < 365,$ the triangle is acute (Theorem 6-5).

Answer: $h = 12,$ $x = 9$; the triangle is acute.

The two legs are equal (tick marks), so the altitude $x$ cuts the triangle into two right triangles that are congruent by HL; it halves the base: each half is $8.$

In the left right triangle, $x$ is opposite the $55^\circ$ angle and $8$ is adjacent:

The angles of the right triangle add up to $180^\circ$: $y = 180 - 90 - 55 = 35.$

Answer: $x \approx 11.4,$ $y = 35.$

The diagonals of a rhombus are perpendicular and bisect each other. Let them meet at $O$: $\triangle AOD$ has a right angle at $O,$ $OD = \dfrac{12}{2} = 6$ and hypotenuse $AD = 2\sqrt{13}.$

Answer: $AC = 8.$

Draw the altitudes $\overline{DH}$ and $\overline{CK}$ to $\overline{AB}$; $DHKC$ is a rectangle, so $HK = 5$ and $CK = DH.$

$\triangle AHD$ is 30-60-90 with hypotenuse $8$: $DH = 4,$ $AH = 4\sqrt{3}$ (Theorem 6-7).

$\triangle BKC$ is 30-60-90 with $CK = 4$ opposite the $60^\circ$ angle: $KB = \dfrac{4}{\sqrt{3}} = \dfrac{4\sqrt{3}}{3}$ and $BC = 2 \cdot KB = \dfrac{8\sqrt{3}}{3}.$

The area $S$ of a trapezoid is half the sum of the bases times the altitude:

Answer: the perimeter is $18 + 8\sqrt{3}$ and the area is $20 + \dfrac{32\sqrt{3}}{3}.$

Each angle of a regular hexagon is $\dfrac{(6 - 2) \cdot 180^\circ}{6} = 120^\circ,$ so a side next to the bottom side rises at $60^\circ$ to the horizontal.

Drop a perpendicular from its upper end to the line of the bottom side: a 30-60-90 triangle with hypotenuse $2,$ so its shorter leg is $1$ and the longer (vertical) leg is $\sqrt{3}$ (Theorem 6-7).

The next side rises the same height to the top side, so $x = \sqrt{3} + \sqrt{3} = 2\sqrt{3}.$

Answer: $x = 2\sqrt{3}$ cm.

The third angle is $180^\circ - 37^\circ - 53^\circ = 90^\circ,$ so the triangle is right at the top: $16$ and $x$ are its legs, and $x$ is opposite the $37^\circ$ angle.

Answer: $x \approx 12.1.$

$\overline{AC} \perp \overline{BD},$ so $\triangle ACB$ and $\triangle ACD$ are right triangles that share the leg $AC.$ In $\triangle ACB,$ $BC = 8$ is opposite the $33^\circ$ angle:

In $\triangle ACD,$ $CD$ is opposite the $48^\circ$ angle:

Answer: $CD \approx 13.7.$

The third angle is $180^\circ - 38^\circ - 52^\circ = 90^\circ,$ so the base $x$ is the hypotenuse, and $9$ is the leg adjacent to the $38^\circ$ angle.

Answer: $x \approx 11.4.$

a. Let $P$ be the person at the top, $F$ the foot of the cliff and $S$ the steamer: $\triangle PFS$ is right at $F,$ $PF = 32$ m, and the angle of elevation at $S$ equals the angle of depression, $6^\circ.$

b. At $P,$ $m\angle FPS = 90^\circ - 6^\circ = 84^\circ,$ and $FS$ is opposite it:

Answer: b. about $304$ m.

Sketch a right triangle with the leg opposite $\angle A$ equal to $7$ and the adjacent leg equal to $4.$ By the Pythagorean Theorem the hypotenuse is $\sqrt{7^2 + 4^2} = \sqrt{65}.$

Answer: $\sin A = \dfrac{7\sqrt{65}}{65},$ $\cos A = \dfrac{4\sqrt{65}}{65}.$

From the table: $\tan 27^\circ \approx 0.5095$ and $\tan 21^\circ \approx 0.3839,$ so $\tan 27^\circ + \tan 21^\circ \approx 0.8934.$

But $\tan (27^\circ + 21^\circ) = \tan 48^\circ \approx 1.1106,$ which is not $0.8934.$

One counterexample is enough: the statement is not correct.

Answer: a. $0.5095$; b. $0.3839$; c. $0.8934$; d. $1.1106$; e. no.