6.7 Using Trigonometric Ratios Sử dụng tỉ số lượng giác

a. Let $B$ be the boat, $T$ the top of the lighthouse and $F$ its foot at sea level. Then $\triangle BFT$ has a right angle at $F,$ $FT = 24$ m and $m\angle TBF = 8^\circ.$

b. At $T,$ $m\angle BTF = 90^\circ - 8^\circ = 82^\circ,$ and the distance $x = BF$ is opposite that angle:

Answer: b. about $171$ m.

a. Let $P$ be the plane, $G$ the point of the ground directly below it and $M$ the marker. $\triangle PGM$ has a right angle at $G$ and $PG = 740$ m.

The horizontal through $P$ is parallel to the ground, so the angle of elevation at $M$ equals the angle of depression: $m\angle PMG = 12^\circ,$ and $m\angle MPG = 90^\circ - 12^\circ = 78^\circ.$

b. $PM$ is the hypotenuse and $PG$ is opposite the $12^\circ$ angle:

c. $GM$ is opposite the $78^\circ$ angle at $P$:

Answer: b. about $3559$ m; c. about $3481$ m.

Let $h = TC,$ the height of $T$ above the line $AB.$ $\triangle TCA$ and $\triangle TCB$ have right angles at $C,$ and $m\angle ATC = 90^\circ - 20^\circ = 70^\circ,$ $m\angle BTC = 90^\circ - 27^\circ = 63^\circ.$ So:

$AB = AC - BC,$ so:

$T$ is $h$ above $A,$ and $A$ is $3850$ m above sea level: $3850 + 1911.1 = 5761.1.$

Answer: about $5761$ m above sea level.

Let $ABCD$ be the rhombus with $AB = 6$ and $m\angle DAB = 70^\circ,$ and let the diagonals meet at $O.$ The diagonals of a rhombus are perpendicular, bisect each other and bisect the angles, so $\triangle AOB$ has a right angle at $O$ and $m\angle OAB = 35^\circ.$

In $\triangle AOB$ the hypotenuse is $AB = 6$:

Answer: the diagonals are about $7$ m and $10$ m.

Sketch a right triangle with the leg adjacent to $\angle A$ equal to $2$ and the hypotenuse equal to $7.$ Let $x$ be the opposite leg. By the Pythagorean Theorem:

Answer: $\sin A = \dfrac{3\sqrt{5}}{7},$ $\tan A = \dfrac{3\sqrt{5}}{2}.$

Sketch a right triangle with the leg opposite $\angle A$ equal to $13$ and the hypotenuse equal to $85.$ Let $x$ be the adjacent leg. By the Pythagorean Theorem:

Answer: $\cos A = \dfrac{84}{85},$ $\tan A = \dfrac{13}{84}.$

Let $\overline{BC}$ be the bottom side, so $K$ is its midpoint and $BK = 50.$ The five triangles at $O$ are congruent (SSS), so $m\angle BOC = \dfrac{360^\circ}{5} = 72^\circ$; $\triangle OBC$ is isosceles, so $\overline{OK} \perp \overline{BC}$ and $m\angle BOK = 36^\circ.$

In right $\triangle OKB,$ $BK = 50$ is opposite the $36^\circ$ angle:

$OJ = OB$ because $O$ is equidistant from the vertices, so:

Answer: $JK \approx 154.$

Let the two consecutive sides meet at the vertex $V,$ with midpoints $M$ and $N,$ so $VM = VN = 15.$ Each angle of a regular decagon is $\dfrac{(10 - 2) \cdot 180^\circ}{10} = 144^\circ.$

$\triangle MVN$ is isosceles, so the perpendicular $\overline{VP}$ to $\overline{MN}$ bisects $\angle MVN$ and $\overline{MN}$: $m\angle MVP = 72^\circ.$ In right $\triangle VPM,$ $MP$ is opposite $72^\circ$:

Answer: about $28.5$ mm.