All Solutions
Section 5-6: The Sine Law
$sin theta=15.5cdot dfrac{sin 105^circ}{20.5}$
$$
theta=sin^{-1}left(15.5cdot dfrac{sin 105^circ}{20.5}right)=47^circ
$$
$sin theta=9cdot dfrac{sin 38^circ}{7}$
$theta=sin^{-1}left(9cdot dfrac{sin 38^circ}{7}right)=52^circ$
Since $theta >90^circ$, this is one of the ambiguous case,
$$
theta=180^circ-52^circ=128^circ
$$
b) $128^circ$
It’s lower than the length of the sides.
c) Since $h<6.6<9.8$, there are two possible measure of the third side.
b) $6.3$ m
c) 2 possible measures of third side
defarraystretch{1.4}%
begin{tabular}{|l|l|l|}
hline
$angle A < 90^circ$ & $a<h$ & no triangle exists \ cline{2-3}
& $a=h$ & one right triangle \ cline{2-3}
& $h<b<a$ & one triangle \ cline{2-3}
& $h<a90^circ$ & $ab$ & one triangle \ hline
end{tabular}
end{table}
Since $h$ is greater than the hypotenuse $b=2.8$, no triangle exists.
Since $h$ is greater than the hypotenuse $c=2.1$, no triangle exists.
Since $a=h$, one right triangle exists.
b) no triangle exists
c) one right triangle
defarraystretch{1.4}%
begin{tabular}{|l|l|l|}
hline
$angle A < 90^circ$ & $a<h$ & no triangle exists \ cline{2-3}
& $a=h$ & one right triangle \ cline{2-3}
& $h<b<a$ & one triangle \ cline{2-3}
& $h<a90^circ$ & $ab$ & one triangle \ hline
end{tabular}
end{table}
$dfrac{sin 120^circ}{12.3}=dfrac{sin theta}{9.1}$
$theta=9.1cdot dfrac{sin 120^circ}{12.3}$
$theta=sin^{-1}left( 9.1 cdot dfrac{sin 120^circ}{12.3}right)=40^circ$
$h=14.2sin 45^circ= 10$
Since $h<RS <QS$, we have two triangles.
$dfrac{sin 45^circ}{10.8}=dfrac{sin theta}{14.2}$
$theta=14.2cdot dfrac{sin 45^circ}{10.8}$
$theta=sin^{-1}left( 14.2 cdot dfrac{sin 45^circ}{10.8}right)=68^circ$
Another triangle has an obtuse angle $theta=180-68=112^circ$
b) $68^circ$, $112^circ$
defarraystretch{1.4}%
begin{tabular}{|l|l|l|}
hline
$angle A < 90^circ$ & $a<h$ & no triangle exists \ cline{2-3}
& $a=h$ & one right triangle \ cline{2-3}
& $h<b<a$ & one triangle \ cline{2-3}
& $h<a90^circ$ & $ab$ & one triangle \ hline
end{tabular}
end{table}
We see that $angle A<90^circ$, calculate $h$ to determine whether we have one or two triangles.
$h=9.3sin 35^circ=5.3$
Since $h<a<b$, there are two possible triangles.
We need to find $c, angle B$ and $angle C$
First Triangle:
$dfrac{sin angle B}{b}=dfrac{sin angle A}{a}$
$angle B=sin^{-1}left(9.3cdot dfrac{sin 35^circ}{7.2}right)=48^circ$
The sum of angles in any triangle is $180^circimplies angle C=180-(35+48)=97^circ$
Use sine law to obtain the remaining side.
$dfrac{sin C}{c}=dfrac{sin A}{a}$
$c=acdot dfrac{sin C}{sin A}=7.2cdot dfrac{sin 97^circ}{sin 35^circ}=12.4$
Second Triangle: $angle B$ must be greater than $90^circ$
$dfrac{sin (180^circ-angle B)}{b}=dfrac{sin angle A}{a}$
$180-angle B=sin^{-1}left(9.3cdot dfrac{sin 35^circ}{7.2}right)=48^circ$
$angle B=180^circ-48^circ=132^circ$
$angle C=180-(35+132)=13^circ$
$dfrac{sin C}{c}=dfrac{sin A}{a}implies c=acdot dfrac{sin C}{sin A}=7.2cdot dfrac{sin 13^circ}{sin 35^circ}=2.8$
$a=7.3$ , $b=14.6$ , $angle A=30^circ$
$h=14.6sin 30^circ=7.3$
Since $a=h$, we have one right triangle.
Therefore,
$angle B=90^circ$
$angle C=180^circ-(90^circ+30^circ)=60^circ$
We can use Pythagorean theorem to solve for the other leg,
$$
c=sqrt{14.6^2-7.3^2}=12.6
$$
$h=2.8sin 33^circ=1.52$
Since $a<h$, no triangle exists.
$angle C=180^circ-(75^circ+43^circ)=62^circ$
$dfrac{sin C}{c}=dfrac{sin A}{a}implies a=ccdot dfrac{sin A}{sin C}$
$a=22.2cdot dfrac{sin 75^circ}{sin 62^circ}=24.3$
$dfrac{sin B}{b}=dfrac{sin A}{a}$
$$
b=adfrac{sin B}{sin A}=24.3cdot dfrac{sin 43^circ}{sin 75^circ}=17.2
$$
second triangle: $c=2.8$ , $angle 132^circ$ , $angle C=13^circ$
b) $c=12.6$ , $angle B=90^circ$, $angle C=60^circ$
c) no triangle exists
d) $a$ = 24.3 cm, $b$ = 17.2 cm, $angle C=62^circ$
Using sine law
$dfrac{sin theta}{20}=dfrac{sin(180-78)^circ}{35}$
$sin theta=20cdot dfrac{sin(102^circ)}{35}$
$theta=sin^{-1}left(dfrac{20sin 102^circ}{35}right)=34^circ$
Therefore, the rope makes $34^circ$ with the ground.
34^circ
$$
$angle RPQ=180-40=140^circ$
Since the sum of angles for a triangle is $180^circ$, we get
$angle PRQ=180-(140+32)=8^circ$
Now, we can find $PR$ using sine law
$dfrac{sin 32^circ}{PR}=dfrac{sin 8^circ}{PQ}$
$PR=dfrac{105sin 32^circ}{sin 8^circ}=400$ m
Now, we can find $h$ as
$sin theta=dfrac{text{opposite side}}{text{hypotenuse}}$
$sin 40^circ=dfrac{h}{PR}$
$h=PRsin 40^circ=400sin 40^circ=257$ m
Therefore, the building is $257$ m high
Using sine law
$dfrac{sin 13^circ}{x}=dfrac{sin 32^circ}{9750}$
$x=dfrac{9750sin13^circ}{sin32^circ}=4139$ m
From the sketch, we see that
$angle BCA=71^circ$
Thus, we can calculate $BC$ as
$cos 71^circ=dfrac{270}{BC}implies BC=829.32$ m
$theta=71^circ-54^circ=17^circ$
Now, we can use sine law to find $x$ as
$dfrac{sin17^circ}{x}=dfrac{sin 54^circ}{829.32}$
$x=dfrac{829.32sin17^circ}{sin 54^circ}=299.7$ m
Therefore, the aqueduct is $299.7$ m long.
In the problem, it says one end of the lake is 270 m from the balloon. The exact interpretation of this would be in the diagram below.
Using sine law
$dfrac{x}{sin(71-54)^circ}=dfrac{270}{sin 54^circ}$
$x=dfrac{270cdot sin(71-54)^circ}{sin 54^circ}=97.6$ m
$angle ADB=180^circ-(34^circ+90^circ)=56^circ$
$angle DAC=34^circ-13^circ=21^circ$
Now, we can use sine law
$dfrac{sin 21^circ}{x}=dfrac{sin 56^circ}{30}$
$x=dfrac{30sin 21^circ}{sin 56^circ}=13$ m
Therefore, the height of the tower is $13$ m
To find the height of the tower $x$, we must find $BD$ and subtract it with $BC$
From the sketch, we can find $BC$ as
$sin 13^circ=dfrac{BC}{30}implies BC=30sin 13^circ=6.75$
We can find $AB$ as
$cos 13^circ=dfrac{AB}{30}implies AB=30cos 13^circ=29.23$
Now, we can also find $BD$ as
$tan 34^circ=dfrac{BD}{AB}implies BD=29.23tan 34^circ=19.72$
Therefore, the height of the tower is
$x=BD-BC=19.72-6.75=13$ m
a) To find the height of the kite, we must find the hypotenuse $BD$ and use sine to find the altitude.
We can see that $angle ABD=180^circ-50^circ=130^circ$
Since the sum of the angles in a triangle is $180^circ$
$theta=180^circ-(35^circ+130^circ)=15^circ$
Now, we can use sine law to find $BD$
$dfrac{sin35^circ}{BD}=dfrac{sin 15^circ}{11}$
$BD=dfrac{11sin35^circ}{sin 15^circ}=24.4$ m
Thus, the height of the kite is
$h=24.4sin 50^circ=boxed{text{19; m}}$
b) The length of the string is $boxed{BDtext{ = 24;m}}$
c) We shall use another sine law to find $BC$
We can see that $angle BCD=180-66=114^circ$ implying that
$angle BDC=180^circ-(50^circ+114^circ)=16^circ$
$dfrac{sin 16^circ}{BC}=dfrac{sin 114^circ}{BD}$
$BC=dfrac{BD sin 16^circ}{sin 114^circ}$
$$
BC=dfrac{24sin 16^circ}{sin 114^circ}=boxed{7 text{ m}}
$$
a) To find the height of the kite, we must find the hypotenuse $BD$ and use sine to find the altitude.
We can see that $angle BCD=180^circ-66^circ=114^circ$
Since the sum of the angles in a triangle is $180^circ$
$theta=180^circ-(50^circ+114^circ)=16^circ$
Now, we can use sine law to find $BD$
$dfrac{sin114^circ}{BD}=dfrac{sin 16^circ}{11}$
$BD=dfrac{11sin114^circ}{sin 16^circ}=36$ m
Thus, the height of the kite is
$h=36sin 50^circ=boxed{text{28; m}}$
b) The length of the string is $boxed{BDtext{ = 36;m}}$
c) We shall use another sine law to find $AB$
We can see that $angle ABD=180-50=130^circ$ implying that
$angle ADB=180^circ-(35^circ+130^circ)=15^circ$
$dfrac{sin 15^circ}{AB}=dfrac{sin 35^circ}{BD}$
$AB=dfrac{36 sin 15^circ}{sin 35^circ}=boxed{16 text{m}}$
Friend 2 is in other side of the kite and Carol is 11m from Friend 1.
Similar procedure will be done in the succeeding cases.
$angle ABD=180^circ-50^circ=130^circ$
$theta=180^circ-(35^circ+130^circ)=15^circ$
b) length of string: $BD=dfrac{11sin 35^circ}{sin 15^circ}=24.4approx 24$m
a) height of kite: $h=BDsin 50^circ=24.4sin 50^circ=24.4sin 50^circ=19$ m
c) distance of other friend to Carol:
$dfrac{BC}{sin(180-50-66)^circ}=dfrac{24.4}{sin 66^circ} implies BC=24$ m
Friend 2 is in other side of the kite and Carol is 11m from Friend 2.
$theta=180^circ-(50+66)^circ=64^circ$
b) length of string:
$dfrac{BD}{sin 66^circ}=dfrac{11}{sin 64^circ}implies BD=11.2=11$ m
a) height of kite: $h=BDsin 50^circ=11.2sin 50^circ=9$ m
c) distance of other friend to Carol:
$angle ABD=180-50=130^circ$ ; $angle BDA=180-(35+130)-15^circ$
$dfrac{AB}{sin 15^circ}=dfrac{11.2}{sin 35^circ}implies 5$ m
Carol is in other side of the kite and Carol is 11m from Friend 2.
$theta=180^circ-(50+66)^circ=64^circ$
b) length of string:
$dfrac{DC}{sin 66^circ}=dfrac{11}{sin 64^circ}implies BD=11.2=11$ m
a) height of kite: $h=BDsin 50^circ=11.2sin 50^circ=9$ m
c) distance of other friend to Carol:
$angle ABD=180-66=114^circ$ ; $angle BDA=180-(35+114)=31^circ$
$angle ADC=31^circ+64^circ=95^circ$
$dfrac{AC}{sin95^circ}=dfrac{11.2}{sin 35^circ}implies AC=19$ m
Carol is in other side of the kite and Carol is 11m from Friend 1.
$angle ADC=180-(50+35)=95^circ$
b) length of string:
$dfrac{11}{sin 95^circ}=dfrac{DC}{sin 35^circ}implies DC=6.33=6$ m
a) height of kite: $h=DCsin 50^circ=6.3sin 50^circ=5$ m
c) distance of other friend to Carol:
$angle BDC=180-(66+50)=64^circ$
$dfrac{BC}{sin 64^circ}=dfrac{6.3}{sin 66^circ}implies BC=6$ m
Friend 1 is in other side of the kite and Carol is 11m from Friend 2.
$angle ABD=180-66=114^circ$
$theta=180-(50+114)=16^circ$
b) length of string:
$dfrac{AD}{sin 114^circ}=dfrac{11}{sin 16^circ}implies AD=36$ m
a) height of the kite:
$h=ADsin 50^circ=37sin 50^circ=28$ m
c) distance of Carol from the other friend
$angle ADC=180-(50+35)=95^circ$
$dfrac{AC}{sin 95^circ}=dfrac{36}{sin 35^circ}implies AC=63$ m
Friend 1 is in other side of the kite and Carol is 11m from Friend 1.
$angle ADC=180-(50+35)=95^circ$
b) length of string:
$dfrac{AD}{sin 35^circ}=dfrac{11}{sin 95^circ}implies 6.3=6$ m
a) height of the kite:
$h=ADsin 50^circ=6.3sin 50^circ=5$ m
c) distance of Carol from the other friend
$angle ABD=180-66=114^circ$
$theta=180-(114+50)=16^circ$
$dfrac{AB}{sin 16^circ}=dfrac{6.3}{sin 114^circ}implies AB=2$ m
defarraystretch{1.3}%
begin{tabular}{|c|c|c|c|c|}
hline
Cases & Description & a) & b) & c) \ hline
1 & begin{tabular}[c]{@{}c@{}}All friends are on the same side of the kite.\ Carol is $11$ m from Friend 1end{tabular} & 19 m & 24 m & 7m \ hline
2 & begin{tabular}[c]{@{}c@{}}All friends are on the same side of the kite.\ Carol is $11$ m from Friend 2end{tabular} & 28 m & 36 m & 16 m \ hline
3 & begin{tabular}[c]{@{}c@{}}Friend 2 is in other side of the kite.\ Carol is 11m from Friend 1.end{tabular} & 19 m & 24 m & 24 m \ hline
4 & begin{tabular}[c]{@{}c@{}}Friend 2 is in other side of the kite.\ Carol is 11m from Friend 2end{tabular} & 9 m & 11 m & 5 m \ hline
5 & begin{tabular}[c]{@{}c@{}}Carol is in other side of the kite.\ Carol is 11m from Friend 2end{tabular} & 9 m & 11 & 19 m \ hline
6 & begin{tabular}[c]{@{}c@{}}Carol is in other side of the kite.\ Carol is 11m from Friend 1.end{tabular} & 5 m & 6 m & 6 m \ hline
7 & begin{tabular}[c]{@{}c@{}}Friend 1 is in other side of the kite.\ Carol is 11m from Friend 2.end{tabular} & 28 m & 36 m & 63 m \ hline
8 & begin{tabular}[c]{@{}c@{}}Friend 1 is in other side of the kite. \ Carol is 11m from Friend 1.end{tabular} & 5 m & 6 m & 2 m \ hline
end{tabular}
end{table}
This also implies that the angles opposite to these sides are equal,
$angle B = angle A$
Now, from the small triangle, we can see that
$angle A=180-(90+2.8)=87.2^circ=angle B$
We can obtain $angle C=180-87.2-87.2=5.6^circ$
Using sine law
$dfrac{x}{sin 87.2^circ}=dfrac{47}{sin 5.6}$
$x=(sin 87.2)cdot dfrac{47}{sin 5.6^circ}=481$ m
Therefore, the top and bottom of the tower is equidistant at a point 481 m from the base of the tower.
$430$ must be the opposite of $35^circ$
$dfrac{sin 35^circ}{430}=dfrac{sin angle C}{110}$
$angle C=sin^{-1}left(110cdot dfrac{sin 35^circ}{430}right)=8.44^circ=8^circ$
$angle B=180^circ-(8.44+35)^circ=136.56^circ=137^circ$
Use another sine law to find $AC$
$dfrac{AC}{sin 136.56^circ}=dfrac{430^circ}{sin 35^circ}$
$AC=dfrac{430sin 136.56^circ}{sin 35^circ}=515$ m
condition: $l> m$
relationship between sides: $dfrac{sin L}{l}=dfrac{sin M}{m}$
height of the triangle: $h=msin L =lsin M$
If $m=lsin M=h$, we get one right triangle.
condition: $msin L < l<m$
relationship between sides: $dfrac{sin L}{l}=dfrac{sin M}{m}$
height of the triangle: $h=msin L =lsin M$
condition: $l< msin L$
relationship between sides: $dfrac{sin L}{l}=dfrac{sin M}{m}$
height of the triangle: $h=msin L =lsin M$
b) condition: $msin L < l<m$
relationship between sides: $dfrac{sin L}{l}=dfrac{sin M}{m}$
height of the triangle: $h=msin L =lsin M$
c) condition: $l< msin L$
See sketches inside.
Remember that bearing is an angle starting from the north (positive $y$-axis) going clockwise.
$45^circ+theta=237^circ-180^circ$
$theta=237^circ-180^circ-45^circ=12^circ$
Using sine law
$dfrac{y}{sin 12^circ}=dfrac{11}{sin(285-237)^circ}$
sailor’s distnace from lighthouse A:
$y=dfrac{11sin12^circ}{sin(285-237)^circ}=3$ km
sailor’s distance from lighthouse B:
$phi=180-12-(285-237)=120^circ$
$dfrac{x}{sin 120^circ}=dfrac{11}{sin(285-237)^circ}$
$x=12.82=13$ km
Therefore, the sailor is 3 km from Light House A and 13 km from Light House B.
$phi = 120^circ$
The shortest distance of the sailor to the shore must be perpendicular to the shore.
From the right triangle in the figure, we can write
$sin(180^circ-phi)=dfrac{z}{y}$
$sin(180^circ-120^circ)=dfrac{z}{3}$
$z=3sin(180^circ-120^circ)=2.6 text{km} =3$ km
b) $3$ km
Using sine law,
$dfrac{x}{sin 160^circ}=dfrac{222.5}{sin 12^circ}$
$x=dfrac{222.5sin 160^circ}{sin 12^circ}=366$ m
$sin 8^circ=dfrac{y}{x}$
$y=xsin 8^circ=366sin 8^circ=51$ m
Since $51>40$, you won’t get swamped.
b) no
length of lower wire: $sin 36^circ=dfrac{155}{x}implies x=dfrac{155}{sin 36^circ}=264$ m
length of higher wire: $sin 59^circ=dfrac{350}{y}implies y=dfrac{350}{sin 59^circ}=408$ m
higher guy wire: 408 m
