$c^2 = a^2 + b^2 – 2ab cos C$

$c^2=7^2+4^2-2(7)(4)cos 62$

$c^2=49+16-26.3$

$c^2=38.7$

$c=sqrt{38.7}$

$$
c=6.2
$$

$c^2 = a^2 + b^2 – 2ab cos C$

$c^2=12^2+9^2-2(12)(9)cos 125$

$c^2=144+81-(-123.8)$

$c^2=348.9$

$c=sqrt{348.9}$

$$
c=18.7
$$

To solve for the angle, we can rewrite cosine law as

$cos C=dfrac{(a^2+b^2)-c^2}{2ab}implies C=cos^{-1}left[dfrac{(a^2+b^2)-c^2}{2ab}right]$

where $c$ is the opposite side of $angle C$ and $a$ and $b$ are the other legs.

a) $theta=cos^{-1}left[ dfrac{(6^2+7^2)-4^2}{2(6)(7)}right]=35^circ$

b) $theta=cos^{-1}left[ dfrac{(4.7^2+5.2^2)-3.4^2}{2(4.7)(5.2)}right]=40^circ$

a) $35^circ$

b) $40^circ$

Apply cosine law in each case

a) $w^2=10.5^2+11.2^2-2(10.5)(11.2)cos 43^circ$

$w=sqrt{10.5^2+11.2^2-2(10.5)(11.2)cos 43^circ}=8.0$

b) $4^2=3^2+2^2-2(3)(2)cos theta$

$cos theta=dfrac{3^2+2^2-4^2}{2(3)(2)}$

$$
theta=cos^{-1}left(dfrac{3^2+2^2-4^2}{2(3)(2)}right)=104^circ
$$

c) $a^2=b^2+c^2-2bccos Aimplies cos A=dfrac{b^2+c^2-a^2}{2bc}$

$A=cos^{-1}left( dfrac{8.3^2+6.6^2-11.5^2}{2(8.3)(6.6)}right)=100^circ$

d) $p^2=q^2+r^2-2qrcos P$

$p=sqrt{q^2+r^2-2qrcos P}$

$$
p=sqrt{25.1^2+71.3^2-2(25.1)(71.3)(1/4)}=69.4
$$

a) $8.0$

b) $104^circ$

c) $100^circ$

d) $69.4$

a) Use cosine law to find $LN$

$LN^2=7.5^2+11.2^2-2(7.5)(11.2)cos 105^circ$

$LN=sqrt{.5^2+11.2^2-2(7.5)(11.2)cos 105^circ}$

$LN=m=15.0$ cm

Use sine law to find $angle L$

$dfrac{sin angle L}{11.2}=dfrac{sin 105^circ}{15}$

$L=sin^{-1}left(dfrac{11.2sin 105^circ}{15}right)=46^circ$

$$
angle N=180^circ-(105^circ+46^circ)=29^circ
$$

b) $angle R=180^circ-(120^circ+28^circ)=32^circ$

Use sine law to find $RS$

$dfrac{RS}{sin 28^circ}=dfrac{25.6}{sin 120^circ}$

$t=RS=dfrac{25.6sin 28^circ}{sin 120^circ}=13.9$ cm

Find $ST$ using cosine law

$ST^2=13.9^2+25.6^2-(2)(13.9)(25.6)cos 32^circ$

$ST=sqrt{13.9^2+25.6^2-(2)(13.9)(25.6)cos 32^circ}$

$ST=r= 15.7$ cm

c) Use cosine law to find $angle A$

$10^2=8^2+5^2-2(8)(5)cos A$

$A=cos^{-1}left(dfrac{8^2+5^2-10^2}{2(8)(5)}right)=97.9^circ=98^circ$

Use sine law to find $angle B$

$dfrac{sin B}{5}=dfrac{sin 97.9}{10}$

$B=sin^{-1}left(dfrac{5sin 97.9}{10}right)=29.68^circ=30^circ$

$$
angle C=180^circ-(98+30)^circ=52^circ
$$

d) $X=180^circ-(21+35)^circ=124^circ$

Use sine law to find $XY$

$dfrac{XY}{sin 35^circ}=dfrac{18.7}{sin 124^circ}$

$XY=z=dfrac{18.7sin 35^circ}{sin 124^circ}=12.94=12.9$ cm

Use cosine law to find $XZ$

$XZ=y=sqrt{12.94^2+18.7^2-2(12.94)(18.7)cos 21^circ}=8.1$ cm

a) $m=15.0$ cm , $angle L=46^circ$, $angle N=29^circ$

b) $angle R=32^circ$, $t=13.9$ cm, $r=15.7$ cm

c) $angle = 98^circ$ , $angle B=30^circ$, $angle C=52^circ$

d) $angle X=124^circ$ , $y=8.1$ cm , $z=12.9$ cm

Use cosine law

$2^2=8^2+6.5^2-2(8)(6.5)cos theta$

$cos theta=dfrac{8^2+6.5^2-2(8)(6.5)}{2(8)(6.5)}$

$theta=cos^{-1}left(dfrac{8^2+6.5^2-2(8)(6.5)}{2(8)(6.5)}right)=11^circ$

The shot must be at $11^circ$ angle.

$$
11^circ
$$

Use cosine law

$t^2=h^2+b^2-2hbcos T$

$t^2=160^2+270^2-2(160)(270)cos 23^circ$

$t=sqrt{160^2+270^2-2(160)(270)cos 23^circ}$

$t=138$ m

$138$ m

Refer to the figure in your textbook.

We can see that

$angle ADC=180^circ-angle ADB$

$angle ADC=180^circ- 45^circ=135^circ$

Since the sum of angles in a triangle is always $180^circ$

$angle CAD=180^circ-135^circ-30^circ=15^circ$

Use sine law to find $AC$

$dfrac{AC}{sin 135^circ}=dfrac{1.0}{sin 15^circ}$

$AC=dfrac{1.0sin 135^circ}{sin 15^circ}=2.73$

We can now find $AB$ using cosine law

$AB^2=BC^2+AC^2-2(BC)(AC)cos C$

$AB^2=2^2+2.73^2-2(2)(2.73)cos 30^circ$

$AB=sqrt{2^2+2.73^2-2(2)(2.73)cos 30^circ}=1.4$ m

$AB=1.4$ m

From the figure

$angle FAB=70^circ-25^circ=45^circ$

$angle ABF=345^circ-(70+180)^circ=95^circ$

$$
angle AFB=180^circ-(95^circ+45^circ)=40^circ
$$

Use sine law to find $BF$

$dfrac{BF}{sin 45^circ}=dfrac{20.3}{sin 40^circ}$

$BF=dfrac{20.3sin 45^circ}{sin 40^circ}=22.33=22.3$ km

We can find $AF$ by either cosine or sine law. Here we’ll use cosine law.

$AF=sqrt{20.3^2+22.33^2-2(20.3)(22.33)cos 95^circ}=31.5$ km

Therefore, the fire is 31.5 km from tower A and 22.3 km from tower B.

a)

$bold{Question}$ Two roads intersect at an angle of $15^circ$ as shown in the figure above. Darryl is standing at point D which is 270 m from the intersection. If his friend Mark is at point B and $angle ADB=110^circ$, how far is Mark from the intersection to the nearest meter?

$bold{Answer}$

The sum of angles in a triangle is always $180^circ$

$angle ABD=180^circ-(15^circ+110^circ)=55^circ$

Use sine law to find $AB$

$dfrac{AB}{sin 110^circ}=dfrac{270}{sin 55^circ}$

$AB=dfrac{270sin 110^circ}{sin 55^circ}=309.73=310$ m

Therefore, Mark is 310 m from the intersection.

a)

$bold{Question}$ Two roads intersect at an angle of $15^circ$ as shown in the figure above. Darryl is standing at point D which is 270 m from the intersection. If his friend Mark is at point B and is 300 m from the intersection, how far is Mark from Darryl? Round your answers to the nearest meter.

$bold{Answer}$

Use cosine law to find $BD$

$BD^2=270^2+300^2-2(270)(300)cos 15^circ$

$BD=sqrt{270^2+300^2-2(270)(300)cos 15^circ}$

$BD=80$ m

Therefore, Mark is 80 m away from Darryl.

a) Two roads intersect at an angle of $15^circ$ as shown in the figure above. Darryl is standing at point D which is 270 m from the intersection. If his friend Mark is at point B and $angle ADB=110^circ$, how far is Mark from the intersection to the nearest meter?

b) Two roads intersect at an angle of $15^circ$ as shown in the figure above. Darryl is standing at point D which is 270 m from the intersection. If his friend Mark is at point B and is 300 m from the intersection, how far is Mark from Darryl? Round your answers to the nearest meter.

Use cosine law

$d^2=55.9^2+90^2-2(55.9)(90)cos(90-5.5)^circ$

$d=sqrt{55.9^2+90^2-2(55.9)(90)cos(90-5.5)^circ}$

$d=101.3$ m

$d=101.3$ m

a) The minimum information to use cosine law

(1) two sides and one included angle, such as: $a,b,$ and $angle C$

$c=sqrt{a^2+b^2-2abcos C}$

$b^2=a^2+c^2-2accos Bimplies B=cos^{-1}left(dfrac{a^2+c^2-b^2}{2ac}right)$

$A=180^circ-(B+C)$

(2) three sides, $a$, $b$, $c$

$C=cos^{-1}left(dfrac{a^2+b^2-c^2}{2ab}right)$

$B=cos^{-1}left(dfrac{a^2+c^2-b^2}{2ac}right)$

$$
A=180^circ-(B+C)
$$

b) The minimum information to use sine law

(1) one angle and two sides, such as, $A, a, b$

$dfrac{sin A}{a}=dfrac{sin B}{b}implies B=sin^{-1}left(dfrac{bsin A}{a}right)$

$C=180-(A+B)$

$dfrac{c}{sin C}=dfrac{a}{sin A}implies c=dfrac{asin C}{sin A}$

(2) two angles and one side, $A$, $B$, and $c$

$C=180^circ-(A+B)$

$dfrac{sin C}{c}=dfrac{sin A}{a}implies a=dfrac{csin C}{sin A}$

$dfrac{b}{sin B}=dfrac{a}{sin A}implies b=dfrac{asin B}{sin A}$

Let $x$ be the shortest side. The side opposite to the smallest angle must be $x$ and the side opposite to the largest angle must be $x+10$

Use sine law to find $x$

$dfrac{x}{sin 20^circ}=dfrac{x+10}{sin 120^circ}$

$x(sin 120^circ)=(x+10)(sin 20)$

$0.866x=(x+10)(0.342)$

$x=dfrac{10(0.342)}{0.866-0.342}=6.53$ cm
$x+10=6.53+10=16.53$ cm

We can find the remaining side $s$ by cosine law

$s^2=x^2+(x+10)^2-2(x)(x+10)cos 40^circ$

$s^2=6.53^2+16.53^2-2(6.53)(16.53)cos 40^circ$

$s=sqrt{6.53^2+16.53^2-2(6.53)(16.53)cos 40^circ}=12.27$

The perimeter is

$6.53+16.53+12.27=35.33=35$ cm

$35$ cm

begin{table}[]
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}