$sin 35^circ =dfrac{1}{h_1}$.

Step 2: Find the hypotenuse for the bigger triangle $h_2$

$sin 35^circ=dfrac{45}{h_2}$

Step 3: The length of the line is $h_1+h_2=dfrac{1}{sin 35^circ}+dfrac{45}{sin 35^circ}$

Step 2: Find the hypotenuse for the bigger triangle $h_2$

Step 3: Find $h_1+h_2$

For sine law:

Since it is isosceles, $angle A = angle C= dfrac{180-80}{2}=50^circ$

The side opposite to $80^circ$ is $4.0+2(0.5)=5.0$

$dfrac{x}{sin 50^circ}=dfrac{5.0}{sin 80^circ}implies x=dfrac{5.0sin 50^circ}{sin 80^circ}=3.89$ m

For cosine law

The side opposite to $80^circ$ is $4.0+2(0.5)=5.0$

$5^2=x^2+x^2-2(x)(x)cos 80^circ$

$25=2x^2-2x^2cos 80^circ$

$25=2x^2(1-cos 80^circ)$

$x=sqrt{dfrac{25}{2(1-cos80^circ)}}=3.89$ m

Both methods have the same answer.

He can calculate $x$ using sine.

$sin40^circ=dfrac{2.5}{x}implies x=dfrac{2.5}{sin 40^circ}=3.89$ m

a) Find $DF$ using tangent.

$tan 35^circ=dfrac{DF}{15}$

$DF=15tan 35^circ=10.5$

Find $x$ using sine.

$sin 45^circ=dfrac{DF}{x}$

$x=dfrac{10.5}{sin 45^circ}=15$ cm

Find $BC$ using cosine law.

Here, $DB=DC=$ 15 cm, so

$BC=sqrt{15^2+15^2-2(15)(15)cos 70^circ}=17.2$ cm

Find $x$ using sine. The opposite of $27^circ$ is BC.

$sin 27^circ=dfrac{BC}{x}$

$x=dfrac{BC}{sin 27^circ}=dfrac{17.2}{sin 27^circ}=38$ cm

We know that $angle ABC=180^circ$ so we can obtain $angle DBC$ as

$angle DBC=180^circ-(90+60)^circ=30^circ$

Now that we have $angle DBC=30^circ$, we can use sine law to find $BD$

$dfrac{BD}{sin 115^circ}=dfrac{10}{sin 30^circ}$

$BD=dfrac{10sin 115^circ}{sin 30^circ}=18.13$ cm

With $BD=18.13$, we can use sine law to find $BE$

$dfrac{BE}{sin 55^circ}=dfrac{BD}{sin 65^circ}$

$BE=dfrac{18.13sin 55^circ}{sin 65^circ}=16.39$ cm

With $BE=16.39$, we can use sine to find $x$

$sin 70^circ=dfrac{BE}{x}$

$x=dfrac{BE}{sin 70^circ}=dfrac{16.39}{sin 70^circ}=17.44=17$ cm

$BC=sqrt{18^2+14^2-2(18)(14)cos 95^circ}=23.75$ cm

We can find $AB$ and $AC$ using Pythagorean formula

$AB=sqrt{18^2+15^2}=23.43$

$AC=sqrt{14^2+15^2}=20.52$

Now, we can calculate $theta$ using cosine law

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

$theta=cos^{-1}left(dfrac{AB^2+AC^2-BC^2}{2(AB)(AC)}right)$

$theta=cos^{-1}left(dfrac{23.43^2+20.52^2-23.75^2}{2(23.43)(20.52)}right)=65^circ$

*Note that the answer at the back of the book is $93^circ$ which is incorrect.

b) $38$ cm

c) $17$ cm

d) $65^circ$

$angle D=180^circ-(50+66)^circ=64^circ$

Calculate $DC$ using sine law

$dfrac{DC}{sin 50^circ}=dfrac{175.0}{sin 64^circ}$

$DC=dfrac{175.0sin 50^circ}{sin 64^circ}=149.153$ m

Calculate $h$ using tangent.

$tan 74^circ=dfrac{h}{DC}$

$h=DCtan 74^circ=(149.153)(tan 74^circ)=520.2$ m

$angle CAD=180-(74+90)=16^circ$

$dfrac{h}{sin 74^circ}=dfrac{DC}{sin (16)^circ}$

$dfrac{h}{sin 74^circ}=dfrac{149.153}{sin 16^circ}$

$h=dfrac{149.153sin 74^circ}{sin 16^circ}=520.2$ m

b) Yes. Use sine law in $triangle ADC$

$tan 2^circ=dfrac{226}{AB}$

$AB=dfrac{226}{tan 2^circ}=6471.8$ m

Similarly,

$AV=dfrac{226}{tan 3^circ}=4312.3$ m

We can calculate $BV$ using cosine law

$BV=sqrt{(AB)^2+(AV)^2-2(AB)(AV)cos (angle BAV)}$

$BV=sqrt{6471.8^2+4312.3^2-2(6471.8)(4312.3)cos 80^circ}$

$BV=7127$ m

The distance between Beamville and Vineland is $7127$ m

We can obtain $angle BCA=250-180=70^circ$

We can solve for $BC$ and $AC$ using tangent.

$tan 34^circ=dfrac{166}{BC}$

$BC=dfrac{166}{tan 34^circ}=246.1$

Similarly,

$AC=dfrac{166}{tan 40^circ}=197.8$

Now, we can calculate $AB$ using cosine law.

$AB=sqrt{BC^2+AC^2-2(BC)(AC)cos (angle BCA)}$

$AB=sqrt{246.1^2+197.8^2-(246.1)(197.8)cos(70^circ)}$

$AB=258$ m

Therefore, the distance between boat A and boat B is $258$ m.

Let $B$ be the point at the bottom of the balcony.

$tan 20^circ=dfrac{h}{BR}implies BR=dfrac{h}{tan 20^circ}$

Similarly,

$tan 18^circ=dfrac{h}{BP}implies BP=dfrac{h}{tan 18^circ}$

Now, we can use cosine law

$100^2=BR^2+BP^2-2(BR)(BP)cos (angle RBP)$

Since Romeo is facing North and Paris is facing west, $angle RBP =90^circ$

Thus, the cosine law becomes the Pythagorean formula since $cos 90^circ=0$

$100^2=left(dfrac{h}{tan 20^circ}right)^2+left(dfrac{h}{tan 18^circ}right)^2$

$100^2=h^2 left(dfrac{1}{tan^2 20^circ}+dfrac{1}{tan^2 18^circ}right)$

$h=sqrt{dfrac{100^2}{dfrac{1}{tan^220^circ}+dfrac{1}{tan^218^circ}}}$

$h=24$ m

$dfrac{AS}{sin 35^circ}=dfrac{800}{sin 68^circ}$

$AS=dfrac{800sin 35^circ}{sin 68^circ}=494.9$ m

The distance from the helicopter to the sailboat is $HS$ which can be obtained using sine law

$dfrac{HS}{sin 73^circ}=dfrac{AS}{sin 40^circ}$

$HS=dfrac{494.9sin 73^circ}{sin 40^circ}$

$HS=736.3$ m

Therefore, the helicopter is $736$ m away from the sailboat.

$tan 51^circ=dfrac{8}{AB}implies AB=dfrac{8}{tan 51^circ}=6.47$ m

Since the girls are equidistant from the boat, $AT=6.47$ m

We can use cosine law to find $theta$

$6.47^2=6.47^2+8.8^2-2(6.47)(8.8)cos theta$

$$
theta=cos^{-1}left(dfrac{6.47^2+8.8^2-6.47^2}{2(6.47)(8.8)}right)=47.15^circ=47^circ
$$

Truck 1:
$sqrt{6.0^2+2.6^2}=6.54$
$sqrt{2.6^2+2.1^2}=3.34$
$sqrt{6^2+2.1^2}=6.35$

Truck 2:
$sqrt{4^2+2.1^2}=4.52$
$sqrt{2.1^2+2.5^2}=3.26$
$sqrt{4^2+2.5^2}=4.72$

Truck 3:
$sqrt{4.5^2+1.6^2}=4.78$
$sqrt{4.5^2+1.8^2}=4.85$
$sqrt{1.6^2+1.8^2}=2.41$

$triangle ABC$: $AB=sqrt{4.5^2+2^2-2(4.5)(2)cos 130^circ}=6$ m

$triangle GHI:$ all given

$triangle JKL:$ $LK=sqrt{4.7^2+4.0^2-2(4.7)(4)cos 42^circ}=3.2$ m

$triangle GHI$ cannot fit in Truck 3 because $2.41$ m is less than the minimum side so it must go to Truck 2.

$triangle JKL$ cannot fit in Truck 3 because its minimum diagonal is 2.41 and all sides of $triangle JKL$ are more than $2.41$ so it must go to Truck 2.

We can use sine law to find $AC$

$dfrac{AC}{sin 30^circ}=dfrac{80}{sin 65^circ}$

$AC=dfrac{80sin 30^circ}{sin 65^circ}=44.14$ m

Now, we can use tangent to find $h$

$tan 28^circ=dfrac{h}{AC}$

$h=ACtan28^circ=44.14tan 28^circ=23$ m

Therefore, the tree is 23 m tall.

We can not perform any cosine law or sine law with the given information.

b) The additional information needed to find $x$ are the following.

(i) height of the balloon $h$

(ii) either of the following: $angle ADC$, $angle DAC$, $angle ACD$

After two hours, the fast car is $100 text{km/h} times 2text{h} = 200$ km away from the intersection A.

Similarly, the slow car is $80;text{km/h}times 2text{ h}=160$ km away from A

We can find the distance between two cars $FS$ using cosine law

$FS=sqrt{200^2+160^2-2(200)(160)cos 34^circ}=112.0$ km

$FH=sqrt{FS^2+HS^2-2(FS)(HS)cos 20^circ}$

$FH=sqrt{112^2+100^2-2(112)(100)cos 20^circ}=38.66$ km

$FH=39$ km

The distance of the fast car from the helicopter is $39$ km

$sin 20^circ=dfrac{h}{100}$

$h=100sin 20^circ=34.2$ km

The altitude of the helicopter is $34$ km

We can use tangent to find $AB$ and $BC$

$tan 35^circ=dfrac{1515}{AB}implies AB=dfrac{1515}{tan 35^circ}=2163.6$ m

$tan 31^circ=dfrac{1515}{BC}implies BC=dfrac{1515}{tan 31^circ}=2521.4$ m

Now, we can use cosine law to find the length of the tunnel $x$

$x=sqrt{AB^2+BC^2-(AB)(BC)cos (angle ABC)}$

$x=sqrt{2163.6^2+2521.4-2(2163.6)(2521.4)cos 12^circ}=605$ m

Note that the answer at the back of your book is 524 m which is incorrect and has been corrected in the published errata.

Now, we can calculate the distance between two planes using cosine law.

$PQ=sqrt{120^2+180^2-2(120)(180)cos 55^circ}$

$PQ=148.4$ km

Therefore, the two airplanes are 148.4 km apart. Note that we do not need the altitudes of the plane to solve this problem.

Notice that $MB=CA$ and $BT=AD$

Using cosine,

$$
cos 55^circ=dfrac{CA}{44.5}implies CA=44.5cos 55^circ=25.52
$$

$cos 34^circ=dfrac{AD}{71.0}implies AD=71.0cos 34^circ=58.86$

$MB+BT=25.52+58.86=84.4$ m

b) The straight-line distance between both windows is $MT$

$MT=sqrt{(CA)^2+(AD)^2}=sqrt{25.52^2+58.85^2}=64.2$ m

b) 64.2 m