Trigonometry

Let the temple be $AB$ ($25$ m) and the house be $CD$ ($13$ m), with the man at $A$ on top of the temple, looking at the house roof $C$.

Draw $CE\perp AB$ with $E$ on $AB$. Then

$AE=25-13=12\text{ m},\qquad \angle ACE=\text{angle of depression}=45^\circ$

(alternate angles with the horizontal through $A$).

In right-angled $\triangle AEC$:

$\tan45^\circ=\frac{AE}{CE} \Rightarrow 1=\frac{12}{CE} \Rightarrow CE=12\text{ m}.$

The distance between the house and the temple is $BD=CE=$ 12 m.

Let the horizontal distance between the house and the temple be $d$ m.

The vertical drop from the temple roof to the house roof is $20 - 10 = 10$ m.

The angle of depression is $45^\circ$, so the angle of elevation from the house roof is also $45^\circ$:

$\tan 45^\circ = \frac{10}{d} \Rightarrow 1 = \frac{10}{d} \Rightarrow d = 10$

The distance between the house and the temple is 10 m.

Let the distance between the house and the temple be $d$ m.

The man is on the roof of the temple (30 m), so the roof of the house (10 m) is $30 - 10 = 20$ m below his eye level.

The angle of depression equals the angle of elevation from the house roof to the man, so in the right-angled triangle

$$\tan 60^\circ = \frac{20}{d} \Rightarrow \sqrt{3} = \frac{20}{d} \Rightarrow d = \frac{20}{\sqrt{3}} = \frac{20\sqrt{3}}{3} \approx 11.55 \text{ m}$$

Answer: the distance between the house and the temple is $\dfrac{20\sqrt3}{3} \approx 11.55$ m

Let the man be at the temple roof $A$ (height 22 m) and the house roof $B$ (height 10 m). Drop the horizontal from $A$ to the house wall at $C$.

Vertical difference $AC = 22 - 10 = 12$ m. The angle of depression is $30^\circ$, so $\angle CBA = 30^\circ$ (alternate angle), and $BC$ is the required distance $d$.

$\tan 30^\circ = \frac{AC}{BC} \Rightarrow \frac{1}{\sqrt{3}} = \frac{12}{d} \Rightarrow d = 12\sqrt{3} \approx 20.78 \text{ m}$

The distance between the house and the temple is about 20.78 m.

Let the temple be $AB = 15$ m and the tree $CD = 5$ m, the man being at the top $A$ of the temple. Drawing a horizontal from $A$ to meet the vertical through the tree-top, the vertical drop is

$15 - 5 = 10 \text{ m}.$

Let the distance between the temple and the tree be $d$ m. The angle of depression equals the angle of elevation from the tree-top to $A$, so

$\tan 60^\circ = \frac{10}{d} \Rightarrow \sqrt{3} = \frac{10}{d} \Rightarrow d = \frac{10}{\sqrt{3}} = \frac{10\sqrt{3}}{3} \approx 5.77 \text{ m}.$

The distance between the temple and the tree is $\dfrac{10\sqrt{3}}{3}$ m $\approx 5.77$ m.

Let the house be $AB = 30$ ft, the tree be $CD$, and $BD = 10\sqrt{3}$ ft the distance between them. Draw $AE$ parallel to $BD$ (horizontal) to meet $CD$ at $E$, so $AE = BD = 10\sqrt{3}$ ft and $\angle EAC = 30^\circ$ (angle of depression).

In right-angled $\triangle AEC$ (with $C$ the top of the tree):

$\tan 30^\circ = \frac{EC}{AE} \Rightarrow \frac{1}{\sqrt{3}} = \frac{EC}{10\sqrt{3}} \Rightarrow EC = 10 \text{ ft}$

The top of the tree is $10$ ft below the roof, so

$\text{Height of tree} = 30 - 10 = 20 \text{ ft}.$

Let the house be $AB = 60$ ft, the tree $CD = h$ ft, and $BD = 20\sqrt{3}$ ft. The vertical drop from the roof to the top of the tree is $(60 - h)$ ft.

$\tan 30^\circ = \frac{60 - h}{20\sqrt{3}} \Rightarrow \frac{1}{\sqrt{3}} = \frac{60 - h}{20\sqrt{3}}$ $60 - h = 20 \Rightarrow h = 40$

The height of the tree is $40$ ft.

Angle of depression: the angle formed below the horizontal line of sight when an observer at a higher level looks down at an object at a lower level.

(b) $AE = AB - EB = AB - CD = 21.5 - 1.5 = 20$ m (since $EB = CD$).

(c) $\angle FAC = 30^\circ$, so $\angle ACE = 30^\circ$ (alternate angles, $AF \parallel EC$).

$\tan 30^\circ = \frac{AE}{EC}\;\Rightarrow\;\frac{1}{\sqrt3}=\frac{20}{EC}\;\Rightarrow\; EC = 20\sqrt3\approx 34.64\text{ m}$

Distance between tower and house $BD = EC \approx 34.64$ m.

(d) If $AE = EC$, then $\tan\theta = \dfrac{AE}{EC}=1$, so $\theta = 45^\circ$.

Comparison: $45^\circ - 30^\circ = 15^\circ$. The angle of depression would be $15^\circ$ more than $30^\circ$.

Angle of elevation: the angle formed above the horizontal line of sight when an observer at a lower level looks up at an object at a higher level.

(b) In right triangle $AEC$: $\sin 60^\circ = \dfrac{AC}{AE}$

$AC = 140\sin60^\circ = 140\times\frac{\sqrt3}{2}=70\sqrt3\approx 121.24\text{ m}$

(c) Height of the kite from the ground $= AC + CB = AC + DE = 121.24 + 1.2 = 122.44$ m.

(d) For $\angle AEC = 45^\circ$:

$AC' = 140\sin45^\circ = 140\times\frac{1}{\sqrt2}=70\sqrt2\approx 98.99\text{ m}$

Fall in height $= 121.24 - 98.99 = 22.25$ m.

The kite would fly about 22.25 m lower than before.

Angle of elevation: the angle formed above the horizontal line of sight when an observer at a lower level looks up at an object at a higher level.

(b) Figure (AB is the standing part, BC the ground distance, AC the broken part resting on the ground):

        A
        |\
        | \   broken part AC
        |  \
        |   \
   AB   |    \
        |     \
        B------C
          9√3 m     ∠ACB = 30° at C

(c) Let $B$ be the foot, $A$ the break point and $C$ the point where the top touches the ground; $BC = 9\sqrt3$ m and $\angle ACB=30^\circ$. The broken part is $AC$.

$\cos30^\circ=\frac{BC}{AC}\;\Rightarrow\;\frac{\sqrt3}{2}=\frac{9\sqrt3}{AC}\;\Rightarrow\; AC = 18\text{ m}$

The broken part is 18 m long.

(d) If the remaining part $AB = 9\sqrt3$ m, then $AB = BC$, so triangle $ABC$ is an isosceles right triangle.

$\tan\theta=\frac{AB}{BC}=\frac{9\sqrt3}{9\sqrt3}=1\;\Rightarrow\;\theta=45^\circ$

The top makes an angle of $45^\circ$ with the ground, because the two perpendicular sides are equal.

Angle of elevation: the angle formed above the horizontal line of sight when an observer at a lower level looks up at an object at a higher level.

(b) Let $AC=20$ m be the broken part, $B$ the foot, $\angle ACB=60^\circ$. Distance between top and bottom after breaking $=BC$:

$\cos60^\circ=\frac{BC}{AC}\;\Rightarrow\;BC=20\times\frac12=10\text{ m}$

(c) Remaining standing part:

$AB = AC\sin60^\circ = 20\times\frac{\sqrt3}{2}=10\sqrt3\approx17.32\text{ m}$

Original height $= AB + AC = 10\sqrt3+20\approx 37.32$ m.

(d) If $BC = AB$, then in right triangle $ABC$:

$\tan\theta=\frac{AB}{BC}=1\;\Rightarrow\;\theta=45^\circ$

The top would make $45^\circ$ with the ground, since the two legs of the right triangle are equal (isosceles right triangle).

Angle of elevation: the angle formed above the horizontal line of sight when an observer at a lower level looks up at an object at a higher level.

(b) Circumference $=2\pi r = 88$:

$r=\frac{88\times7}{2\times22}=14\text{ ft}$

(c) The pole is at the centre, so the horizontal distance is $r=14$ ft and the pole height is 14 ft.

$\tan\theta=\frac{14}{14}=1\;\Rightarrow\;\theta=45^\circ$

The rope makes $45^\circ$ with the ground.

(d) Length of each rope (hypotenuse):

$\ell=\sqrt{14^2+14^2}=14\sqrt2\approx19.8\text{ ft}$

With 5 ft extra for the knot, each rope needs $19.8+5=24.8$ ft.

Since $24.8\text{ ft} < 45\text{ ft}$, a rope of 45 ft is sufficient for each rope. (If both ropes were to be cut from one 45 ft piece, they would need $2\times24.8=49.6$ ft, which is more than 45 ft, so it would not be enough.)

(a) It is an angle of elevation, because the man looks upward from a lower level to the top of the tower.

(b) Figure (E is the man's eye, AB the tower, EC horizontal):

            A
           /|
          / |
         /  |  h
        /   |
   E ──30°──── C
   |         |
   M ─────── B
    30√3 m

Man MD = 1.68 m, so tower = h + 1.68 m.

(c) Horizontal distance $=30\sqrt3$ m. Height of tower above the man's eye:

$\tan30^\circ=\frac{h}{30\sqrt3}\;\Rightarrow\;h=30\sqrt3\times\frac1{\sqrt3}=30\text{ m}$

Height of tower $=30+1.68=31.68$ m.

(d) For $45^\circ$, the distance $d$ satisfies $\tan45^\circ=\dfrac{30}{d}$, so $d=30$ m.

Distance to move $=30\sqrt3-30=30(\sqrt3-1)\approx21.96$ m.

The man should move about 21.96 m towards the tower.

Angle of elevation: the angle formed above the horizontal line of sight when an observer at a lower level looks up at an object at a higher level.

(b) Figure:

        A
        |\
 6√3 m  | \  wire
        |  \
        |___60°\
       B         C

AB is the pole, C is the point where the wire is fixed.

(c) In right triangle $ABC$:

$\sin60^\circ=\frac{AB}{AC}\;\Rightarrow\;\frac{\sqrt3}{2}=\frac{6\sqrt3}{AC}\;\Rightarrow\;AC=12\text{ m}$

The supporting wire is 12 m long.

(d) At $60^\circ$: $BC=\dfrac{6\sqrt3}{\tan60^\circ}=\dfrac{6\sqrt3}{\sqrt3}=6$ m.

At $45^\circ$: $BC'=\dfrac{6\sqrt3}{\tan45^\circ}=6\sqrt3\approx10.39$ m.

Difference $=6\sqrt3-6=6(\sqrt3-1)\approx4.39$ m.

The wire should be fixed about 4.39 m farther from the foot of the pole.

Angle of depression: the angle formed below the horizontal line of sight when an observer at a higher level looks down at an object at a lower level.

(b) Let tower $AB=75$ ft, house $CD$, distance $BD=25$ ft. Draw $CE\parallel DB$ with $E$ on the tower; $EC=25$ ft, $\angle ACE=45^\circ$.

$\tan45^\circ=\frac{AE}{EC}\;\Rightarrow\;AE=25\text{ ft}$

House height $CD=EB=75-25=50$ ft.

(c) The tower is higher than the house by $AE=75-50=25$ ft.

(d) From the roof C, the point P (on the tower, below the top A) is seen at elevation $30^\circ$: $\angle PCE=30^\circ$.

$\tan30^\circ=\frac{PE}{25}\;\Rightarrow\;PE=\frac{25}{\sqrt3}\approx14.43\text{ ft}$

Distance of P below the top $=AE-PE=25-14.43=10.57$ ft.

The point viewed is about 10.57 ft below the top of the tower.

(a) An angle of elevation is formed (looking upward from the roof $B$ to the tower top $D$).

(b) $DE = CD - CE = CD - AB = 15\sqrt3-10\sqrt3=5\sqrt3\approx8.66$ m.

(c) In right triangle $BED$:

$\tan30^\circ=\frac{DE}{BE}\;\Rightarrow\;\frac1{\sqrt3}=\frac{5\sqrt3}{BE}\;\Rightarrow\;BE=15\text{ m}$

Distance between house and tower $=AC=BE=15$ m.

(d) From the base $A$ of the house to the top $D$ of the tower: horizontal distance $AC=15$ m, height $CD=15\sqrt3$ m.

$\tan\theta=\frac{15\sqrt3}{15}=\sqrt3\;\Rightarrow\;\theta=60^\circ$

The angle of elevation is $60^\circ$.

(a) Diagram:

 A ─────────────────  horizontal line at roof level
 |  \ 30°
 |    \  line of sight
 E ───── C   (pole top, 9 ft)
 |        |  pole CD = 9 ft
 B ───────D   (ground)

House AB = 25 ft, pole CD = 9 ft, angle of depression of the pole top C from A is $30^\circ$.

(b) House is higher than the pole by $25-9=16$ ft.

(c) Draw a horizontal line from the pole top to the house at E; $AE=16$ ft and the angle at C is $30^\circ$ (alternate angle).

$\tan30^\circ=\frac{AE}{EC}\;\Rightarrow\;\frac1{\sqrt3}=\frac{16}{d}\;\Rightarrow\;d=16\sqrt3\approx27.71\text{ ft}$

Distance between house and pole $=16\sqrt3\approx27.71$ ft.

(d) Height of the new observation point $=25+32=57$ ft. Vertical drop to the pole top $=57-9=48$ ft; horizontal distance $=16\sqrt3$ ft.

$\tan\theta=\frac{48}{16\sqrt3}=\sqrt3\;\Rightarrow\;\theta=60^\circ$

The angle of depression would be $60^\circ$.

Angle of elevation: the angle formed above the horizontal line of sight when an observer at a lower level looks up at an object at a higher level.

(b) The angle of elevation of the top of the tower from the roof of the house equals the angle of depression (alternate angles): $30^\circ$.

(c) $EC=BD=10\sqrt3$ m and $\angle ACE=30^\circ$:

$\tan30^\circ=\frac{AE}{EC}\;\Rightarrow\;AE=10\sqrt3\times\frac1{\sqrt3}=10\text{ m}$

Height of house $CD=EB=AB-AE=30-10=20$ m.

(d) If $AE=EC$, triangle $AEC$ is right-angled and isosceles, so $\angle ACE=45^\circ$. Since $AH\parallel EC$, $\angle CAH=\angle ACE=45^\circ$ (alternate angles).

$\angle CAH=$ $45^\circ$.

(a) It is the angle of elevation (the observer looks up from a lower level to the building top).

(b) Figure (E is Himali's eye, AB the building):

            A
           /|
          / |
         /  |  h
        /   |
   E ─30°─── C
   |         |
   H ─────── B
      36 m

HE = 1.22 m.

(c) Height above her eye level:

$\tan30^\circ=\frac{h}{36}\;\Rightarrow\;h=\frac{36}{\sqrt3}=12\sqrt3\approx20.78\text{ m}$

Height of the building $=20.78+1.22=22$ m (approx.).

(d) For $60^\circ$: distance $d$ with $\tan60^\circ=\dfrac{12\sqrt3}{d}$, so $d=\dfrac{12\sqrt3}{\sqrt3}=12$ m.

Distance to walk $=36-12=24$ m.

She should walk 24 m nearer to the building, because a larger angle of elevation needs a shorter horizontal distance.

Angle of elevation: the angle formed above the horizontal line of sight when an observer at a lower level looks up at an object at a higher level.

(b) $AE=AB-EB=AB-CD=24.5-4.5=20$ m.

(c) In right triangle $AED$ with $\angle ADE=30^\circ$:

$\tan30^\circ=\frac{AE}{DE}\;\Rightarrow\;\frac1{\sqrt3}=\frac{20}{DE}\;\Rightarrow\;DE=20\sqrt3\approx34.64\text{ m}$

Distance between tower and house $BC=DE\approx34.64$ m.

(d) If $AE=ED$, then $\tan\theta=1$, so $\theta=45^\circ$.

Comparison: $45^\circ-30^\circ=15^\circ$. The angle of elevation would be $15^\circ$ more.

Angle of elevation: the angle formed above the horizontal line of sight when an observer at a lower level looks up at an object at a higher level.

(b) $PT=PQ-TQ=PQ-RS=18-1.5=16.5$ m.

(c) In right triangle $PTR$ with $\angle PRT=30^\circ$:

$\tan30^\circ=\frac{PT}{TR}\;\Rightarrow\;TR=16.5\sqrt3\approx28.58\text{ m}$

Distance between pole and man $SQ=TR\approx28.58$ m.

(d) If $PT=TR$, then $\tan\theta=1$, so $\theta=45^\circ$.

Compared with $30^\circ$, the angle of elevation would be $15^\circ$ more.

(a) The angle $\angle RPT$ is the angle of elevation of the tower top R from the roof P.

(b) $TR=RS-TS=RS-PQ=72-30=42$ m.

(c) In right triangle $PTR$:

$\tan60^\circ=\frac{TR}{PT}\;\Rightarrow\;\sqrt3=\frac{42}{PT}\;\Rightarrow\;PT=\frac{42}{\sqrt3}=14\sqrt3\approx24.25\text{ m}$

Distance between house and tower $QS=PT\approx24.25$ m.

(d) The point is 28 m below the top, so its height is $72-28=44$ m. Its height above the roof level $=44-30=14$ m, with horizontal distance $14\sqrt3$ m.

$\tan\theta=\frac{14}{14\sqrt3}=\frac1{\sqrt3}\;\Rightarrow\;\theta=30^\circ$

Yes, the angle of depression of the roof from that point is $30^\circ$, because $\tan\theta=\dfrac1{\sqrt3}$.

Angle of elevation: the angle formed above the horizontal line of sight when an observer at a lower level looks up at an object at a higher level.

(b) Tree height $=x$ m, broken at 6 m. The standing part is 6 m, so the broken part is

$x-6\text{ m}$

(c) In the right triangle formed (standing part 6 m, broken part as hypotenuse, angle $30^\circ$ at the ground):

$\sin30^\circ=\frac{6}{x-6}\;\Rightarrow\;\frac12=\frac6{x-6}\;\Rightarrow\;x-6=12\;\Rightarrow\;x=18\text{ m}$

The tree was 18 m high before breaking.

(d) Let the tree break at height $h$. The broken part is $18-h$ and makes $45^\circ$ with the ground:

$\sin45^\circ=\frac{h}{18-h}\;\Rightarrow\;18-h=h\sqrt2\;\Rightarrow\;h=\frac{18}{1+\sqrt2}=18(\sqrt2-1)\approx7.46\text{ m}$

The tree should break at about 7.46 m above the ground.

(a) The angle of elevation is the angle formed above the horizontal line of sight when an observer looks up at an object. Here, the angle of elevation of the roof is $60^\circ$.

(b) Figure (AB is the house, C the man's position):

        A
        |\
   h    | \
        |  \
        |___60°\
       B  10 m  C

(c) In right triangle $ABC$:

$\tan60^\circ=\frac{AB}{BC}\;\Rightarrow\;\sqrt3=\frac{h}{10}\;\Rightarrow\;h=10\sqrt3\approx17.32\text{ m}$

(d) After walking 20 m farther away, the distance from the foot is $10+20=30$ m.

$\tan\theta=\frac{10\sqrt3}{30}=\frac1{\sqrt3}\;\Rightarrow\;\theta=30^\circ$

The angle of elevation will be $30^\circ$.

Angle of elevation: the angle formed above the horizontal line of sight when an observer at a lower level looks up at an object at a higher level.

(b) Tower is higher than the man by $41.3-1.3=40$ m.

(c) With $\theta=45^\circ$:

$\tan45^\circ=\frac{40}{d}\;\Rightarrow\;d=40\text{ m}$

(d) For $30^\circ$: $\tan30^\circ=\dfrac{40}{d'}\Rightarrow d'=40\sqrt3\approx69.28$ m.

Distance to walk backward $=40\sqrt3-40=40(\sqrt3-1)\approx29.28$ m.

The man should walk back about 29.28 m.

Angle of elevation: the angle formed above the horizontal line of sight when an observer at a lower level looks up at an object at a higher level.

(b) Figure (P is the pole top, O the centre of the pond, M the man at the edge, ME is his height):

          P
         /|
        / |
       /  |  h
      /45°|
   E ─────T      T is at eye level above O
   |      |
   M ──── O   (radius 21 m)

(c) Horizontal distance $=$ radius $=21$ m. Height of the pole above the man's eye:

$\tan45^\circ=\frac{h}{21}\;\Rightarrow\;h=21\text{ m}$

Height of pole above water level $=21+1.4=22.4$ m.

(d) For $30^\circ$:

$h'=21\tan30^\circ=\frac{21}{\sqrt3}=7\sqrt3\approx12.12\text{ m}$

New height above water $=12.12+1.4=13.52$ m.

Reduction $=22.4-13.52=8.88$ m (that is, $21-7\sqrt3$).

The pole must be reduced by about 8.88 m, since a smaller angle of elevation needs a smaller pole at the same distance.

(a) Figure (P pole top, O centre, B bank, depth 1.5 m below the water):

            P  (pole top)
           /|
          / |
         /  |  50 m above water
        /   |
    θ  /    |
   B ─────── O   (centre of pond)
       50 m (radius)

Radius $=50$ m.

(b) Height of the pole above water level $=51.5-1.5=50$ m.

(c) Radius $=\dfrac{100}{2}=50$ m.

$\tan\theta=\frac{50}{50}=1\;\Rightarrow\;\theta=45^\circ$

(d) If $\theta=30^\circ$ with the pole height above water $50$ m:

$\tan30^\circ=\frac{50}{r}\;\Rightarrow\;r=50\sqrt3\approx86.60\text{ m}$

The radius of the pond would be about 86.6 m.

(a) Example of the angle of depression: the angle formed when a person standing on top of a tower looks down at a boat on a lake (or a bird looking down at the ground).

(b) Figure (K is the kite, string KM $=105\sqrt3$ m, M the man's hand):

            K
           /|
  string  / |
         /  | h
        /   |
   M ─60°── C
   |        |
   G ────── B

GM $=1.75$ m.

(c) Height of the kite above the man's hand:

$\sin60^\circ=\frac{h}{105\sqrt3}\;\Rightarrow\;h=105\sqrt3\times\frac{\sqrt3}{2}=157.5\text{ m}$

Height of the kite from the ground $=157.5+1.75=159.25$ m.

(d) At $45^\circ$:

$h'=105\sqrt3\sin45^\circ=\frac{105\sqrt3}{\sqrt2}=\frac{105\sqrt6}{2}\approx128.60\text{ m}$

Height from the ground $=128.60+1.75=130.35$ m (approx.).

Angle of elevation: the angle formed above the horizontal line of sight when an observer at a lower level looks up at an object at a higher level.

(b) Figure (E is the boy's eye, AB the tower):

            A
           /|
          / |
         /  |  h
        /   |
   E ─60°── C
   |        |
   G ────── B
      d

GE $=1.2$ m.

(c) Height of the tower above the boy's eye $h=53.2-1.2=52$ m.

$\tan60^\circ=\frac{52}{d}\;\Rightarrow\;d=\frac{52}{\sqrt3}\approx30.02\text{ m}$

(d) At $45^\circ$: $d'=\dfrac{52}{\tan45^\circ}=52$ m.

Difference $=52-30.02=21.98$ m.

At $45^\circ$ the boy is about 21.98 m farther from the tower than at $60^\circ$ (a smaller angle needs a larger distance).

(a) The angle formed is the angle of elevation (from the roof C of the house to the top A of the tower).

(b) $EC=BD=15$ m and $\angle ACE=60^\circ$ (alternate angle to the angle of depression).

$\tan60^\circ=\frac{AE}{EC}\;\Rightarrow\;AE=15\sqrt3\approx25.98\text{ m}$

(c) Height of the house $CD=EB=AB-AE=45-15\sqrt3\approx19.02$ m.

(d) For a depression of $45^\circ$ the vertical drop from the observer to the roof level equals the horizontal distance:

$AE'=15\tan45^\circ=15\text{ m}$

Distance to come down $=AE-AE'=15\sqrt3-15=15(\sqrt3-1)\approx10.98$ m.

The observer must come down about 10.98 m from the top.

Let the tower be $AB = 75$ ft, the house be $CD$, and $BD = 25\sqrt{3}$ ft the distance between them. Draw $AE$ horizontally to meet $CD$ at $E$, so $AE = BD = 25\sqrt{3}$ ft and $\angle EAC = 30^\circ$ (angle of depression).

In right-angled $\triangle AEC$ (with $C$ the roof of the house):

$\tan 30^\circ = \frac{EC}{AE} \Rightarrow \frac{1}{\sqrt{3}} = \frac{EC}{25\sqrt{3}} \Rightarrow EC = 25 \text{ ft}$

The roof is $25$ ft below the top of the tower, so

$\text{Height of house} = 75 - 25 = 50 \text{ ft}.$

(a) The angle of elevation is the angle formed above the horizontal line of sight when an observer looks up at an object that is higher than the observer's eye level.

(b)

          T
          |\
          | \
   12 m   |  \
          |   \
          |    \
        B +--x°-- O
           12 m

$BT$ is the tower (12 m), $O$ is the observer 12 m from the foot $B$, and $x^\circ$ is the angle of elevation at $O$.

(c) $\tan x^\circ = \dfrac{BT}{OB} = \dfrac{12}{12} = 1 \Rightarrow x = 45^\circ$.

(d) For an angle of elevation of $30^\circ$ from a new point at distance $d$ from the foot:

$\tan 30^\circ = \frac{12}{d} \Rightarrow d = 12\sqrt{3} \approx 20.78 \text{ m}$

Distance to move farther $= 12\sqrt{3} - 12 = 12(\sqrt{3} - 1) \approx 8.78$ m.

Let the height of the man be $h$ m. The line of sight rises through $(36.1 - h)$ m over a horizontal distance of 20 m.

$\tan 60^\circ=\frac{36.1-h}{20}$

$\sqrt3=\frac{36.1-h}{20} \Rightarrow 36.1-h=20\sqrt3=20\times1.732=34.64$

$h=36.1-34.64=1.46\ \text{m}$

The height of the man is 1.46 m.

(a) The angle of elevation is the angle formed above the horizontal line of sight when an observer looks up at an object higher than the observer's eye level.

(b) $AB-CD=20.5-1.5=19$ m. The man is 19 m shorter than the pole.

(c) Since $CE\parallel DB$ and $CD\parallel EB$, $EB=CD=1.5$ m, so $AE=20.5-1.5=19$ m. In right-angled $\triangle AEC$:

$\tan45^\circ=\frac{AE}{CE}\Rightarrow 1=\frac{19}{CE}\Rightarrow CE=19\text{ m}.$

The man is 19 m from the pole.

(d) For an angle of elevation of $30^\circ$, let the new distance be $d$:

$\tan30^\circ=\frac{19}{d}\Rightarrow d=19\sqrt3\approx 32.91\text{ m}.$

Since $32.91>19$, he must move backward (away from the pole) by $32.91-19\approx 13.91$ m.

(a) The angle $\angle PAR$ is the angle of elevation.

(b) $PR = PQ - RQ = PQ - AB = 11.6 - 1.6 = 10$ m.

(c) In right-angled $\triangle PAR$:

$\tan 60^\circ = \frac{PR}{AR} \Rightarrow \sqrt{3} = \frac{10}{AR} \Rightarrow AR = \frac{10}{\sqrt{3}} \approx 5.77 \text{ m}$

So the man is about $5.77$ m from the house.

(d) $PR = AR$ when the angle of elevation is $45^\circ$, because then $\tan\angle PAR = \dfrac{PR}{AR} = \tan 45^\circ = 1$.

Let the height of the house be $h$ m. The horizontal distance is $100\sqrt3$ m and the angle of elevation is $45^\circ$.

$\tan45^\circ=\frac{h}{100\sqrt3}\Rightarrow 1=\frac{h}{100\sqrt3}$

$h=100\sqrt3\approx\textbf{173.2 m}$

The pole stands at the centre of the pond, on its bottom, which is 3 m below the water surface.

Height of the pole above the water $= 48-3 = 45$ m.

Radius of the pond $= \frac{90}{2} = 45$ m, so the point on the circumference is 45 m horizontally from the pole.

Let $\theta$ be the angle of elevation of the top of the pole from that point (on the water level).

$\tan\theta = \frac{\text{height}}{\text{distance}} = \frac{45}{45} = 1 \implies \theta = 45^\circ$

The angle of elevation is 45°.

The eye of the person is 1.5 m above the ground, so the height of the tree above the eye level is

$41.5-1.5=40\text{ m}.$

Let the distance between the person and the tree be $d$ m. In the right triangle formed,

$\tan45^\circ=\frac{40}{d}\Rightarrow1=\frac{40}{d}\Rightarrow d=40\text{ m}.$

The distance between the tree and the person is 40 m.

(a) $AB=AF-BF=21-2=\mathbf{19}$ m.

(b) The ratio $\tan\theta=\dfrac{\text{perpendicular}}{\text{base}}=\dfrac{AB}{BC}$ is used, because $AB$ (perpendicular) is known and the distance $BC$ (base) is needed.

(c) $\tan30^\circ=\dfrac{AB}{BC}\Rightarrow\dfrac1{\sqrt3}=\dfrac{19}{BC}\Rightarrow BC=19\sqrt3\approx\mathbf{32.9}$ m.

(d) $\sin30^\circ=\dfrac{AB}{AC}\Rightarrow\dfrac12=\dfrac{19}{AC}\Rightarrow AC=\mathbf{38}$ m.

(a) The angle of depression is the angle between the horizontal line through the observer's eye and the line of sight, when the object observed is below the horizontal level.

(b) Figure: draw a vertical tower $AB = 50$ m and a vertical house $CD = 34.64$ m on level ground. From the top $A$ draw a horizontal line $AX$ and the line of sight $AC$ to the top of the house, with $\angle XAC = 60^\circ$. Draw $CE \perp AB$ with $E$ on the tower; then $AE = 50 - 34.64$ and $CE$ is the horizontal distance.

(c) Height of the tower above the house top: $AE = 50 - 34.64 = 15.36$ m.

In right-angled $\triangle AEC$, $\angle ACE = 60^\circ$ (alternate angles).

$\tan 60^\circ = \dfrac{AE}{CE} \Rightarrow CE = \dfrac{15.36}{\sqrt{3}} = 8.87 \text{ m}$

The horizontal distance is about 8.87 m.

(d) For an angle of depression of $45^\circ$ the vertical drop from the observer to the house top equals the horizontal distance, since $\tan 45^\circ = 1$. So the observer must be 8.87 m above the house top.

Distance to come down $= 15.36 - 8.87 = 6.49$ m.