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⬅ Back to Trigonometry
📐 Key Concepts: Oblique Triangles

Law of Sines:

$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$

Law of Cosines:

$$c^2 = a^2 + b^2 - 2ab\cos C$$

Use Law of Sines when you know AAS or ASA. Use Law of Cosines when you know SAS or SSS.

Problem: Height of Tower (Right and Oblique Triangles)

Towers A and B stand on a horizontal plane, tower B being 150 m high. The angle of elevation of the top of tower A as seen from point C (in the same vertical plane) is 50°. The angle of depression of C viewed from the top of tower B is 27.5°. The angle subtended at the top of tower B by the top of tower A and C is 50°. Find the height of tower A.

Tower Problem – Diagram Tower Problem – Diagram Tower Problem – Diagram
Solution Solution Solution Solution

Step 1 — Find BC (horizontal distance from tower B to point C) using the depression angle from top of B:

$$BC = \frac{BT_B}{\tan27.5°} = \frac{150}{0.5206} \approx 288.1 \text{ m}$$

Length $CT_B = \sqrt{BC^2 + 150^2} = \sqrt{288.1^2 + 150^2} \approx 325.0$ m.

Step 2 — Set up triangle $C$–$T_B$–$T_A$. From $C$: elevation angle to $T_B = 27.5°$; elevation angle to $T_A = 50°$. Since the towers are on the same side of $C$, angle at $C$ in the triangle $= 50° - 27.5° = 22.5°$. The angle at $T_B$ between lines to $C$ and $T_A = 50°$ (given). Therefore angle at $T_A = 180° - 50° - 22.5° = 107.5°$.

Step 3 — Law of Sines:

$$\frac{CT_A}{\sin50°} = \frac{CT_B}{\sin107.5°} \Rightarrow CT_A = \frac{325.0 \times \sin50°}{\sin107.5°} = \frac{325.0 \times 0.7660}{0.9537} \approx 261.0 \text{ m}$$

Step 4 — Height of tower A:

$$h_A = CT_A \times \sin50° = 261.0 \times 0.7660 \approx \boxed{200 \text{ m}}$$

Problem: Elevation of Mountain Peak B

Mountain peak A is 1,040 m above sea level. From peak A, the angle of elevation of peak B is 10°. A pilot flying directly over peak A at 2,820 m above sea level measures the angle of depression to peak B as 46°12'. Determine the elevation of peak B.

Mountain Peak – Diagram Mountain Peak – Diagram Mountain Peak – Diagram
Solution Solution Solution Solution

Let $d$ = horizontal distance between peak A and peak B, and $h_B$ = elevation of peak B.

From peak A (at 1040 m), angle of elevation to B = 10°:

$$h_B - 1040 = d\tan10° \quad \cdots (1)$$

From the pilot (at 2820 m directly above A), angle of depression to B = 46°12' = 46.2°:

$$2820 - h_B = d\tan46.2° \quad \cdots (2)$$

Add (1) and (2):

$$1780 = d(\tan10° + \tan46.2°) = d(0.1763 + 1.0538) = 1.2301\,d$$
$$d = \frac{1780}{1.2301} \approx 1447 \text{ m}$$

Elevation of peak B:

$$h_B = 1040 + 1447\tan10° = 1040 + 1447(0.1763) \approx 1040 + 255 = \boxed{1295 \text{ m above sea level}}$$

Problem: Distance Across a Building (Law of Cosines)

Two survey points A and B are on opposite sides of a building and cannot be seen from each other. A third point C is chosen such that CA = 100 m, CB = 120 m, and the angle ACB = 55°. Find the distance AB.

Apply the Law of Cosines with $a = CA = 100$ m, $b = CB = 120$ m, included angle $C = 55°$, solving for side $c = AB$:

$$c^2 = a^2 + b^2 - 2ab\cos C = 100^2 + 120^2 - 2(100)(120)\cos 55°$$
$$c^2 = 10000 + 14400 - 24000(0.5736) = 24400 - 13766 = 10634$$
$$c = \sqrt{10634} \approx \boxed{103.1 \text{ m}}$$

Problem: Finding a Missing Angle (Law of Sines)

In triangle ABC, side $a = 15$ m, side $b = 20$ m, and angle $A = 32°$. Find angle B, angle C, and side $c$.

Step 1 — Find angle B using Law of Sines:

$$\frac{\sin B}{b} = \frac{\sin A}{a} \Rightarrow \sin B = \frac{20\sin 32°}{15} = \frac{20(0.5299)}{15} = 0.7066$$
$$B = \arcsin(0.7066) \approx 44.9°$$

Step 2 — Find angle C:

$$C = 180° - A - B = 180° - 32° - 44.9° = 103.1°$$

Step 3 — Find side c:

$$c = \frac{a\sin C}{\sin A} = \frac{15\sin 103.1°}{sin 32°} = \frac{15(0.9744)}{0.5299} \approx \boxed{27.6 \text{ m}}$$

Problem: Law of Cosines Side

Two sides of a triangle are 20 m and 35 m with included angle 52°. Find the opposite side.

$$c^2=20^2+35^2-2(20)(35)\cos52^\circ$$
$$c=27.6\text{ m}$$

Answer: The opposite side is about 27.6 m.

Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: q281

MSTE - Geometry and Trigonometry / Theorems on Triangles and Circles / Engr. Janclyde Espinosa (Clidez)

The angle of elevation of the top of a building from point A on a ground is 24º. From point B, which is 44.5 meters closer, the angle of elevation is 34.5º. What is the height of the building?

Answer:

  1. 56.3m
  2. 52.4m
  3. 38.9m
  4. 62.8m
Let $h$ be the building height and let the farther distance be $x$. Then:
$\tan24^\circ=\frac{h}{x}$
$\tan34.5^\circ=\frac{h}{x-44.5}$
So $x=\frac{h}{\tan24^\circ}$ and $x-44.5=\frac{h}{\tan34.5^\circ}$. Subtract:
$44.5=h(\cot24^\circ-\cot34.5^\circ)$
$\boxed{h=56.3\text{ m}}$

Question Bank: q479

MSTE - Geometry and Trigonometry / Trigonometry / Engr. Janclyde Espinosa (Clidez)

The angle of elevation of the top of a building from point A on the ground is 24.2˚. From point B, which is 44.5 ft. closer, the angle of elevation is 38.1˚. What is the height of the building?

Answer:

  1. 46.9
  2. 47.2
  3. 46.1
  4. 47.8
Let $h$ be the building height. The farther distance is $h/\tan24.2^\circ$ and the closer distance is $h/\tan38.1^\circ$. Their difference is 44.5 ft:
$44.5=h(\cot24.2^\circ-\cot38.1^\circ)$
$\boxed{h=46.9}$

Question Bank: t348

MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3

In measuring the distance AB, the tape was not accurately levelled and that one end is 0.85 m lower than the other end. If the measured distance is 28.63 m, what is the correct distance?

  1. 28.617 m
  2. 28.643 m
  3. 29.127 m
  4. 28.357 m
The measured length is the slope distance ($28.63$ m) with a vertical difference of $0.85$ m between ends.
The correct horizontal distance $=\sqrt{28.63^2-0.85^2}=\sqrt{819.677-0.7225}=\sqrt{818.954}=28.617$ m.
$\boxed{28.617\text{ m}}$

Question Bank: t367

MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3

A right triangle ABC has its right angle at B. If AC = $x + y$ and AB = $x - y$, what is the value of side BC?

  1. $2xy$
  2. $4\sqrt{xy}$
  3. $4xy$
  4. $2\sqrt{xy}$
With the right angle at $B$, $AC$ is the hypotenuse. By the Pythagorean theorem, $BC^2=AC^2-AB^2$.
$BC^2=(x+y)^2-(x-y)^2=4xy$.
$BC=2\sqrt{xy}$.
$\boxed{2\sqrt{xy}}$

Question Bank: t370

MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3

A fisherman at sea observed that the angle of elevation of a cliff is $12^\circ$. If the height of the cliff above sea level is 30 m, how far is he from the shore?

  1. 141.14 m
  2. 163.52 m
  3. 124.87 m
  4. 104.58 m
The cliff height and horizontal distance form a right triangle with $\tan12^\circ=\dfrac{30}{d}$.
$d=\dfrac{30}{\tan12^\circ}=\dfrac{30}{0.2126}=141.14$ m.
$\boxed{141.14\text{ m}}$

Question Bank: t371

MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3

In order to measure the height of a flagpole, an instrument is setup at A and the angle of elevation to the top of the flagpole was measured as $27.5^\circ$. The instrument is then setup at B, which is 56.3 m closer to the flagpole and on the same vertical plane as point A and the flagpole, and measured the angle of elevation as $62.3^\circ$. What is the height of the flagpole?

  1. 45.89 m
  2. 50.91 m
  3. 42.28 m
  4. 40.33 m
Let $x$ be the horizontal distance from $B$ to the pole. Then $h=x\tan62.3^\circ=(x+56.3)\tan27.5^\circ$.
$x(\tan62.3^\circ-\tan27.5^\circ)=56.3\tan27.5^\circ\Rightarrow x(1.9047-0.5206)=29.31$, so $x=21.17$ m.
$h=21.17\tan62.3^\circ=40.33$ m.
$\boxed{40.33\text{ m}}$

Question Bank: t373

MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3

A flagpole stands on the edge of the top of a building. At a point 200 m from the building the angles of elevation of the top and bottom of the pole are $32^\circ$ and $30^\circ$ respectively. Calculate the height of the flagpole.

  1. 8.60 m
  2. 12.30 m
  3. 9.50 m
  4. 10.20 m
From 200 m, the height to the bottom of the pole is $200\tan30^\circ=115.47$ m and to the top is $200\tan32^\circ=124.97$ m.
Flagpole height $=124.97-115.47=9.50$ m.
$\boxed{9.50\text{ m}}$

Question Bank: t374

MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3

An observer finds the angle of elevation of a building from X on a level ground as $27^\circ$. After moving 120 m closer to the building, the angle of elevation was $42^\circ$. Find the height of the building.

  1. 152.78 m
  2. 148.56 m
  3. 167.34 m
  4. 140.85 m
Let $d$ be the closer distance. Then $h=d\tan42^\circ=(d+120)\tan27^\circ$.
$d(\tan42^\circ-\tan27^\circ)=120\tan27^\circ\Rightarrow d(0.9004-0.5095)=61.14$, so $d=156.4$ m.
$h=156.4\tan42^\circ=140.85$ m.
$\boxed{140.85\text{ m}}$

Question Bank: t375

MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3

From a building across, the angle of depression of the base of the front edifice is $23.5^\circ$ and the angle of elevation of its top is $52.78^\circ$. The height observation is 14 m.

What is the height of the edifice?

  1. 56.39 m
  2. 68.74 m
  3. 76.58 m
  4. 42.23 m

How far is the building from the said edifice?

  1. 38.7 m
  2. 24.8 m
  3. 45.6 m
  4. 32.2 m

What is the angle of elevation of the top of the edifice from the foot of the building?

  1. $65.32^\circ$
  2. $78.96^\circ$
  3. $60.27^\circ$
  4. $54.75^\circ$

Part 1.

The observation point is 14 m high. The horizontal distance is $D=\dfrac{14}{\tan23.5^\circ}=32.2$ m.
The top rises above the observer by $D\tan52.78^\circ=32.2(1.317)=42.4$ m.
Total edifice height $=14+42.4=56.39$ m.
$\boxed{56.39\text{ m}}$

Part 2.

The angle of depression of $23.5^\circ$ to the base, from the 14 m observation height, gives $\tan23.5^\circ=\dfrac{14}{D}$.
$D=\dfrac{14}{\tan23.5^\circ}=32.2$ m.
$\boxed{32.2\text{ m}}$

Part 3.

The foot of the building is at the base level, horizontal distance $D=32.2$ m from the edifice, whose full height is $56.39$ m.
$\tan\theta=\dfrac{56.39}{32.2}=1.751\Rightarrow\theta=\tan^{-1}(1.751)=60.27^\circ$.
$\boxed{60.27^\circ}$

Question Bank: t378

MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3

An inaccessible cliff was sighted from point A. The angle of elevation was observed to be $18.45^\circ$. The surveyor went closer to the edge of the cliff at point B. Point B is 425 m nearer but 25.6 m lower than A. The angle of elevation was observed to be $36.3^\circ$ from B. If the elevation of A is 602 m, what is the elevation of the top of the cliff?

  1. 885.325 m
  2. 883.075 m
  3. 881.455 m
  4. 880.325 m
Let $x$ be the horizontal distance from $A$ to the cliff and $h_A$ the height of the top above $A$. Then $h_A=x\tan18.45^\circ$.
$B$ is 425 m nearer and 25.6 m lower, so $h_A+25.6=(x-425)\tan36.3^\circ$.
$0.3337x+25.6=0.7346(x-425)\Rightarrow 0.4010x=337.8$, so $x=842.5$ m and $h_A=281.1$ m.
Elevation of top $=602+281.1=883.075$ m.
$\boxed{883.075\text{ m}}$

Question Bank: t379

MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3

Two stations A and B were setup to determine the height of a mountain. The angles of elevation of the top of the mountain measured from stations A and B were $26.5^\circ$ and $31.25^\circ$, respectively. Station A is 75 m above station B. Station B is 450 m closer to the mountain. If the elevation of station A is 563.54 m, what is the elevation of the top of the mountain in meters?

  1. 2166.93
  2. 2214.65
  3. 2114.78
  4. 2365.45
Let $x$ be the horizontal distance from $B$ to the mountain and $h_B$ the top's height above $B$: $h_B=x\tan31.25^\circ$.
$A$ is 450 m farther and 75 m higher: $h_B-75=(x+450)\tan26.5^\circ$.
$0.6069x-75=0.4986(x+450)\Rightarrow 0.1082x=299.4$, so $x=2766$ m and $h_B=1679$ m.
Station $B$ elevation $=563.54-75=488.54$; top $=488.54+1679\approx2166.93$ m.
$\boxed{2166.93}$

Question Bank: t380

MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3

Points A and B on opposite sides of a building are 124 m apart. The angles of depression from the top of the building to A and B are $54.3^\circ$ and $68.5^\circ$, respectively. Determine the following:

The height of the building.

  1. 128.57 m
  2. 98.75 m
  3. 115.65 m
  4. 111.46 m

Distance of point A from the building.

  1. 43.91 m
  2. 80.09 m
  3. 41.55 m
  4. 82.45 m

Distance of point B from the building.

  1. 80.09 m
  2. 82.45 m
  3. 43.91 m
  4. 41.55 m

Part 1.

A and B are on opposite sides, so their horizontal distances add to 124: $\dfrac{h}{\tan54.3^\circ}+\dfrac{h}{\tan68.5^\circ}=124$.
$h(0.7187+0.3939)=124\Rightarrow h(1.1126)=124$.
$h=111.46$ m.
$\boxed{111.46\text{ m}}$

Part 2.

$A$ has the smaller depression angle, so it is farther: $d_A=\dfrac{h}{\tan54.3^\circ}=\dfrac{111.46}{1.3915}=80.09$ m.
$\boxed{80.09\text{ m}}$

Part 3.

$B$ has the larger depression angle, so it is closer: $d_B=\dfrac{h}{\tan68.5^\circ}=\dfrac{111.46}{2.5387}=43.91$ m.
(Check: $80.09+43.91=124$ m.)
$\boxed{43.91\text{ m}}$

Question Bank: t383

MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3

From a window 7.4 m above the ground, the angle of elevation of the top of a nearby edifice is $54.25^\circ$ and the angle of depression of its base is $21.42^\circ$. What is the height of the edifice?

  1. 36.5 m
  2. 33.6 m
  3. 34.7 m
  4. 32.1 m
From the window (7.4 m high), the depression to the base gives the horizontal distance: $D=\dfrac{7.4}{\tan21.42^\circ}=18.86$ m.
The part of the edifice above the window is $D\tan54.25^\circ=18.86(1.389)=26.2$ m.
Total height $=7.4+26.2=33.6$ m.
$\boxed{33.6\text{ m}}$

Question Bank: t385

MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3

The side of a hill makes an angle of $12^\circ$ with the horizontal. A wire is to be run from the top of a 60-m tower on the top of the hill to a stake located 36 m down the hillside from the base of the tower. What length of wire is needed in meters?

  1. 79.65
  2. 76.12
  3. 72.54
  4. 82.45
Place the tower base at the origin; the tower top is at $(0,60)$. The stake is 36 m down the $12^\circ$ slope at $(-36\cos12^\circ,-36\sin12^\circ)=(-35.21,-7.48)$.
Wire length $=\sqrt{35.21^2+(60+7.48)^2}=\sqrt{1239.7+4553.5}=\sqrt{5793}=76.12$ m.
$\boxed{76.12}$

Question Bank: t387

MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3

From a window "O" of a tall building 92 meters above a level ground, the angle of depression of the top of a nearby tower is $38^\circ$. From the base of the building, the angle of elevation of the top of the tower is $22^\circ$. Determine the following:

The height of the tower in meters.

  1. 28.63
  2. 31.36
  3. 25.47
  4. 35.65

The distance from the tower to the building in m.

  1. 68.54
  2. 75.85
  3. 77.62
  4. 96.32

The angle subtended by the tower from point O in degrees.

  1. 12.54
  2. 13.86
  3. 11.85
  4. 10.54

Part 1.

Let $h$ be the tower height and $D$ the building–tower distance. From the base: $h=D\tan22^\circ$. From window O (92 m): $92-h=D\tan38^\circ$.
Adding: $92=D(\tan22^\circ+\tan38^\circ)=D(1.1853)$, so $D=77.62$ m.
$h=77.62\tan22^\circ=31.36$ m.
$\boxed{31.36}$

Part 2.

Using $92=D(\tan22^\circ+\tan38^\circ)$ from the elevation at the base and depression at O:
$D=\dfrac{92}{\tan22^\circ+\tan38^\circ}=\dfrac{92}{0.4040+0.7813}=77.62$ m.
$\boxed{77.62}$

Part 3.

From O, the depression to the tower top is $38^\circ$. The depression to the tower base (at ground, 92 m below O) is $\tan^{-1}\dfrac{92}{77.62}=49.85^\circ$.
Angle subtended $=49.85^\circ-38^\circ=11.85^\circ$.
$\boxed{11.85}$

Question Bank: t396

MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3

The side of a mountain slopes upward at an angle of $21^\circ 14'$. At a point on the mountain side a mine tunnel is constructed at an angle of $15^\circ 27'$ downward from the horizontal. Find the vertical distance to the surface of the mountain from a point 250 m down the tunnel.

  1. 142.89 m
  2. 187.54 m
  3. 160.23 m
  4. 198.63 m
The tunnel point lies a horizontal distance $x=250\cos15^\circ27'=241.0$ m into the mountain and $250\sin15^\circ27'=66.6$ m below the entrance.
Directly above that point, the mountain surface is $x\tan21^\circ14'=241.0(0.3886)=93.7$ m above the entrance level.
Vertical distance to surface $=93.7+66.6=160.23$ m.
$\boxed{160.23\text{ m}}$

Question Bank: t397

MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3

The distances of a point C from two points A and B, which cannot be measured directly, are required. The line CA is continued through A for a distance of 175 m to D, the line CB is continued through B for 225 m to E, and the distances AB = 300 m, DB = 326 m, and DE = 488 m are measured.

What is the distance AC in meters?

  1. 129.3
  2. 164.3
  3. 145.4
  4. 132.8

What is the distance BC in meters?

  1. 412.1
  2. 350.8
  3. 376.3
  4. 389.9

What is the measure of angle ACB in degrees?

  1. $57.9^\circ$
  2. $68.3^\circ$
  3. $51.1^\circ$
  4. $76.3^\circ$

Part 1.

Triangle ABD ($AB=300$, $AD=175$, $BD=326$): $\angle BAD=\cos^{-1}\dfrac{300^2+175^2-326^2}{2(300)(175)}=82.15^\circ$, so $\angle BAC=180^\circ-82.15^\circ=97.85^\circ$. Also $\angle ABD=32.18^\circ$.
Triangle BDE ($BD=326$, $BE=225$, $DE=488$): $\angle DBE=123.64^\circ$. Then $\angle ABE=\angle ABD+\angle DBE=155.8^\circ$, so $\angle ABC=180^\circ-155.8^\circ=24.2^\circ$.
$\angle ACB=180^\circ-97.85^\circ-24.2^\circ=57.9^\circ$. By the Law of Sines, $AC=\dfrac{300\sin24.2^\circ}{\sin57.9^\circ}=145.4$ m.
$\boxed{145.4}$

Part 2.

Using the angles found in triangle ABC ($\angle BAC=97.85^\circ$, $\angle ABC=24.2^\circ$, $\angle ACB=57.9^\circ$) and $AB=300$:
$BC=\dfrac{AB\sin(\angle BAC)}{\sin(\angle ACB)}=\dfrac{300\sin97.85^\circ}{\sin57.9^\circ}=350.8$ m.
$\boxed{350.8}$

Part 3.

The angles of triangle ABC sum to $180^\circ$. With $\angle BAC=97.85^\circ$ (from triangle ABD) and $\angle ABC=24.2^\circ$ (from triangle BDE):
$\angle ACB=180^\circ-97.85^\circ-24.2^\circ=57.9^\circ$.
$\boxed{57.9^\circ}$

Question Bank: t400

MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3

From the top of 120-m tall tower, the angle of depression of point A south of the tower is $31.3^\circ$ and the angle of depression point B east of the tower is $54.2^\circ$. Points A and B are on the same elevation at the base of the tower. Find the distance AB.

  1. 215.5 m
  2. 236.7 m
  3. 198.3 m
  4. 243.1 m
Horizontal distances from the base: $d_A=\dfrac{120}{\tan31.3^\circ}=197.3$ m (south), $d_B=\dfrac{120}{\tan54.2^\circ}=86.4$ m (east).
Since A is south and B is east, they are perpendicular: $AB=\sqrt{197.3^2+86.4^2}=\sqrt{46{,}397}=215.5$ m.
$\boxed{215.5\text{ m}}$

Question Bank: t401

MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3

The obelisk of a certain Rizal Monument rises to some height above its dais. The angles of elevation of the top and bottom of the obelisk from two stations A and B on the same horizontal plane as the base of the dais are $48^\circ$ and $30^\circ$, respectively. The corresponding horizontal angles to the common center of both dais and obelisk from the ends of the base line A-B, 25 meters long are $75^\circ$ and $60^\circ$, respectively. Find the height of the obelisk in meters.

  1. 16.32
  2. 14.29
  3. 12.58
  4. 18.74
Horizontal triangle ABC: $\angle A=75^\circ$, $\angle B=60^\circ$, so $\angle C=45^\circ$. By the Law of Sines with $AB=25$:
$AC=\dfrac{25\sin60^\circ}{\sin45^\circ}=30.62$ m, $BC=\dfrac{25\sin75^\circ}{\sin45^\circ}=34.15$ m.
Top height (from A) $=AC\tan48^\circ=34.00$ m; bottom height (from B) $=BC\tan30^\circ=19.72$ m.
Obelisk height $=34.00-19.72=14.29$ m.
$\boxed{14.29}$

Question Bank: t402

MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3

The angle of elevation of the top point D of a tower from A is $24.35^\circ$. From another point B the angle of elevation of the top of the tower is $56.21^\circ$. The points A and B are 287.6 m apart on the same horizontal plane as the foot (point C) of the tower. The horizontal angle subtended by A and B at the foot of the tower is $90^\circ$. Determine the following:

The distance AC in meters.

  1. 65.4
  2. 275.3
  3. 216
  4. 83.4

The distance BC in meters.

  1. 275.3
  2. 65.4
  3. 83.4
  4. 216

The height of the tower in meters.

  1. 124.57
  2. 85.42
  3. 97.73
  4. 108.75

Part 1.

Let $h$ be the tower height. Then $AC=\dfrac{h}{\tan24.35^\circ}$ and $BC=\dfrac{h}{\tan56.21^\circ}$. Since $\angle ACB=90^\circ$, $AB^2=AC^2+BC^2$.
$287.6^2=h^2\left(\dfrac{1}{\tan^2 24.35^\circ}+\dfrac{1}{\tan^2 56.21^\circ}\right)=5.329\,h^2\Rightarrow h=124.57$ m.
$AC=\dfrac{124.57}{\tan24.35^\circ}=275.3$ m.
$\boxed{275.3}$

Part 2.

With $h=124.57$ m (from the previous part):
$BC=\dfrac{h}{\tan56.21^\circ}=\dfrac{124.57}{1.4932}=83.4$ m.
$\boxed{83.4}$

Part 3.

From $AB^2=AC^2+BC^2$ with $AC=h/\tan24.35^\circ$ and $BC=h/\tan56.21^\circ$:
$287.6^2=h^2(4.880+0.448)=5.329\,h^2$.
$h=\sqrt{287.6^2/5.329}=124.57$ m.
$\boxed{124.57}$

Question Bank: t405

MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3

From the top of a building, the angle of depression of a point A located due West of the building is $22.5^\circ$. From the mid-height of the building, the angle of depression of a point B located due South of the building is $18.5^\circ$. A and B are on the same horizontal plane as the base of the building and are 300 m apart.

What is the height of the building?

  1. 112.41 m
  2. 93.57 m
  3. 123.63 m
  4. 105.66 m

How far is the point B from the building?

  1. 157.89 m
  2. 169.87 m
  3. 141.21 m
  4. 136.58 m

Part 1.

Let $H$ be the height. To A (from the top): $d_A=\dfrac{H}{\tan22.5^\circ}=2.4142H$. To B (from mid-height $H/2$): $d_B=\dfrac{H/2}{\tan18.5^\circ}=1.4943H$.
A is west and B is south (perpendicular), so $d_A^2+d_B^2=300^2$: $(2.4142^2+1.4943^2)H^2=8.061H^2=90{,}000$.
$H=\sqrt{90{,}000/8.061}=105.66$ m.
$\boxed{105.66\text{ m}}$

Part 2.

B is seen from mid-height $H/2=52.83$ m at depression $18.5^\circ$:
$d_B=\dfrac{H/2}{\tan18.5^\circ}=\dfrac{52.83}{0.3346}=157.89$ m.
$\boxed{157.89\text{ m}}$

Question Bank: t407

MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3

At a point A south of a tower the angle of elevation of the top of the tower is $50^\circ$. At another point B, 200 meters east of A, the angle of elevation is $22^\circ$. Find the height of the tower.

  1. 58.7 m
  2. 98.7 m
  3. 65.4 m
  4. 85.9 m
Let $h$ be the height. A is due south of the base at $d_A=\dfrac{h}{\tan50^\circ}$; B is 200 m east of A at $d_B=\dfrac{h}{\tan22^\circ}$, with $d_B^2=200^2+d_A^2$.
$\dfrac{h^2}{\tan^2 22^\circ}-\dfrac{h^2}{\tan^2 50^\circ}=200^2\Rightarrow h^2(6.127-0.704)=40{,}000$.
$h=\sqrt{40{,}000/5.422}=85.9$ m.
$\boxed{85.9\text{ m}}$

Question Bank: w24

MSTE - Geometry and Trigonometry / Trigonometry / MSTE May 2019

The angle of elevation of the top of a building from point $A$ on the ground is 24°. From point $B$, which is 44.5 meters closer, the angle of elevation is 34.5°. What is the height of the building?

  1. 52.4 m
  2. 38.9 m
  3. 56.3 m
  4. 62.8 m
In $\triangle ABD$: $\alpha = 180^\circ - 24^\circ - 145.5^\circ = 10.5^\circ$.
$\frac{BD}{\sin 24^\circ} = \frac{44.5}{\sin 10.5^\circ} \Rightarrow BD = 99.32\text{ m}$
$H = 99.32\,\sin 34.5^\circ = \boxed{56.26\text{ m}}$

Question Bank: w69

MSTE - Geometry and Trigonometry / Trigonometry / MSTE November 2019

A mountain peak $A$ is 1040 m above sea level. From mountain peak $A$, the angle of elevation of mountain peak $B$ was 10 degrees. The pilot of Emirates Airline, upon flying directly over peak $A$, took the angle of depression of peak $B$ equal to $46^\circ 12'$ and reads his altimeter to be 2,820 m above sea level. Determine the elevation of mountain peak $B$.

  1. 1325.68 m
  2. 1297.45 m
  3. 1254.25 m
  4. 1186.98 m
Let $A$ (at +1040 m) and $C$ (the plane, at +2820 m) be on the same vertical, so $AC = 2820 - 1040 = 1780$ m.
In triangle $ABC$: angle at $A = 90^\circ - 10^\circ = 80^\circ$, angle at $C = 90^\circ - 46^\circ 12' = 43^\circ 48'$, so angle at $B = 56^\circ 12'$.
Sine law: $\dfrac{AB}{\sin 43^\circ 48'} = \dfrac{1780}{\sin 56^\circ 12'}$, giving $AB = 1482.60$ m.
Rise from $A$ to $B$: $y = AB\sin 10^\circ = 1482.60\sin 10^\circ = 257.45$ m.
Elevation of $B = 1040 + 257.45$
$\boxed{\text{Elev }B = 1297.45\text{ m}}$

Question Bank: w70

MSTE - Geometry and Trigonometry / Trigonometry / MSTE November 2019

Immediately after Mayon volcano showed tell-tale signs of activity, PHIVOLCS set up stations to monitor the volcano. Two of such stations were located at points $A$ and $B$, 7 km apart and on the same horizontal plane as the base of the volcano. From station $A$, the angle of elevation of the top of Mayon is $8^\circ$. When the volcano erupted, the volcanic ash emitted above its crater subtended an angle of $64^\circ$ from each station $A$ and $B$. Assuming that the points of observation are on the same vertical plane, find the height of the ash.

  1. 32.3 km
  2. 36.8 km
  3. 30.8 km
  4. 28.4 km
With $\alpha = 8^\circ$ (elevation of the crater base), $\theta = 64^\circ$ (subtended ash angle) and base $s = AB = 7$ km, the height of the ash column is:
$h = s\tan\theta\,\dfrac{1 + \tan\alpha\tan 2\alpha}{1 - \tan\theta\tan 2\alpha}$
$h = 7\tan 64^\circ\,\dfrac{1 + \tan 8^\circ\tan 16^\circ}{1 - \tan 64^\circ\tan 16^\circ}$
$\boxed{h = 32.3\text{ km}}$

Question Bank: w71

MSTE - Geometry and Trigonometry / Trigonometry / MSTE November 2019

In hilly or mountainous terrain it is often difficult to determine heights. A geologist took measurements at two points, $B$ and $C$, 928 m apart. The angle of elevation from $B$ to the top of the mountain at $A$ is $47^\circ 10'$. The angle of elevation from $C$ to $B$ is $8^\circ 40'$, and the angle of elevation from $C$ to $A$ is $32^\circ 30'$. It is known that the elevation at $C$ is 1537 m to the nearest meter. What is the elevation at $A$?

  1. 2903 m
  2. 2943 m
  3. 2763 m
  4. 2843 m
In triangle $CBA$: $\delta = 32^\circ 30' - 8^\circ 40' = 23^\circ 50'$, $\beta = (180^\circ - 47^\circ 10') + 8^\circ 40' = 141.5^\circ$, so $\alpha = 180^\circ - 23^\circ 50' - 141.5^\circ = 14.67^\circ$.
Sine law: $\dfrac{AC}{\sin 141.5^\circ} = \dfrac{928}{\sin 14.67^\circ}$, so $AC = 2281.11$ m.
Vertical rise from $C$ to $A$: $H = AC\sin 32^\circ 30' = 2281.11\sin 32^\circ 30' = 1225.64$ m.
Elevation of $A = 1537 + 1225.64 = 2762.64$ m
$\boxed{\text{Elev }A \approx 2763\text{ m}}$