TRIGONOMETRY

QUESTIONS CHOICES

1

2

3 ( cos A )^4 - ( sin A )^4 is equal to _____.

4

5

6

7

Sin (B – A) is equal to ____, when B = 270° and A is an acute angle.

a. - cos A b. cos A c. - sin A d. sin A

If sec² A is 5/2, the quantity 1- sin² A is equivalent to a. 2.5b. 1.5c. 0.4d. 0.6

a. cos 4Ab. cos 2Ac. sin 2Ad. sin 4A

Of what quadrant is A, if sec A is positive and csc A is negative?

a. IVb. IIc. IIId. I

Angles are measured from the positive horizontal axis, and the positive direction counterclockwise. What are the values of sin B and cos B in the 4th quadrant?

a. sin B > 0 and cos B < 0b. sin B < 0 and cos B < 0c. sin B > 0 and cos B > 0d. sin B < 0 and cos B > 0

Csc 520° is equal to a. cos 20°b. csc 20°c. tan 45°d. sin 20°

Solve for Ə in the following equation: Sin 2Ə = cos Ə a. 30°b. 45°c. 60°d. 15°

8 If sin 3A = cos 6B, then

9 Solve for x, if tan 3x = 5 tan x.

### If sin x cos x + sin 2x = 1, what are the values of x?

### Solve for G is csc ( 11G – 16° ) = sec ( 5G + 26° ).

a. A + B = 90°b. A + 2B = 30°c. A + B = 180°d. None of these

a. 20.705°b. 30.705°c. 35.705°d. 15.705°

a. 32.2° , 69.3°b. -20.67° , 69.3°c. 20.90° , 69.1°d. -32.2° , 69.3°

a. 7°b. 5°c. 6°d. 4°

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### Find the value of sin ( arc cos 15/17 ).

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Find the value of A between 270° and 360° if 2 sin² A – sin A = 1.

a. 300°b. 320°c. 310°d. 330°

If cos 65° + cos° = cos Ə, find Ə in radians. a. 0.765b. 0.087c. 1.213d. 1.421

a. 8/11b. 8/19c. 8/15d. 8/17

The sine of a certain angle is 0.6, calculate the cotangent of the anlge.

a. 4/3b. 5/4c. 4/5d. ¾

### If sec 2A = 1/ sin 13A, determine the angle A in degrees.

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### Find the value of y in the given: y = ( 1 + cos 2Ə ) tan Ə

### Find the value of sinƏ + cosƏ tanƏ/ cosƏ

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### Simplify the expression sec Ə - ( secƏ ) sin²Ə

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a. 5°b. 6°c. 3°d. 7°

If tan x = 1/2, tan y = 1/3, what is the value of tan ( x + y )?

a. 1/2b. 1/6c. 2d. 1

a. sin Əb. cos Əc. sin 2Əd. cos 2Ə

a. 2 sin Əb. 2 cos Əc. 2 tan Əd. 2 cot Ə

Simplify the equation sin²Ə ( 1+ cot²Ə ) a. 1b. sin²Əc. sin²Ə sec²Əd. sec²Ə

a. cos² Əb. cos Əc. sin² Əd. sin Ə

Arc tan [ 2 cos ( arc sin [ ( 3^1/3 ) /2 ) is equal to] a. π/3b. π/4c. π/16d. π/2

### Evaluate arc cot [ 2 cos (arc sin 0.5 ) ]

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### Solve for A for the given equation cos² A = A = 1 – cos² A.

a. 30°b. 45°c. 60°d. 90°

Solve for x in the given equation: Arc tan ( 2x ) + arc tan ( x ) = π/4

a. 0.149b. 0.281c. 0.421d. 0.316

Solve for x in the equation: arc tan ( x + 1 ) + arc tan ( x – 1 ) = arc tan ( 12 ).

a. 1.5b. 1.34c. 1.20d. 1.25

a. 45, 125, 225, 335 degreesb. 45, 125, 225, 315 degreesc. 45, 135, 225, 315 degreesd. 45, 150, 220, 315 degrees

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Evaluate the following: ( sin 0° + sin 1° + sin 2° + ...+ sin 89° + sin 90° / cos0° + cos 1° + cos 2° +...+ cos 89° + cos 90° )

a. 1b. 0c. 45.5d. 10

Simplify the following: ( cos A + cos B / sin A – sin B ) + ( sin A + sin B / cos A – cos B )

a. 0b. sin Ac. 1d. cos A

Evaluate: ( 2 sinƏ cosƏ – cosƏ / 1 – sinƏ + sin²Ə – cos²Ə )

a. sin Əb. cos Əc. tan Əd. cot Ə

Solve for the value of “ A “ when sin A = 3.5x and cos A = 5.5x.

a. 32.47°b. 33.68°c. 34.12°d. 35.21°

If sin A = 2.511x, cos A = 3.06x and sin 2A = 3.939x, find the value of x?

a. 0.265b. 0.256c. 0.562d. 0.625

### If coversed sin Ə = 0.134, find the value of Ə.

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a. 30°b. 45°c. 60°d. 90°

A man standing on a 48.5m building high, has an eyesight height of 1.5m from the top of the building, took a depression reading from the top of another nearby wall, which are 50° and 80° respectively. Find the height of the nearby building in meters. The man is standing from the edge of the building and both buildings lie on the same horizontal plane.

a. 39.49b. 35.50c. 30.74d. 42.55

Point A and B 1000m apart are plotted on a straight highway running East and West. From A, the bearing of a tower C is 32°W of N and from B the bearing of C is 26° N of E, Approximate the shortest distance of tower cC to the highway.

a. 364 mb. 374 mc. 384 md. 394 m

Two triangles have equal bases. The altitude of one triangle is 3 units more than its base and the altitude of the other triangle is 3unites less than its base. Find the altitudes, if the areas of the triangles differ by 21 square units.

a. 6 and 12b. 3 and 9c. 5 and 11d. 4 and 10

A ship started sailing S 42°35' W at the rate of 5 kph. After 2 hours, ship B started at the same port going N 46°20' W at the rate of 7 kph. After how many hours will the second ship be exactly north of ship A?

a. 3.68b. 4.03c. 5.12d. 4.83

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An aerolift airplane can fly at an airspeed of 300 mph. If there is a wind blowing towards the cast at 50 mph, what should be the plane's compass heading in order for its course to be 30°? What will be the plane's ground speed if it flies in this. a. 19.7°, 307.4 mph

b. 20.1°, 309.4 mphc. 21.7°, 321.8 mphd. 22.3°, 319.2 mph

A man finds the angle of elevation of the top of a tower to be 30°. He walks 85m nearer the tower and finds its angle of elevation to be 60°. What is the height of the tower.

a. 76.31 mb. 73.31 mc. 73.16 md. 73.61 m

A pole cast a shadow 15m long when the angle of elevation of the sun is 61°. If the pole is leaned 15° from the vertical directly towards the sun, determine the length of the pole.

a. 54.23 mb. 48.23 mc. 42.44 md. 46.21 m

A wire supporting a pole is fastened to it 20 ft from the ground and to the ground 15 ft from the pole. Determine the length of the wire and the angle it makes with the pole.

a. 24 ft, 53.13°b. 24 ft, 36.87°c. 25 ft, 53.13°d. 25 ft, 36.87°

The angle of the elevation of the top of the tower B from the top of the tower A is 28° and the angle of elevetion of the top of the tower A from the base of the tower B is 46°. The two towers lie in the same horizontal plane. If the height of tower B is 120m, find the hieght of tower A.

a. 66.3 mb. 79.3 mc. 87.2 md. 90.7 m

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Points A and B are 100m apart and are of the same elevation as the foot of a building. The angles of elevation of the top of the building from points A and B are 21° and 32° respectively. How far is A from the building in meters.

a. 259.28b. 265.42c. 271.64d. 277.29

The Captain of a ship views the top of a lighthouse at an angle of 60° with the horizontal at an elevation of 6m above the sea level. Five minutes later, the same Captain of the ship views the top of the same lighthouse at an angle of 30° with the horizontal. Determine the speed of the ship if the lighthouse is known to be 50m above sea level.

a. 0.265 m/sec b. 0.155 m/secc. 0.169 m/secd. 0.210 m/sec

An observer wishes to determine the height of a tower. He takes sight at the top of the tower from A and B, which are 50ft apart, at the same elevation on a direct line with the tower. The vertical angle at point A is 30° and point B is 40°. What is the height of the tower?

a. 85.60 ftb. 92.54 ftc. 110.29 ftd. 143.97 ft

A PLDT tower and a monument stand on a level plane. The angles of depression of the top and the bottom of the monument viewed from the top of the PLDT tower at 13° and 35° respectively. The height of the tower is 50m. Find the height of the monument.

a. 29.13 mb. 30.11 mc. 32.12 md. 33.51 m

If an equilateral triangles is circumscribed about a circle of radius 10 cm, determine the side of the triangle.

a. 34.64 cmb. 64.12 cmc. 36.44 cmd. 32.10 cm

The two legs of a triangle are 300 and 150 m each, respectively. The angle opposite the 150 m side is 26°. What is the third side?

a. 197.49 mb. 218.61 mc. 341.78 md. 282.15 m

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The sides of the triangular lot are 130 m., 180 m and 190 m. The lot is to be didvided by a line bisecting the longest side and drawn from the opposite vertex. Find the length of the line.

a. 120 mb. 130 mc. 125 md. 128 m

The sides of the triangleare 195, 157 and 210, respectively. What is the area of the triangle?

a. 73,250 sq. unitsb. 10,250 sq. unitsc. 14,586 sq. unitsd. 11260 sq. units

The sides of a triangle are 8, 15 and 17 units. If each side is doubled, how many square units will the area of the new trianglebe?

a. 240b. 420c. 320d. 200

TRIGONOMETRY

ANSWER DISCUSSION

a. - cos A

c. 0.4

b. cos2A

a. IV

b. csc 20°

a. 30°

sin ( 270° - A ) = sin 270° cos A - sin A cos 270 = ( - 1 ) cos A - sin A ( 0 ) sin ( 270° - A ) = - cos A

sin² A + cos² A = 1 1 - sin² A = cos² A sec² A = 5/2NOTE: cos A = 1/secA, thus cos² A = 1/sec² ASUBSTITUE: ( 2 ) IN ( 1 ): 1 - sin² A = 1/sec² A = 1/ ( 5/2 ) = 0.4

cos^4 A - sin^4 A = cos² A cos² A - sin ² A sin² A =cos² A (1 - sin² A) - sin² A (1 - cos² A) = cos² A - cos² sin² A - sin² A + sin² A cos² A = cos² A - sin² A = cos 2ANOTE: cos 2A = cos² A - sin² A

In the fourht quadrant:

sec Ə = hypotanuse/adjacent side = c/a

csc Ə = hypotanuse/opposite side = - c/b

d. sin B < 0 and cos B > 0In the fourth quadrant:

sin B = opposite side/hypotanuse = - b/c

cos B = adjacent side/hypotanuse = a/c

Thus, sin B < 0 and cos B > 0

csc 520° = csc ( 520° - 360° )csc 520° = csc 160°

csc 160° = csc ( 180° - 160° )csc 160° = csc20°

Thus csc 520° = csc 20°

sin 2Ə = cosƏ eqn 1 note: sin 2Ə = 2 sin Ə cosƏ eqn2 substitute: (2) in (1) 2 sinƏ cosƏ = cos Ə 2 sinƏ = 1 sinƏ = 0.5 Ə = 30°

b. A + 2B = 30°

a. 20.705°

c. 20.90°, 69.1°

sin 2Ə = cosƏ eqn 1 note: sin 2Ə = 2 sin Ə cosƏ eqn2 substitute: (2) in (1) 2 sinƏ cosƏ = cos Ə 2 sinƏ = 1 sinƏ = 0.5 Ə = 30°

sin 3A = cos 6B eqn 1NOTE: cos 6B = sin ( 90° - 6B ) eqn 2Substitute: ( 2 ) in ( 1 ) sin 3A = sin ( 90° - 6B ) sin 3A = 90° - 6B 3A = 90° - 2B A = 30° - 2B A + 2B = 30°

tan 3x = 5 tan x eqn 1tan 3x = tan ( 2x + x ) = tan2x + tan x/1 - tan2xtanx eqn 2 Substitute: ( 2 ) in ( 1 ) tan2x + tan x/1/tan2x + tan x = 5 tan x tan 2x + tan x = 5 tan x - 5 tan 2x tan² x tan 2x = 4 tan x - 5 tan 2x tan² x tan 2x ( 1 + 5 tan² x ) = 4 tan x eqn 3 tan 2x = 2 tan x/1 - tan² x eqn 4 Substitute ( 4 ) in ( 3 ): ( 2 tan x/1 - tan² x )( 1 + 5 tan² x ) = 4 tan x 2 tan x ( 1+ 5 tan x + 10 tan³x = 4 tan x - 4 tan³x 14 tan³x = 2 tan x tan² x = 0.142857 tan x = 0.3779642 x = 20.705°

Sin x cos x + sin 2x = 1 eqn 1 NOTE: 2 sin x cos x = sin 2x sin x cos x 0.5 sin 2x eqn 2Substitute: ( 2 ) in ( 1 ) 0.5 sin 2x sin 2x = 1 1.5 sin 2x = 1 sin 2x = 0.6667 2x = 41.8 x = 20.9°NOTE: Complementary angles have the same values of their sine functions Thus, the other angle is equal to: 90° - 20.9° = 69.1°

b. 5° Csc ( 11G – 16 ° ) = sec ( 5G + 26° ) 1/sin ( 11G – 16° ) = 1/cos (11G – 16° ) cos ( 5G + 26° ) = sin ( 11G – 16° ) eqn 1 Note: sin Ə = cos ( 90° - Ə ) let: Ə = 11G – 16° sin ( 11G – 16° = cos [ 90° - ( 11G – 16° ) ] sin ( 11G – 16° = cos ( 106° - 11G ) eqn 2

Substitute: ( 2 ) in ( 1 ) cos ( 5G + 26° ) = cos ( 106 – 11G ) 5G + 26° = 106° - 11G G = 5°

d. 330°

b. 0.087

d. 8/17

a. 4/3

Csc ( 11G – 16 ° ) = sec ( 5G + 26° ) 1/sin ( 11G – 16° ) = 1/cos (11G – 16° ) cos ( 5G + 26° ) = sin ( 11G – 16° ) eqn 1 Note: sin Ə = cos ( 90° - Ə ) let: Ə = 11G – 16° sin ( 11G – 16° = cos [ 90° - ( 11G – 16° ) ] sin ( 11G – 16° = cos ( 106° - 11G ) eqn 2

Substitute: ( 2 ) in ( 1 ) cos ( 5G + 26° ) = cos ( 106 – 11G ) 5G + 26° = 106° - 11G G = 5°

2 sin² A – sin A = 1 sin² A – 0.5 sin A = 0.5

By completing the square: ( sin A – 0.25 )² = 0.5 + ( 0.25 )² ( sin A – 0.25 )² = 0.5625 sin A – 0.25 = ± 0.75Take the minus sign: sin A = 0.25 – 0.75 = - 0.5 A = - 30° or A = - 30° = 360° = 330°

Cos 65° + cos 55° = cos Ə cos Ə = 0.99619 Ə = 5° x ( 2π radians/360° ) Ə = 0.087 radian

x = sin [ cos^-1 ( 15/17 ) ] Let: Ə = cos^-1 ( 15/17 ) cos Ə = 15/17 b = 1√ c² – a² = √ (17)² – (15)² b = 8

x = sin Ə = opposite side/hypotanuse = b/c

x = 8/17

Let: Ə = angle sin Ə = 0.6 = 3/5 a = √ c² – b² = √ (5)² – (3)² = 4

cot Ə = adjacent side/opposite side = a/b cot Ə = 4/3

b. 6°

d. 1

c. sin 2Ə

c. 2 tan Ə

a. 1

b. cos Ə

b. π/ 4

Sec 2A = 1/sin 13A 1/cos 2A = 1/sin 13A cos 2A = sin 13A eqn 1Note: sin Ə = cos ( 90° – Ə ) Let: Ə = 13A sin 13A = cos ( 90° - 13A ) eqn 2Substitute: ( 2 ) in ( 1 ) cos 2A = cos 90° - 13A ) 2A = 90° - 13A A = 6°

Tan ( x + y ) = ( tan x + tan y ) / ( 1 – tan x tan y ) = [ (½) + (1/3)] / [ 1- (½ )(1/3) } = 1

Y = ( 1 + cos 2Ə ) tan Ə eqn 1cos 2Ə = cos² Ə – sin² Əcos 2Ə = ( 1 – sin² Ə ) - sin² Əcos 2Ə = 1 – 2 sin² Ə eqn 2

Substitute: ( 2 ) in ( 1 ) y = ( 1 + 1 – 2 sin² Ə ) tan Ə = ( 2 – 2 sin² Ə ) tan Ə = 2 ( 1 – sin² Ə tan Ə ) = 2 ( cos² Ə ) ( sinƏ/cosƏ ) = 2 cos Ə sinƏ y = sin 2Ə

x = ( sinƏ + cosƏ tanƏ/cosƏ ) = [ (sinƏ/cosƏ) + ( cosƏ tanƏ/cosƏ ) ] = tanƏ + tanƏ x = 2 tanƏ

x = sin² Ə ( 1 + cot² Ə ) = sin² Ə [ 1 + ( cosƏ/sinƏ )² ] = sin² Ə [ sin² Ə +cos² Ə /sin² Ə ] = sin² Ə /sin² Ə = 1

x = sec Ə - ( secƏ ) sin² Ə = sec Ə ( 1- sin² Ə ) = sec Ə ( cos² Ə ) = (1/cosƏ ) ( cos²Ə ) x = cos Ə

x = tan^-1 { 2 cos [ sin^-1( √3 / 2 ) ]} x = tan^-1 ( 2cos 60°Thus, x = tan^-1 ( 2 cos 60° ) = tan^-1 (1) = 45° ( 2π radians/360° ) x = π/4 radians

a. 30°

b. 0.281

b. 1.34

c. 45, 135, 225, 315 degrees

x = cot^-1 { 2 cos [sin^-1 (0.5) ]} = cot^-1 ( 2 cos 30° ) = cot^-1 ( 1.732 )cot x = 1.732

1/tanx = 1.732 tan x = 1/1.732 = 0.57736 x = 30°

tan ^-1 (2x) + tan^-1 x = π/4 eqn 1Let: tan A = 2x A = tan^-1 (2x) eqn 2 tan B = x B = tan^-1 x eqn 3Substitute ( 2 ) in ( 1 ): A + B = π/4 = 45°tan ( A + B ) = tan 45°( tan A + tan B ) / ( 1- tan A tan B ) = 1 [ 2x + x / 1 – 2x(x) ] = 1 3x = 1 – 2x² 2x² + 3x -1 = 0Using the quadratic formula; x = [ -3 ±√ (3)² – 4(2)(-1) ] / 2(2) = -3 ± 4.123 x = -3 + 4.123 /4 x = 0.28

Arc tan ( x + 1 ) + arc tan ( x – 1 ) = arc tan ( 12 ) eqn 1Let: tan A = x + 1 A = tan^-1 ( x + 1 ) eqn 2 tan B = x – 1 B = tan^-1 ( x – 1 ) eqn 3Substitute ( 2 ) and ( 3 ) in ( 1 ): A + B = tan^-1 ( 12 ) Tan ( A + B ) = tan ( tan^-1 ( 12 ) tan A + tan B / 1 – tan A tan B = 12 ( x + 1 ) + ( x – 1 ) / 1 - ( x+ 1)( x – 1 ) = 12 2x = 12 – 12 ( x² – x + x – 1 ) 2x = 12 – 12x² + 12 12x² + 2x – 24 = 0Using the quadratic formula; x = -2 ± √ (2)² - 4(12)(-24) / 2(12) = -2 ± 34 / 24 = -2 + 34 / 24 x = 1.34

cos² A = 1 - cos² A 2 cos² A = 1 cos² A = 0.5 cos A = ± 0.707

If cos A = + 0.707 A = 45° or 315°

If cos A = - 0.707 A = 135° or 225°

a. 1

a. 0

d. cot Ə

a. 32.47°

b. 0.256

cos² A = 1 - cos² A 2 cos² A = 1 cos² A = 0.5 cos A = ± 0.707

If cos A = + 0.707 A = 45° or 315°

If cos A = - 0.707 A = 135° or 225°

( sin² Ə + sin² 1 + sin² 3 …sin² 89 + sin² 90 ) / cos² Ə + cos² 1 + cos² 3 … cos² 89 + cos² 90 )NOTE: sin² A + cos² B = 1 and cos² A + cos² B = 1,provided A and B are complementary angles, ( A + B = 90 ).

Thus, the equation can be written as = { [ (sin² 0 + sin² 90 ) + ( sin² 1 + sin² 89 ) .......( sin² 44 + sin² 46 ) (sin² 45 )] / [( cos² 0 + cos² 90 ) + ( cos² 1 + cos² 89 ) ... ( cos² 44 + cos² 46 ) ( cos² 45) ]} = 1

[( cos Ə + cos B ) / (cos A - sin B )] + [( sin A + sin B ) / cos A - cos B )] = {[ ( cos A + cos B ) ( cos A + cos B ) + ( sin A - sin B ) ( sinA + sin B )] / [ ( sin A - sinB ) ( cos A - cis B )]} = { [ (cos² A - cos² B + sin² A - sin² B )] / [ ( sin A - sin B ) ( cos A - cos B )]} = 0

x = 2 sin Əcos Ə / 1 - sin Ə + sin² Ə - cos² Ə = [cos Ə ( 2 sin Ə - 1 )] / [( 1- cos² Ə ) + sin Ə - sin Ə] = [cos Ə ( 2 sin Ə -1)] / [ sin² Ə + sin² Ə - sin Ə] = [ cos Ə ( 2 sin Ə - 1 )] / [ 2 sin² Ə - sin Ə] = [ cos Ə ( 2 sin Ə - 1 )] / [ sin Ə ( 2 sin Ə - 1 )] = cos Ə / sin Ə x = cot Ə

sin A = 3.5 x eqn 1 cos A = 5.5 x eqn 2Divided ( 1 ) by ( 2 ): sin A / cos A = 3.55 x / 5.55 x tan A = 0.63636 A = 32.47° \

sin A = 2.511 x ; cos A = 3.06 x ; sin 2A = 3.939 xNote: sin 2A = 2 sin A cos A Substitute: 3.939x = 2 (2.511x)(3.06x) 3.939x = 15.367x² x = 0.256

c. 60°

a. 39.49

b. 374 m

d. 4 and 10

b. 4.03

\

sin A = 2.511 x ; cos A = 3.06 x ; sin 2A = 3.939 xNote: sin 2A = 2 sin A cos A Substitute: 3.939x = 2 (2.511x)(3.06x) 3.939x = 15.367x² x = 0.256

coversed sin Ə = 0.134 eqn 1Note: coversed sin Ə = 1 - sin Ə eqn 2Substitute: ( 2 ) in ( 1 ) 0.134 = 1 - sin Ə sin Ə = 1 - 0.134 sin Ə = 0.866 Ə = 60°

tan 80° = 50/x x = 8.816m tan 50° = 50 - h / 8.816 10.506 = 50 - h h = 39.49 m

Ə = 180° - ( 26° + 58° ) = 96° By sine law: sin 96° / 1000 = sin58° / BC BC = 852.719 m sin 26° = d / BC d = BC sin 26° = 852.719 sin 26° d = 374 m

h1 = b + 3 eqn 1 h2 = b - 3 eqn 2 A1 = A2 + 21 ½ bh1 = ½ bh2 + 21 eqn 3Substitute ( 1 ) and ( 2 ) in ( 3 ): ½b ( b + 3) = ½b ( b - 3 ) + 21 {[ (b² + 3b) / (2)] = [( b² - 3b ) / 2 ] + 21} 2 b² + 3b = b² - 3b +42 6b = 42 b = 7 Thus, h1 = 7 + 3 = 10 units h2 = 7 - 3 = 4 units

Note: 7t = total distance traveled by ship B 10 + 5t = total distance traveled by ship ABy sin law; sin 42°35' / 7t = sin 46°20' / 10 + 5t (10 + 5t ) [ sin 42°35' / sin 46°20' ] = 7t 9.354 +4.677t = 7t 2.323t = 9.354 t = 4.03 hrs.

c. 21.7°, 321.8 mph

d. 73.61 m

a. 54.23 m

d. 25 ft, 36.87°

b. 79.3 m

Note: 7t = total distance traveled by ship B 10 + 5t = total distance traveled by ship ABy sin law; sin 42°35' / 7t = sin 46°20' / 10 + 5t (10 + 5t ) [ sin 42°35' / sin 46°20' ] = 7t 9.354 +4.677t = 7t 2.323t = 9.354 t = 4.03 hrs.

By sine law: ( 50 / sin β ) = ( 300 / sin 60° ) β = 8.3° α = 30° - Ə = 30° - 8.3° α = 21.7°

§ + 60° β = 180° § + 60° + 80° = 180° § = 111.7°By sine law: ( sin 111.7° / V ) = ( sin 38° / 50 ) V = 321.8 mph

tan 30° = ( h / 85 + x ) h = ( 85 + x ) tan 30° eqn 1 tan 60° = h/x h = x tan 60° eqn 2Equate (1) to (2): ( 85 x ) tan 30° = x tan 60° 85 + x = 3x x = 42.5 m

Substitute x = 42.5 in ( 2 ): h = 42.5 tan60° h = 73.61 m

Ə + 61° + 90° + 15° = 180° Ə = 14°By sine law: ( sin 14° / 15 ) = ( sin 16° /x ) x = 54.23 m.

By Phytagorean theorem: x = √ (15)² + (20)² = 25 ft.

tan Ə = 15/20 Ə = 36.87°

tan 28° = 120 – h / x x = ( 120 – h ) / tan 28° eqn 1 tan 46° = h/x eqn 2 x = h / tan 46°Equate (1) to (2): ([ 120 – h) / (tan 28° )] = ( h / tan 46° ) 120 – h = 0.513 h h = 79.3 m

a. 259.28

c. 0.169 m/sec

b. 92.54 ft

d. 33.51 m

a. 34.64 cm

c. 341.78 m

tan 28° = 120 – h / x x = ( 120 – h ) / tan 28° eqn 1 tan 46° = h/x eqn 2 x = h / tan 46°Equate (1) to (2): ([ 120 – h) / (tan 28° )] = ( h / tan 46° ) 120 – h = 0.513 h h = 79.3 m

tan 21° = h / 100 + x h = ( 100 + x ) tan 21° eqn 1 tan 32° = h / x h = x tan 32° eqn 2Equate (1) to (2): ( 100 + x ) = tan 21° = x tan 32° 100 + x = 1.6278 x x = 159.286 m.Thus, the distance of point A from the building is = 100 + 159.286 = 259.286 m.

tan 60° = 44 / x x = 25.4 m. tan 30° = [ 44 / ( s + x ) ] s + x = 76.21 s + x = 76.211 s = 50.81 m

V = s / t = [50.81 / 5(60) ] V = 0.169 m/sec

tan 40° = h /x x = h /tan 40° eqn 1 tan 30° = [ h / ( 50 + x ) ] x = [h / (tan30°) - 50] eqn 2 Equate (1) to (2): h/ tan 40° = [h / (tan 30°) - 50] 1.19175 h = 1.73205 h – 50 h = 92.54 ft.

tan 35° = 50 / x x = 71.407 m. tan 13° = [(50 – h) / x] tan 13° = [ (50 – h) / 71.407 ] h = 33.51 m.

Note: Since equilateral triangle, A = B = C = 60° tan 30° = r / 0.5x x = 34.64 cm.

By sine law: 150 / sin 26° = 300 / sin B B = 61.25° 26° + 61.25° + C = 180° C = 92.75°By sine law: 150 / sin 26° = C / sin 92.75° C = 341.78 m.

c. 125 m

c. 14,586 sq. units

a. 240

By sine law: 150 / sin 26° = 300 / sin B B = 61.25° 26° + 61.25° + C = 180° C = 92.75°By sine law: 150 / sin 26° = C / sin 92.75° C = 341.78 m.

By cosine law: b² = a² + c² – 2ac cos B (180)² =(130)² + (190)² – 2(130)(190) cos B B = 65.35°

By sine law: x² = a² + (c/2)² – 2(a)(c/2) cos B x² = (130)² + (95)² – 2(130)(95) cos 65.35° x = 125 m.

Using Hero's formula: a = 195: b = 157: c = 210 s = (a + b + c) / 2 s = ( 195 + 157 + 210 ) / 2 s = 281

A = √ s(s – a)(s – b)(s – c) = √ 281(281 – 195)(281 – 157)(281 – 210) A = 14,586.2 square units

Using Heron's formula: a = 16: b = 30: c = 34 s = ( a + b + c ) / 2 = ( 16 + 30 + 34 ) / 2 s = 40

A = √ s(s – a)(s – b)(s – c) = √ 40(40 – 16)(40 – 30)(40 – 34) A = 240 square units