ECAT Mathematics
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ECAT Mathematics
Chapter 12: Application of Trigonometry
72 Questions
90 Minutes
Pass: 60%+
ECAT Prep
0
The Law of Sines states: a/sinA = b/sinB =
A
c/sinC
B
sinC/c
C
cosC/c
D
c/cosC
1
The Law of Cosines: a² =
A
b²+c²−2bc cosA
B
b²+c²+2bc cosA
C
b²−c²+2bc cosA
D
b²+c²
2
The area of triangle with sides b, c and included angle A:
A
(1/2)bc sinA
B
bc sinA
C
(1/2)bc cosA
D
bc cosA
3
The angle of elevation is measured:
A
Below the horizontal
B
Above the horizontal
C
At 90°
D
Along the horizontal
4
The angle of depression is measured:
A
Above the horizontal
B
Below the horizontal
C
At 45°
D
Along the vertical
5
In triangle ABC, if a=7, b=5, A=30°: sinB =
A
5sin30°/7=5/14
B
7/5
C
sin30°
D
1/2
6
Heron's formula: Area = √(s(s−a)(s−b)(s−c)) where s =
A
a+b+c
B
(a+b+c)/2
C
abc
D
(a+b−c)/2
7
If a=3, b=4, c=5 (right triangle), area =
A
6
B
12
C
15
D
7.5
8
A person stands 50m from tree, angle of elevation 45°. Height of tree:
A
50m
B
25m
C
100m
D
50√3m
9
A ladder 10m long leans against wall, making 60° with ground. Height reached =
A
5m
B
5√3m
C
10/√3m
D
10m
10
A ship travels N30°E for 20km. Northward component =
A
20sin30°=10km
B
20cos30°=10√3km
C
20km
D
10√3km
11
Eastward component of N30°E, 20km:
A
20sin30°=10km
B
20cos30°=10√3km
C
20km
D
10√3km
12
Two angles of triangle are 45° and 60°. Third angle:
A
75°
B
90°
C
30°
D
60°
13
In ambiguous case (SSA): there may be:
A
Only one triangle
B
Two triangles or no triangle
C
No solution always
D
Three triangles
14
Law of cosines: c² = a²+b²−2ab cosC. If C=90°:
A
c²=a²+b² (Pythagoras)
B
c²=a²−b²
C
c²=a²+b²+2ab
D
c²=(a+b)²
15
For solving triangle: if all three sides are known, use:
A
Law of Sines
B
Law of Cosines
C
Heron's Formula
D
All of the above
16
From top of cliff 80m high, angle of depression to boat = 30°. Horizontal distance =
A
80√3m
B
80/√3m
C
80m
D
40m
17
Bearing N45°W means:
A
45° east of north
B
45° west of north
C
45° north of west
D
45° south of west
18
If sinB = b sinA/a and b>a, sinB>1:
A
Two solutions
B
No triangle possible
C
Unique triangle
D
Right triangle
19
If b sinA < a, and B is acute or obtuse:
A
No solution
B
One solution
C
Two solutions
D
Three solutions
20
Triangle with a=8, b=5, C=60°. Find c²:
A
c²=64+25−40=49
B
c²=64+25+40=129
C
c²=64−25=39
D
c²=89
21
So c=
A
7
B
√49=7
C
9
D
8
22
Tangent formula: tan((A−B)/2) = ?
A
(a−b)/(a+b) × cot(C/2)
B
(a+b)/(a−b) × cot(C/2)
C
(a−b)/c
D
cotC
23
The circumradius R of triangle: R = a/(2sinA) comes from:
A
Law of cosines
B
Law of sines
C
Heron's formula
D
Pythagorean theorem
24
Area using circumradius: Area =
A
abc/4R
B
abc/(2R)
C
a²bc/(4R)
D
2Rabc
25
Inradius r = Area/s where s is:
A
Area of triangle
B
Perimeter
C
Semi-perimeter
D
a+b
26
From window 12m above ground, angle of elevation of top of building = 60°, angle of depression of base = 45°. If d = horizontal distance, d=
A
12m
B
12/√3m
C
12√3m
D
6m
27
Height of building above window = 12tan60° =
A
12√3
B
12
C
6√3
D
24
28
Total height of building = 12+12√3 =
A
12(1+√3)
B
24√3
C
12√3
D
12+12
29
For triangle with sides 5,12,13: it is
A
Acute
B
Right (5²+12²=13²)
C
Obtuse
D
Equilateral
30
Law of cosines used when we know:
A
SSS or SAS
B
AAS or ASA
C
Only angles
D
Only one side
31
Law of sines used when we know:
A
SSS
B
SAS
C
AAS or ASA
D
All sides
32
Angle of elevation from point 100m away to top of tower = arctan(1). Height =
A
100m
B
50m
C
200m
D
100√3m
33
Half-angle formula for sides: sin(A/2) = √((s−b)(s−c)/(bc)):
A
True (one form of half-angle formula for triangles)
B
False
C
Only for equilateral triangles
D
Only for right triangles
34
Two ships from port: first goes N30°E 10km, second goes N60°W 10km. Angle between paths = ?
A
60°
B
90°
C
30°
D
120°
35
Distance between ships (using cosine rule):
A
10√2km
B
20km
C
10km
D
10√3km
36
A flagpole stands on top of a tower. From a point, angle of elev. to top of pole=45°, to top of tower=30°. Tower height h, pole length p. Then h/tan30°=
A
Distance to point d
B
h only
C
p+h
D
tan45°×d
37
Also (h+p)/tan45°=d→h+p=d=h√3. So p=
A
h(√3−1)
B
h√3−h
C
h
D
Both A and B
38
Area of equilateral triangle with side a:
A
a²√3/2
B
a²√3/4
C
a²/2
D
√3a
39
A tower 100m tall. From its base, angle to distant mountain = 45°. From its top, angle = 30°. Distance to mountain:
A
100/(tan45°−tan30°)m
B
100m
C
100√3m
D
200m
40
sin rule: a/sinA = 2R means 2R is:
A
Diameter of circumscribed circle
B
Radius
C
Area
D
Perimeter
41
The angle of depression from top of building to a car is 30°. Building is 50m. Distance of car from base:
A
50tan30°
B
50/tan30°=50√3
C
50sin30°
D
50m
42
For any triangle, A+B+C =
A
90°
B
180°
C
270°
D
360°
43
If A=B=C=60°, triangle is:
A
Isosceles
B
Equilateral
C
Right
D
Scalene
44
Projection formula: a = b cosC + c cosB comes from:
A
Law of sines only
B
Law of cosines (projection rule)
C
Law of tangents
D
Heron formula
45
cosine rule: cosA = (b²+c²−a²)/(2bc). If b=c=a, cosA =
A
1/2
B
0
C
1
D
−1/2
46
For triangle with a=6,b=8,c=10: cosA =
A
(64+100−36)/160=(128)/160=4/5
B
(36+100−64)/120=72/120=3/5
C
1/2
D
√3/2
47
So angle A =
A
arccos(4/5)≈36.87°
B
arccos(3/5)≈53.13°
C
30°
D
45°
48
Bearing of 270° means:
A
North
B
East
C
South
D
West
49
Bearing of 090° means:
A
North
B
East
C
South
D
West
50
Sine rule to find missing angle B: sinB = b sinA/a. Number of solutions when b<a:
A
0
B
1
C
2
D
Infinite
51
Sine rule: number of solutions when b>a (and A acute):
A
0
B
1
C
2 if bsinA<a else 0
D
Always 2
52
For solving triangle with three sides (SSS), first find angle using:
A
sinA=a/b
B
cosA=(b²+c²−a²)/2bc
C
tanA=a/b
D
Area formula only
53
Navigation: a ship sails 150km on bearing 040°, then 200km on bearing 160°. Find final distance from start using cosine rule. Angle between paths:
A
120°
B
60°
C
80°
D
40°
54
d²=150²+200²−2(150)(200)cos(120°)=22500+40000+30000=92500. d=
A
√92500≈304km
B
√62500=250km
C
√32500≈180km
D
None
55
Heron's formula: a=5,b=6,c=7. s=
A
9
B
18
C
12
D
7
56
Area by Heron's: s=9,a=5,b=6,c=7. s−a=4,s−b=3,s−c=2. Area=
A
√(9×4×3×2)=√216=6√6
B
6√6
C
√216
D
Both B and C
57
The sine rule to find a side: a = b sinA/sinB. Given b=10, A=45°, B=60°:
A
a=10sin45°/sin60°=10(√2/2)/(√3/2)=10√2/√3
B
a=10√2/√3≈8.16
C
Both
D
a=10sin60°/sin45°
58
The triangle is impossible if the sum of any two sides is:
A
Greater than third
B
Less than third
C
Equal to third
D
Zero
59
Angle A=60°, b=c=8: find a using cosine:
A
a²=64+64−64=64; a=8
B
a²=128; a=8√2
C
a²=64; a=8
D
a²=128−64=64; a=8
60
So triangle is:
A
Scalene
B
Equilateral
C
Isosceles
D
Right
61
The area of right triangle with legs 6 and 8:
A
24
B
48
C
12
D
√(6²+8²)
62
Two observers A and B on same side of cliff 200m apart. Angles of elevation to cliff top: A sees 50°, B sees 40°. Using sine rule in triangle formed: height can be found.This technique uses:
A
Heron's formula
B
Direct measurement
C
Law of sines applied to triangle
D
Law of cosines only
63
The in-radius formula: r = (s−a)tan(A/2):
A
True
B
False
C
Only for right triangles
D
Only for equilateral triangles
64
Area of triangle = rs where r is inradius and s is:
A
Perimeter
B
Semi-perimeter
C
Any side
D
Area/r
65
From two points on same horizontal level, angles of elevation to balloon are 60° and 30°. Points are 100m apart. Height of balloon:
A
25√3m
B
50√3m
C
75m
D
100m
66
Cosine rule: b² = a²+c²−2ac cosB. If B=0° (degenerate):
A
b²=(a−c)²
B
b²=(a+c)²
C
b=a+c
D
b=a−c
67
In an isosceles triangle, the two base angles are equal. If apex=40°, each base angle=
A
60°
B
70°
C
80°
D
40°
68
If one angle of triangle is obtuse, the triangle is:
A
Acute
B
Obtuse
C
Right
D
Equilateral
69
The formula for area in terms of diagonals d1, d2 of rhombus: Area =
A
d1+d2
B
d1×d2/2
C
(d1+d2)/2
D
2(d1+d2)
70
The angle between two diagonals of a rhombus is always:
A
45°
B
60°
C
90°
D
180°
71
Sine of an obtuse angle B=120° is:
A
−sin60°=−√3/2
B
sin60°=√3/2
C
−sin120°
D
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