EXCEL REVIEW CENTER
ECE REVIEW FOR APRIL 2020 BOARD EXAM
Answer key to Take Home Trigonometry
1.
6.
B. 2.45 cm
c = 14
B. 374 m
tan 28 
C

x
32°
d
tan 46 
58°
26°
B
A
1000
tan 28
h
S

120 - h
d
s  s - a  s - b  s - c 
A  39.19 cm 2
A  rs
39.19  r (16 )
r  2.45 cm
tan 46

 h

h  79.31 m
852.719
d  852.719 (sin 26 )
 16 cm
2
A  16 16 - 8 16 -10 16 -14 
 Eq .2
 tan 28
120 - h  

 tan 46
120 - h  0.513h
BC
8  10  14

2
h

tan 28
abc
A
h
r
b = 10
 Eq .1
tan 46
Equating Eq.1 to Eq.2:
d
sin 26 
x
120 - h
x
  96
By sine law:
sin 96
sin 58

1000
BC
BC  852.719 m
sin 26 
a=8
120 - h
x
  26  58  180
7.
D. incenter
8.
A. medians
9.
B. 6.96
A
4.
C. 125 m
C
d  374 m
2.
TRIGONOMETRY
A. 39.49
x
b=180
a=130
x=?
80°
50
50-h
c/2=95
B
c/2
A
12 - a
a
b
12 - b
c=190
50
By cosine law:
b 2  a 2  c 2  2 ac cos B
h
180  130  190 - 2 (130 )(190 ) cos B
2
x
x
80°
50°
2
2
B  65.35
By cosine law:
x 2  a 2  ( c / 2 ) 2 - 2 a ( c / 2 )cos B
50-h
50
x 2  130 2  95 2 - 2 (130 )( 95) cos 65.35
x  125 m
tan 80 
50
5.
tan 50 
h  39.49 m
3.
B. 79.3 m
B
A  B  C  60o
a 2  12  a   92  2  9 12  a  cos 60o
2
a  4.2
b 2  12  a   32  2  312  a  cos 60o
2
b  6.43
x  6.96
10. C. 288

x
50 - h
10.506  50 - h
9
x  4.22  6.432  2  4.2  6.43 cos 60o
A. 54.23 m
50 - h
8.816
tan 50 ( 8.816 )  50 - h
A
x 2  a 2  b 2  2ab cos 60o
x
x  8.816 m
tan 50 
C
Let EG = height of the given
figure
x
15°
61°
E
F
15
  61  90  15  180
  14
By sine law:
sin 14 sin 61

15
x
x  54.23 m
D
G
H
C
A
E
B
H
D
DAVAO: 2nd Floor, MERCO Bldg. Rizal St cor Bolton St | BAGUIO: 4th Floor, De Guzman Bldg. Legarda Road
EXCEL REVIEW CENTER
ECE REVIEW FOR APRIL 2020 BOARD EXAM
By Pythagorean theorem:

EH  6 2
2
  3 2 
2
sin 2  
EH  54
Solving triangle EGH:
x 2  20
8x
4
x  40x 2  400
cos 2  
64x 2
2
2
sin   cos   1
cos  
H
G
By Pythagorean theorem:
54
  3 2 
2
2
EG  6
Considering the entire triangular
prism:
 6 2 12 2  6  432
V 
ATB  180o  30o  45o  105o
ACB  180o  60o  45o  75o
 0.25x  x  1 


48x 2

 1
 x 4  40x 2  400 
 

64x 2
Solving for x:
x  8.53
4

0.25x 4  x 2  1
48x 2
From Eq. 1:
E
EG 2 
13. D. 340.7
6.9282x sin   0.5x 2  1
2
TRIGONOMETRY
2
12. C. 994
ATC  180o  30o  ACB  75o
24
AB

o
sin

ATB
sin 45
AB  12  12 3
24
AC

sin ACB sin ATC
AC  24
1
Area   AB  AC  sin 60o  340.7
2
C
1
2
Solving the volume of the two extra
pyramids that we included:
 6 2 3 2 6
  144
V2  2 


3


Solving for the volume of the figure:
V  V1  V2  432  144  288


b

11. C. 8.53 units
14. C. 10.87
a
A
O
D
A
c
B
B
C
BD  CD cot B
AD  CD cot A
AD  BD  c
O
AD  BD  CD  cot A  cot B 
c
D
From the area of a triangle:
1
A   c  CD 
2
2A 
CD 
c
c2
cot A  cot B 
2A 
From the area of a triangle:
1
A   absin C
2
2A 
sin C 
ab
Using cosine law:
cot A  cot B 
In BCP :
Using cosine law:
 42  x 2

2

6

2(4)x
cos



8x cos   x 2  20  Eq.1
In ΔACP:
Using cosine law:
 42  x 2

2

5

2(4)x
cos(60


)


8x cos(60  )  x 2  9
cos 60 cos  
2
8x 
  x 9

sin
60
sin



 4x cos 

2
 6.9282x sin    x  9  Eq.2


From Eq. 1:
x cos  
x 2  20
8
Then we have:
1
4    x 2  20   6.9282x sin   x 2  9
8
D
E
8
8
4
A
20 - x
B
x
4
B
C
D
E
By ratio and proportion:
20
x

8 6 8 6 4 5
x  10.87
15. A. 5.47
A
c  a  b  2ab cos C
2
2
2
c 2  1989c 2  2ab cos C
1988c 2 994c 2

2ab
ab
994c 2
cos C
 cot C  ab
2A 
secC
ab
497c 2
cot C 
A
14
cos C 
497c 2
A
cot C

 994
cot A  cot B
c2
2A 
C
6
13
D
9
E
A
B
BE
13

sin  sin AEB
9
13

sin BAD sin AEB
BE
sin 

9
sin BAD
6
14

sin  sin ACD
CEBU: JRT Bldg. Imus Ave. Cebu City 0917 3239235 | MANILA: CMFFI Bldg. R. Papa St. Sampaloc 09176339235
EXCEL REVIEW CENTER
ECE REVIEW FOR APRIL 2020 BOARD EXAM
15  BE
14

sin CAE sin ACD
15  BE
sin 

14
sin CAE
BE 15  BE

9
14
BE  5.87
BC 
BD   3  9  tan 
A 60

2
7
1
 3  9  BC
2
1
ABD   3  9  BD
2
3
AG 
A
cos 
2




1
3
A
AGF  10  
 sin 
A
2
2
 cos  
2

DCFG  ABC  ABD  AGF
DCFG  148
ABC 
sin P  0.8142
P  sin 1  0.8142 
P  0.9513
cosQ  0.4432
Q  1.111
tan  P  Q   tan  0.9513  1.111
tan  P  Q   1.868
c  181.5
23. A. 33
y
h
19. B. 144
x
B
Using Pythagorean theorem:
h 2  y 2  62 and h 2  x 2  152
y 2  36  x 2  225
x 2  y 2  189
 x  y  x  y   189
P
BC2  36  x 2  6x
C
Let S = area of ABC
Three small triangles are similar to
ABC . The ratio of the square roots of
their area is the same as the ratio of their
corresponding sides.
c=q+p+r
2
q

24.
S c
3
p

S c
7
r

S c
237 pq  r c

 1
c
c
S
As both x and y are integers, (x + y)
and (x – y) must be integral divisors of
189.
The pairs of divisors of 189 are (1,
189), (3, 63), (7, 27), (9, 21).
This yields the four potential sets of
(x, y) as (95, 94), (33, 30), (17, 10)
and (15, 6).
S  12
The last is not a possibility since it
simply degenerates into a line.
s  144
The sum of the three possible
perimeters of ACD is equal to, S:
20. B. angle bisectors
S  3  AC   2  x1  x 2  x 3 
21. A. orthocenter
S  90  2  95  33  17   380
a
37.416°
56.283°
c =?
A = 37°25’ = 37.416°
B = 56°17’ = 56.283°
A  B  C  180
37.416  56.283  C  180
C  86.301
c

A. 315
C
E
84
D
P
40
35
30
Similarly, ACF and BCF share the
same altitude from C, so the ratio of
their areas is the sam as the ratio of their
bases.
b
By sine law:
sin 86.301
x 2  10x  100  x 2  6x  36  196
x  33
B
F
APF and BPF share the same
altitude from P, so the ratio of their
areas is the same as the ratio of their
bases.
C
B
CA 2  100  x 2  10x
By Pythagorean Theorem:
AB2  BC2  CA 2
A
22. B. 181.5 m
18. B. 148
 3 9 
o
A  cos 1 
  71.7
 10  27 
A
  35.85o
2
Using Pythagorean Theorem:
A
C
B
Let x = length of PC
Since APB  BPC  CPA ,
each of them is equal to 120o.
Using cosine law to the smaller
triangles:
AB2  36  100  60  196
D
x
A
 sin 86.301 
a 
 c  0.609 c

 sin 37.416 
1
A triangle  ac sin B
2
1
8346  ( 0.609 c )( c ) sin 56.283)
2
16 , 692  0.5065 c 2
 35
So, area:
16. B. – 1.865
Using calculator:
Radian Mode :
17. D. 380
10  27 2   3  9 2
TRIGONOMETRY
sin 37.416
A
Also, the two pairs of bases are actually
the same, thus in the
same ratio. So,
40 124  APE

30 65  CPD
3APE  4CPD  112
 Eq.1
Applying identical reasoning to the
triangles with bases CD and BD, we ge
CPD APE  CPD  84

35
105
APE  2CPD  84  Eq.2
Using your calculator:
a
DAVAO: 2nd Floor, MERCO Bldg. Rizal St cor Bolton St | BAGUIO: 4th Floor, De Guzman Bldg. Legarda Road
EXCEL REVIEW CENTER
ECE REVIEW FOR APRIL 2020 BOARD EXAM
28. D. 33.51 m
APE  56
35°
CPD  70
So the total area,
A = 56 + 40 + 30 + 35 + 70 + 84 = 315
13°
50-h
50
25. A. 34.64 cm
h
A
x
x
x
x
x
13
35
r
50
B
30°
C
50
TRIGONOMETRY
  tan 1 m
19  5
PQ  tan 1
 126.87o
15  8
7  5
PR  tan 1
 73.74o
1 8
So the inclination of the angle bisector:
PQ  PR
  PR 
2
o
  100.305
11
m  tan   
2
Equation of the line:
y  y1  m  x  x1 
0.5x
Since equilateral triangle, A = B = C =
60°
r
10
tan 30 

r
30
0.5 x 0.5 x
0.5x
x  34.64 cm
tan 35 
h
30
1
 h 
  tan 1  
3
 30 
h
tan    
150
 h 
  tan 1 

 150 
 h  1 1  h 
tan 1 
  tan  
 150  3
 30 
1
h
h



1 
tan 1 
  tan    0
 150  3
 30 
Substitute the choices to determine
which one will satisfy the equation:
Answer: 56.7
tan  3  
y5  
x
50 - h
50 - h

x
tan13 ( 71.407 )  50 - h
11
 x  8
2
11x  2y  78  0
a  c  89
x  71.407 m
tan 13 
26. B. 56.7 ft
50
71.407
32. D. 44162
h  33.514 m
29. D. sin B < 0 and cos B > 0
a
B
-b
c
In the 4th quadrant:
opposite side - b
b
sin B 

hypotenuse
c
c
cos B 
adjacent side

a
1
2
b
c
cos  52  cos  60  400  200 


0
 sin  52  sin  60 

a
hypotenuse
c
Thus, sin B < 0 and cos B > 0
27. C. 0.169 m/sec
30. B. first reflected in the horizontal
axis and then translated
vertically pi/2 units upward
tan 60 
Whereas the fundamental identity for
the trigonometric function is sin2 x +
cos2x = 1, the fundamental identity for
the inverse trigonometric function is
arcsin x + arccos x = /2. Thus arccos x
= /2 – arcsin x. The curve of arcsin x
reflected in the horizontal axis will
represet the curve of – arcsin x. Adding
/2 is geometrically equivalent to
translating the curve vertically /2 units
upward.
44
x
x  25.4 m
31. C. 89
44
tan 30 
S  25.4
S  25.4  76.21
X = 186.808 represents the length XB
from the figure.
Y = 169.979 represents the length ZA
from the figure.
From the right triangle formed by the
hypotenuse XB = 186.808 with base
angle X = 52O, solve the height of the
right triangle using “Rec” function.
S  50.81 m
V
S
t

50.81
 0.169 m / s
5 ( 60 )
CEBU: JRT Bldg. Imus Ave. Cebu City 0917 3239235 | MANILA: CMFFI Bldg. R. Papa St. Sampaloc 09176339235
EXCEL REVIEW CENTER
ECE REVIEW FOR APRIL 2020 BOARD EXAM
TRIGONOMETRY

Now Y=147.2067129 represents the
height of the right triangle formed,
which is also the height of the
trapezoid ABXZ. To solve the area of
ABXZ, use the area of trapezoid:
ab
Area  
 (h)
 2 

Therefore, the radius of the inscribed
circle is 2.45 and the radius of the
circumscribed circle is 7.14.
Therefore, the answer is A.
34. A. 240
b=30
Therefore, the area of the trapezoid
ABXZ is
D. 44162.
a=16
c = 34
S
33. A. r = 2.45, R = 7.14
S
abc
2
16  30  34
2
S  40
A  s ( s - a )( s - b )( s - c )
A  40 ( 40 -16 )( 40 - 30 )( 40 - 34 )
A  240 square units
35. C. 83
MAC  7o
MCB  23o
CMA  180o  7 o  23o
CMA  150o
X
ABC
:
2
D  X(X-A)(X-B)(X-C) :


D ABC
:
X 4D
MCB  106o  23o  83o
If CMB  
So,
CBM  180o  83o    97o  
Using sine law:
sin150o AC

CM
sin 7o
sin 
BC

sin  97    CM
Since AC = BC,
sin150o
sin 

o
sin 7
sin 97o  


  83o
EXCEED EXPECTATIONS

DAVAO: 2nd Floor, MERCO Bldg. Rizal St cor Bolton St | BAGUIO: 4th Floor, De Guzman Bldg. Legarda Road