Unlock a deep understanding of mathematics with our Module and Summary! Clear definitions, comprehensive discussions, relevant example problems, and step-by-step solutions will guide you through mathematical concepts effortlessly. Learn with a systematic approach and discover the magic in every step of your learning journey. Mathematics doesn't have to be complicated—let's make it simple and enjoyable!

1. QUESTIONS AND SOLUTIONS
BASIC TRIGONOMETRY
Written by :
SYAHLUL ERBI SYAPUTRA

2. QUESTIONS AND SOLUTIONS
1. α, β, 𝛾 are the angles in ∆ ABC with α, 𝛾 = 30𝑜
. If sin β =
4
5
. The value of sin α . cos 𝛾
equal…
Solution :
( )
( )
[ ]
( )
5
4
-
180
-
180
180
=
γ
+
α
sin
=
γ
+
α
sin
=
β
sin
γ
+
α
°
=
β
°
=
γ
+
β
+
α
( )
2
1
30
-
-
=
°
sin
=
β
sin
α
cos
γ
cos
.
α
sin
=
γ
α
sin .
+
=
γ
sin
.
α
cos
+
γ
cos
.
α
sin
=
γ
sin
α
cos
γ
cos
.
α
sin
5
4
2
1
-
20
13
2
1
10
13
10
13
2
=
×
=
γ
cos
.
α
sin
=
γ
cos
.
α
sin
.
2. If 𝑡𝑎𝑛 𝐴 =
1
3
and 𝑡𝑎𝑛 𝐵 =
1
2
, so that value of 𝐴 + 𝐵 equal…
Solution :
𝑡𝑎𝑛(𝑎 + 𝑏 ) =
𝑡𝑎𝑛 𝐴 + 𝑡𝑎𝑛 𝐵
1 − 𝑡𝑎𝑛𝐴. 𝑡𝑎𝑛𝐵
=
1
3
+
1
2
1 −
1
3
×
1
2
=
5
6
5
6
= 1
arc tan (1) = 45ᵒ

3. 3. isosceles right triangle, point D is in the middle of side AB. If side BAC = 90𝑜
, so
that value of tan side DCB equal…
Solution :
A D B
BC = AB√2
CD = √(𝐴𝐵) + (
1
2
𝐴𝐵)
2
CD = AB√
5
4
3
𝐵𝐷2
= 𝐶𝐷2
+ 𝐵𝐶2
− 2𝐶𝐷𝐵𝐶 𝑐𝑜𝑠𝐷𝐵𝐶
(
1
2
𝐴𝐵)
2
= (𝐴𝐵√
5
4
)
2
+ (𝐴𝐵√2)
2
− 2 (𝐴𝐵√
5
4
) (𝐴𝐵√2) cos(𝐷𝐶𝐵)
(
1
2
)
2
= (√
5
4
)
2
+ (√2)
2
− 2 (√
5
4
) (√2) cos(𝐷𝐶𝐵)
cos(𝐷𝐶𝐵) =
3
√10
So that tan DCB =
1
3
4. The value from tan 22,5𝑜
– cot 22,5𝑜
equal…
Solution :
𝑡𝑎𝑛
𝜃
2
= √
1 − 𝑐𝑜𝑠𝜃
1 + 𝑐𝑜𝑠𝜃
𝑡𝑎𝑛
45
2
= √
1 − 𝑐𝑜𝑠45
1 + 𝑐𝑜𝑠45
C

4. = √
1 −
1
2 √2
1 +
1
2 √2
= √
1
2
(2 − √2)
1
2
(2 + √2)
= √
2 − √2
2 + √2
= √
2 − √2
2 + √2
×
2 − √2
2 − √2
= √(2 − √2)
2
4 − 2
=
2 − √2
√2
=
√2(√2 − 2)
√2
= √2 − 1
𝑡𝑎𝑛 22,5 = √2 − 1
so that value from
tan 22,5 – cot 22,5 = √2 − 1 −
1
√2 − 1
=
3 − 2√2 − 1
√2 − 1
=
2 − 2√2
√2 − 1
=
2 − 2√2
√2 − 1
×
√2 + 1
√2 + 1
=
2√2 + 2 − 2.2 − 2√2
2 − 1
=
2 − 4
1
= −2

5. 5. If tan
𝐴
2
= x. value from sin A + tan A equal
2
2
1
1
2
1
2
1
2
x
A
x
x
A
x
x
A
+
=
cos
+
=
sin
=
=
tan
( )
2
2
2
2
2
-
1
2
2
-
1
2
2
1
2
1
1
1
2
2
1
2
1
2
x
x
A
A
A
x
x
x
x
x
A
A
A
=
/
tan
tan
=
tan
+
=
+
+
.
=
cos
.
sin
=
sin
( ) ( )
4
4
2
2
2
2
-
1
4
-
1
1
2
-
1
2
-
1
2
1
2
x
x
x
x
x
x
x
x
x
x
x
A
A
=
+
+
=
+
+
=
tan
+
sin
6. The value from
sin 7 𝑥+sin3𝑥
cos 7 𝑥+cos3 𝑥
equal…
Solution :
𝑠𝑖𝑛7𝑥 + 𝑠𝑖𝑛3𝑥
𝑐𝑜𝑠7𝑥 + 𝑐𝑜𝑠3𝑥
=
(2𝑠𝑖𝑛5𝑥𝑐𝑜𝑠2𝑥)
(2𝑐𝑜𝑠5𝑥𝑐𝑜𝑠2𝑥)
=
𝑠𝑖𝑛5𝑥
𝑐𝑜𝑠5𝑥
= 𝑡𝑎𝑛5𝑥
7. The value from sin 50𝑜
+ cos 80𝑜
+ cos 160𝑜
equal…
Solution :
𝑠𝑖𝑛50ᵒ + 𝑐𝑜𝑠80ᵒ + 𝑐𝑜𝑠160ᵒ
𝐶𝑜𝑠 160ᵒ + 𝑐𝑜𝑠80ᵒ + 𝑠𝑖𝑛50ᵒ
𝑠𝑖𝑛𝜃 = 𝑐𝑜𝑠 (
𝜋
2
− 𝜃) = 𝑐𝑜𝑠(90 − 𝜃), so that
𝑠𝑖𝑛50ᵒ = 𝑐𝑜𝑠(90– 50) = 𝑐𝑜𝑠 40ᵒ
𝑐𝑜𝑠160ᵒ + 𝑐𝑜𝑠80ᵒ + 𝑐𝑜𝑠 40ᵒ is

6. 𝐶𝑜𝑠 𝐴 + 𝑐𝑜𝑠𝐵 = 2𝑐𝑜𝑠
1
2
(𝐴 + 𝐵)𝑐𝑜𝑠
1
2
(𝐴 − 𝐵)
𝑐𝑜𝑠160 + 𝑐𝑜𝑠80 = 2𝑐𝑜𝑠
1
2
(160 + 80)𝑐𝑜𝑠
1
2
(160 − 80)
𝑐𝑜𝑠160 + 𝑐𝑜𝑠80 = 2𝑐𝑜𝑠
1
2
(240)𝑐𝑜𝑠
1
2
(80)
𝑐𝑜𝑠160 + 𝑐𝑜𝑠80 = 2 (−
1
2
) 𝑐𝑜𝑠40
𝑐𝑜𝑠160 + 𝑐𝑜𝑠80 = −𝑐𝑜𝑠40
So that (𝑐𝑜𝑠160 + 𝑐𝑜𝑠80) + 𝑐𝑜𝑠40 = (−𝑐𝑜𝑠40 + 𝑐𝑜𝑠40) = 0
8. If tan a + tan b = √6 and cos a cos b =
1
6
√3, so that sin (a+b) equal…
Solution :
𝑡𝑎𝑛 𝑎 =
𝑠𝑖𝑛 𝑎
𝑐𝑜𝑠 𝑎
𝑡𝑎𝑛 𝑏 =
𝑠𝑖𝑛 𝑏
𝑐𝑜𝑠 𝑏
𝑠𝑖𝑛(𝑎 + 𝑏) = 𝑠𝑖𝑛 𝑎 𝑐𝑜𝑠 𝑏 + 𝑐𝑜𝑠 𝑎 𝑠𝑖𝑛 𝑏
𝑡𝑎𝑛 𝑎 + 𝑡𝑎𝑛 𝑏 = √6
sin 𝑎
cos 𝑎
+
sin 𝑏
cos 𝑏
= √6 (𝑘𝑒𝑑𝑢𝑎 𝑟𝑢𝑎𝑠 𝑑𝑖𝑘𝑎𝑙𝑖 cos 𝑎 cos 𝑏)
(sin 𝑎 cos 𝑏 + cos 𝑎 sin 𝑏) = √6 cos 𝑎 cos 𝑏
𝑠𝑖𝑛(𝑎 + 𝑏) = √6 (
1
6
√3)
=
1
6
√18 =
1
6
. 3√2 =
3
6
√2 =
1
2
√2
9. If cos x = sin 150 and for 0≤ 𝑥 ≤ 2𝜋, so that value of x is
( )
( )
60
60
-
150
-
90
150
cos
=
cos
cos
=
cos
=
sin
=
cos
x
x

7. Solution :
360
60
-
360
60
2
1
.
+
=
.
+
=
k
a
k
a
In order to fulfill, so that k =0,1
{ }
{ }
{ }
°
,
°
=
°
=
°
=
300
60
300
60
2
1
a
a
a
10. The value of
sin 270 cos135 tan 135
sin 150 cos225
equal…
Solution :
𝑠𝑖𝑛270 𝑐𝑜𝑠135 𝑡𝑎𝑛135
𝑠𝑖𝑛150 𝑐𝑜𝑠225
=
(−1) (−
1
2 √2) (−1)
(
1
2
) (−
1
2 √2)
=
−
1
2 √2
−
1
4 √2
= −
1
2
× −4 = 2
If A side lancip yang memenuhi persamaan 2 𝑐𝑜𝑠4
𝐴 = 3 𝑠𝑖𝑛 2
𝐴, so that tan A is
( )
( ) x
x
x
x
x
x
2
2
2
2
2
2
2
4
-
1
2
2
2
sin
=
sin
sin
=
cos
sin
=
cos
( )
( )
( )( ) 0
2
-
1
-
2
0
2
3
-
2
2
4
-
2
2
-
1
2
-
1
2
2
2
2
2
2
=
=
+
=
+
=
+
=
=
sin
:
p
p
p
p
p
p
p
p
p
p
p
p
p
x
misal

8. 2
1
1
2
0
1
-
2
=
=
=
p
p
p
2
0
2
-
=
=
p
p
2
1
2
=
=
sin p
x
.
=
=
sin
TM
p
x
2
2
1
2
2
1
=
tan
=
sin
x
x
11. A, B, and C is the sides of a triangle. If A-B = 30𝑜
and sin C =
5
6
, so that A cos B
equal…
Solution :
in a triangle ABC, A +B +C =180, so that
𝐴 + 𝐵 + 𝐶 = 180
𝐶 = 180 − (𝐴 + 𝐵)
sin 𝐶 = 𝑠𝑖𝑛[180 − (𝐴 + 𝐵)]
5
6
= sin 𝐴 cos 𝐵 + cos 𝐴 sin 𝐵
𝑠𝑖𝑛(𝐴 − 𝐵) = sin 𝐴𝑐𝑜𝑠 𝐵 − cos 𝐴 sin 𝐵
𝑠𝑖𝑛30 = sin 𝐴𝑐𝑜𝑠 𝐵 − cos 𝐴 sin 𝐵
1
2
= sin 𝐴𝑐𝑜𝑠 𝐵 − cos 𝐴 sin 𝐵
Elimination
5
6
= sin 𝐴 cos 𝐵 + cos 𝐴 sin 𝐵
1
2
= sin 𝐴𝑐𝑜𝑠 𝐵 − cos 𝐴 sin 𝐵
1
3
= 2 cos 𝐴 sin 𝐵
cos 𝐴 sin 𝐵 =
1
6

9. 12. A ship sails east for 30 miles. Then continue the journey with the direction of 30𝑜
as far as 60 miles. The distance of the ship to the position when the ship departs
is…mil
Solution :
For example, the ship's initial position is at A. It moves eastward 30 miles to B (AB
= 30 miles). Then head to C for 60 miles with direction 30ᵒ (BC = 60 mill). So that
side B is obtained at triangle ABC is 90ᵒ + 30ᵒ = 120ᵒ.
Cos 120ᵒ = cos (180ᵒ–60ᵒ)
= –cos 60ᵒ
=−
1
2
So that the distance of the ship to the position when the ship departs is: with the
cosine rule is obtained:
𝐴𝐶2
= 𝐵𝐶2
+ 𝐴𝐵2
− 2𝐵𝐶. 𝐴𝐵 𝑐𝑜𝑠𝐵
𝐴𝐶2
= 602
+ 302
− 2.60.30𝑐𝑜𝑠120
𝐴𝐶2
= 3600 + 900 − 2.1800 (−
1
2
)
𝐴𝐶2
= 3600 + 900 + 1800
𝐴𝐶2
= 6300
𝐴𝐶 = √6300 = √900.7 = 30√7
13. Know the equation cos 2x + cos x = 0, for 0≤ 𝑥 ≤ 𝜋 value of 𝑥 that satisfies is…
Solution :
cos 2x + cos x = 0
2𝑐𝑜𝑠2
𝑥 − 1 + 𝑐𝑜𝑠𝑥 = 0
(2 cos 𝑥 − 1)(cos 𝑥 + 1) = 0
cos 𝑥 =
1
2
cos 𝑥 = −1
For cos 𝑥 =
1
2
, 0 ≤ 𝑥 ≤ 𝜋
𝑥 =
𝜋
3
For cos 𝑥 = −1,0 ≤ 𝑥 ≤ 𝜋
𝑥 = 𝜋
𝜋
3
𝑑𝑎𝑛 𝜋

10. 14. Known cos (x-y) =
4
5
and sin x. sin y =
3
10
. The value of tan x . tan y is…
Solution :
𝑐𝑜𝑠 (𝑥 − 𝑦) =
4
5
𝑐𝑜𝑠 𝑥 𝑐𝑜𝑠 𝑦 + 𝑠𝑖𝑛 𝑥 𝑠𝑖𝑛 𝑦 =
4
5
𝑐𝑜𝑠 𝑥 𝑐𝑜𝑠 𝑦 =
4
5
−
3
10
=
1
2
𝑡𝑎𝑛 𝑥. 𝑡𝑎𝑛 𝑦 =
𝑠𝑖𝑛 𝑥. 𝑠𝑖𝑛 𝑦
𝑐𝑜𝑠 𝑥. 𝑐𝑜𝑠 𝑦
=
3
10
1
2
=
6
10
=
3
5
15. Shape of
2 𝑡𝑎𝑛𝑥
1+𝑡𝑎𝑛2𝑥
equivalent to …
Solution :
2 𝑡𝑎𝑛𝑥
1 + 𝑡𝑎𝑛2𝑥
=
2 sin 𝑥
cos 𝑥
1 +
𝑠𝑖𝑛2𝑥
𝑐𝑜𝑠2𝑥
=
2 sin 𝑥
cos 𝑥
𝑐𝑜𝑠2𝑥
𝑐𝑜𝑠2𝑥
+
𝑠𝑖𝑛2𝑥
𝑐𝑜𝑠2𝑥
=
2 sin 𝑥
cos 𝑥
𝑐𝑜𝑠2𝑥 + 𝑠𝑖𝑛2𝑥
𝑐𝑜𝑠2𝑥
=
2 sin 𝑥
cos 𝑥
1
𝑐𝑜𝑠2𝑥
=
2 sin 𝑥
cos 𝑥
1
𝑐𝑜𝑠2𝑥
×
𝑐𝑜𝑠2
𝑥
𝑐𝑜𝑠2𝑥
=
2 sin 𝑥 cos 𝑥
1
= sin 2𝑥

11. 16. If α acute angle, and cos 2α =
3
4
, so that sin α is
Solution :
𝑐𝑜𝑠 2𝛼 =
3
4
2𝑐𝑜𝑠2
𝐴 − 1 =
3
4
multiply both sides 4
8𝑐𝑜𝑠2
𝐴 − 4 = 3
8𝑐𝑜𝑠2
𝐴 = 3 + 4
8𝑐𝑜𝑠2
𝐴 = 7
𝑐𝑜𝑠2
𝐴 =
7
8
8𝑠𝑖𝑛2
𝐴 + 𝑐𝑜𝑠2
𝐴 = 1
8𝑠𝑖𝑛2
𝐴 +
7
8
= 1 multiply both sides 8
8𝑠𝑖𝑛2
𝐴 + 7 = 8
8𝑠𝑖𝑛2
𝐴 = 8 − 7
8𝑠𝑖𝑛2
𝐴 = 1
𝑠𝑖𝑛2
𝐴 =
1
8
sin 𝐴 = √
1
8
=
1
√8
×
√8
√8
=
1
8
√8
17. The area of the triangle below is
R
4
P45𝑜
6 Q
Solution :
𝐿 =
1
2
. 𝑃𝑅. 𝑃𝑄. sin 𝑃

12. 𝐿 =
1
2
× 4 × 6 × 𝑠𝑖𝑛45
𝐿 =
1
2
× 24 ×
1
2
√2
𝐿 = 6√2
18. The area of the triangle below is …
M
6 8
50𝑜
70𝑜
K L
Solution :
< 𝐾+< 𝐿+< 𝑀 = 180
50 + 70+< 𝑀 = 180
120+< 𝑀 = 180
< 𝑀 = 180 − 120
< 𝑀 = 60
𝐿 =
1
2
× 6 × 8 × sin 60
𝐿 =
1
2
× 48 ×
1
2
√3
𝐿 = 12√3
19. The area of a regular triangle whose radius is r units is…
Solution :
Area of regular n-side =
1
2
. 𝑛. 𝑟2
. 𝑠𝑖𝑛
360
𝑛
=
1
2
. 𝑛. 𝑟2
. 𝑠𝑖𝑛120
=
1
2
. 3. 𝑟2
.
1
2
√3
=
3
2
. 𝑟2
.
1
2
√3
=
3
4
𝑟2
√3

13. 20. Area of a regular triangle with side length a unit is…
Solution :
Area of regular n-side =
1
2
. 𝑛. 𝑟2
. 𝑠𝑖𝑛
360
𝑛
=
1
2
. 𝑛. 𝑟2
. 𝑠𝑖𝑛120
=
1
2
. 3. 𝑎2
.
1
2
√3
=
3
2
. 𝑎2
.
1
2
√3
=
3
4
𝑎2
√3
21. Area of triangle ABC with side 𝐵 = 100𝑜
side 𝐶 = 30𝑜
and side length a = 8 cm
is …. 𝑐𝑚2
Solution :
< 𝐴+< 𝐵+< 𝐶 = 180
< 𝐴 + 100 + 30 = 180
< 𝐴 + 130 = 180
< 𝐴 = 180 − 130
< 𝐴 = 50
𝐿 =
𝑎2
sin 𝐵. sin 𝐶
2𝑠𝑖𝑛𝐴
=
82
. 𝑠𝑖𝑛100. 𝑠𝑖𝑛30
2𝑠𝑖𝑛50
=
64(0.98)(0,50)
2(0,76)
=
31,36
1,52
= 20,63

14. 22. Around ∆ ABC is 24 cm, If length AB = 8 cm, AC = 9 cm, so that area ∆ ABC
is…𝑐𝑚2
Solution :
C
9
A 8 B
𝐾 = 𝑎 + 𝑏 + 𝑐 𝑠 =
1
2
(𝑎 + 𝑏 + 𝑐)
24 = 𝐴𝐵 + 𝐴𝐶 + 𝐵𝐶 =
1
2
(8 + 9 + 7)
24 = 8 + 9 + 𝐵𝐶 =
1
2
(24)
24 = 17 + 𝐵𝐶 =12
𝐵𝐶 = 7
𝐿 = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)
𝐿 = √12(12 − 7)(12 − 9)(12 − 7)
𝐿 = √12(5)(3)(4)
𝐿 = √720
𝐿 = 12√5
23. The parallelogram PQRS is known as PQ = 11 cm, PS = 10 cm and QS = 13 cm.
area of the parallelogram is…
Solution :
S R
P Q
10
0
11
13

15. 𝑠∆𝑃𝑄𝑆 =
1
2
(11 + 13 + 10)
=
1
2
(34)
= 17
𝐿∆𝑃𝑄𝑆 = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)
= √17(17 − 13)(17 − 10)(17 − 11)
= √17(4)(7)(6)
= √2856
= 53,44
So that the area of the parallelogram is = 2 × 𝐿∆𝑃𝑄𝑆
= 2 × 53,44
= 106,88
24. The following is the formula for the area of any triangle ABC, except …
Solution :
𝐿 =
𝑏2𝑆𝑖𝑛 𝐴 𝑆𝑖𝑛 𝐶
𝑆𝑖𝑛 𝐵
(B)
Because any formula ABC for If is known to have two sides and one side is:
a. 𝐿 =
𝑎2𝑆𝑖𝑛 𝐵 𝑆𝑖𝑛 𝐶
2𝑆𝑖𝑛 𝐴
b. 𝐿 =
𝑏2𝑆𝑖𝑛 𝐴 𝑆𝑖𝑛 𝐶
2𝑆𝑖𝑛 𝐵
c. 𝐿 =
𝑐2𝑆𝑖𝑛 𝐴 𝑆𝑖𝑛 𝐵
2𝑆𝑖𝑛 𝐶
25. In triangle ABC, it is known that side B = 600
, AB
̅̅̅̅ = 12√3, BC
̅̅̅̅ = 4. Area ∆ ABC
equal
Solution :
C
4
A 12√3 B

16. 𝐿 =
1
2
. 𝐴𝐵. 𝐵𝐶. 𝑆𝑖𝑛 𝐵
𝐿 =
1
2
× 12√3 × 4 × 𝑆𝑖𝑛 60
𝐿 =
1
2
× 48√3 ×
1
2
√3
𝐿 = 12 × 3 = 36
26. In triangle PQR, the length of side PQ = 6 cm, side PR = 10 cm and side 𝑅𝑃𝑄 =
1200
, so that length QR equal…
Solution :
R
10
120ᵒ
P 6 Q
cos 𝑃 =
𝑞2
+ 𝑟2
− 𝑝2
2𝑞𝑟
cos 120 =
102
+ 62
− 𝑝2
2.10.6
−
1
2
=
136 − 𝑝2
120
−120 = 272 − 2𝑝2
272 − 2𝑝2
= −120
−2𝑝2
= −120 − 272
−2𝑝2
= −392
𝑝2
=
392
2
𝑝2
= 196
𝑝 = √196 = 14

17. 27. In triangle ABC, it is known that AB = 3 cm, BC = 6 cm and AC = 5 cm , so that the
value of cos B is…
Solution :
C
5 6
A 3 B
cos 𝐵 =
𝑎2
+ 𝑐2
− 𝑏2
2𝑎𝑐
𝑐𝑜𝑠 𝐵 =
62
+ 32
− 52
2 × 6 × 3
cos 𝐵 =
36 + 9 − 25
36
cos 𝐵 =
20
36
=
5
9
28. In the picture below the value of sin c is
C
6
30𝑂
A 4 D B
5
2
20
16
-
36
4
-
6 2
2
=
=
=
=
CD
5
4
5
2
2
1
5
2
30
=
=
=
°
sin
=
sin
BC
BC
BC
BC
CD
B

18. ( ) ( )
15
2
60
60
20
80
5
2
5
4
2
2
2
2
2
2
2
2
=
=
=
+
=
+
=
+
=
BD
BD
BD
BD
BD
CD
BC
6
15
2
6
2
15
2
4
2
1
6
15
2
4
+
=
+
=
sin
=
sin
+
sin
=
sin
C
C
C
c
B
b
29. In triangle ABC, the ratio of the lengths of the sides is 2: 3: 4, the value of the
largest side cosine is…
Solution :
C
3 2
A 4 B
cos 𝐶 =
𝑎2
+ 𝑏2
− 𝑐2
2𝑎𝑏
cos 𝐶 =
22
+ 32
− 42
2.2.3
cos 𝐶 =
4 + 9 − 16
12
= −
3
12
cos 𝐶 =
1
4

19. 30. In triangle ABC applies the relationship 𝑐2
= 𝑎2
+ 𝑏2
− 𝑎𝑏, so that is the side
is…
Solution :
𝑐2
= 𝑎2
+ 𝑏2
− 2𝑎𝑏 cos 𝐶
𝑎2
+ 𝑏2
− 𝑎𝑏 = 𝑎2
+ 𝑏2
− 2𝑎𝑏 cos 𝐶
−𝑎𝑏 = −2𝑎𝑏 𝐶𝑜𝑠𝐶
2𝑎𝑏 𝐶𝑜𝑠𝐶 = 𝑎𝑏
𝐶𝑜𝑠 𝐶 =
𝑎𝑏
2𝑎𝑏
𝐶𝑜𝑠 𝐶 =
1
2
𝑎𝑟𝑐 𝑐𝑜𝑠 (
1
2
) = 60ᵒ
31. In triangle ABC apply 𝑎𝑏 = 𝑐2
− 𝑎2
− 𝑏2
correct question is…
Solution :
𝑐2
= 𝑎2
+ 𝑏2
− 2𝑎𝑏 cos 𝐶
𝑐2
− 𝑎2
− 𝑏2
= −2𝑎𝑏 cos 𝐶
𝑎𝑏 = −2𝑎𝑏 𝐶𝑜𝑠𝐶
𝐶𝑜𝑠 𝐶 = −
𝑎𝑏
2𝑎𝑏
𝐶𝑜𝑠 𝐶 = −
1
2
𝐶 = 𝑎𝑟𝑐 𝑐𝑜𝑠 (−
1
2
)
𝐶 = 120ᵒ
True statement is:
< 𝐴+< 𝐵 = 180− < 𝐶
= 180 − 120
= 60ᵒ

20. 32. Two ships are sailing in the same place forming a side of 120°, ship A has a speed of
80 km/hour and ship B has a speed of 40 km/hour. After 30 minutes the distance
between the two ships is…
Solution :
< (𝐴, 𝐵) = 120ᵒ
𝑠𝐴 = 80𝑘𝑚/𝑗𝑎𝑚 × 30𝑚𝑒𝑛𝑖𝑡
= 80𝑘𝑚/𝑗𝑎𝑚 ×
1
2
𝑗𝑎𝑚 = 40𝑘𝑚
𝑠𝐵 = 40𝑘𝑚/𝑗𝑎𝑚 × 30𝑚𝑒𝑛𝑖𝑡
= 40𝑘𝑚/𝑗𝑎𝑚 ×
1
2
𝑗𝑎𝑚 = 20𝑘𝑚
Distance AB = 𝑥2
𝑥2
= 𝑠𝐴2
+ 𝑠𝐵2
− 2𝑠𝐴. 𝑠𝐵𝐶𝑜𝑠𝛼
=402
+ 202
− 2.40.20. 𝑐𝑜𝑠120
=1600 + 400 − 2.1800 (−
1
2
)
=2000 + 800
=2800
𝑥 = √2800 = 20√7
33. In triangle ABC, 𝑎2
= 𝑏2
+ 𝑐2
√2 bc. The value from tan A is…
Solution :
𝑎2
= 𝑏2
+ 𝑐2
√2 bc
𝑎2
– 𝑏2
– 𝑐2
= √2 bc
Real formula:
𝑎2
= 𝑏2
+ 𝑐2
− 2𝑏𝑐𝐶𝑜𝑠 𝑎
𝑎2
− 𝑏2
− 𝑐2
= −2𝑏𝑐𝐶𝑜𝑠 𝑎
−2𝑏𝑐𝐶𝑜𝑠 𝑎 = √2 bc
𝐶𝑜𝑠 𝑎 = −
𝑏𝑐√2
2𝑏𝑐
𝐶𝑜𝑠 𝛼 = −
1
2
√2

21. 𝑎𝑐𝑟𝑐𝑜𝑠(1/2√2) = 225 so that tan 225 = 1
Look at the following picture.
P
0 𝛼
Q
The radius of the circle above is 2 cm. If PQ = 3 cm so that the value cos 𝛼𝑜
is…
34. In triangle ABC it is known that a = 9, side A = 60𝑜
, side B = 45𝑜
so that b is…
Solution :
𝑎
𝑠𝑖𝑛𝐴
=
𝑏
𝑠𝑖𝑛𝐵
9
𝑠𝑖𝑛60
=
𝑏
𝑠𝑖𝑛45
9
1
2 √3
=
𝑏
1
2 √2
√3𝑏 = 9√2
𝑏 =
9√2
√3
×
√3
√3
𝑏 =
9√6
3
= 3√6
35. In triangle ABC. It is known that side 𝐶 = 105𝑜
, side 𝐴 = 30𝑜
and BC = 4. Length
AC is…
Solution :
C
105ᵒ 4
30ᵒ
A B

22. < 𝐴+< 𝐵+< 𝐶 = 180
30+< 𝐵 + 105 = 180
< 𝐵 + 135 = 180
< 𝐵 = 45ᵒ
𝑎
𝑠𝑖𝑛𝐴
=
𝑏
𝑠𝑖𝑛𝐵
4
𝑠𝑖𝑛30
=
𝑏
𝑠𝑖𝑛45
4
1
2
=
𝑏
1
2 √2
𝑏 = 4√2
36. It is known that triangle PQR with PR = 3 cm and QR =
3√6
2
, side P = 60𝑜
. The size
of the side R is…
Solution :
R
3
3√6
2
60ᵒ
P Q
𝑝
𝑠𝑖𝑛𝑃
=
𝑞
𝑠𝑖𝑛𝑄
3√6
2
𝑠𝑖𝑛60
=
3
𝑠𝑖𝑛𝑄
3√6
2
1
2 √3
=
3
𝑠𝑖𝑛𝑄

23. 3√6
2
sin 𝑄 =
3
2
√3
𝑠𝑖𝑛𝑄 =
3
2
√3
3√6
2
=
3
2
√3 ×
2
3√6
=
√3
√6
×
√6
√6
=
3
6
√2
=
1
2
√2
𝑎𝑟𝑐 sin (
1
2
√2) = 45
So that < 𝑅 is: <P + <Q + <R= 180
60 + 45 + <R = 180
105 + <R = 180
<R = 75ᵒ
37. A triangle ABC and known side A = 45^o, ac = 2 cm. BC = 2√2 cm value of cos side
B is…
Solution :
𝑎
𝑠𝑖𝑛𝐴
=
𝑏
𝑠𝑖𝑛𝐵
2√2
𝑠𝑖𝑛45
=
2
𝑠𝑖𝑛𝐵
2√2
1
2 √2
=
2
𝑠𝑖𝑛𝐵

24. 2
1
2
=
2
𝑠𝑖𝑛𝐵
4 =
2
𝑠𝑖𝑛𝐵
𝑠𝑖𝑛𝐵 =
2
4
=
1
2
𝑎𝑟𝑐 sin (
1
2
) = 30ᵒ
So that 𝑐𝑜𝑠 < 𝐵 = 𝑐𝑜𝑠30ᵒ =
1
2
√3
38. Given a triangle PQR with length QR = √6, PR = √2 and side P = 60𝑜
so that length
PQ is…
Solution :
𝑝
𝑠𝑖𝑛𝑃
=
𝑞
𝑠𝑖𝑛𝑄
√6
𝑠𝑖𝑛60
=
√2
sin 𝑄
√6
1
2 √3
=
√2
𝑠𝑖𝑛𝑄
√6𝑠𝑖𝑛𝑄 =
1
2
√6
𝑠𝑖𝑛 𝑄 =
1
2 √6
√6
sin 𝑄 =
1
2
𝑎𝑟𝑐 sin (
1
2
) = 30ᵒ
So that < 𝑃+< 𝑄+< 𝑅 = 180
60+< 𝑅 + 30 = 180
<R+90=180
<R = 90

25. Q
√6
R √2 P
So that length PQ is: 𝑃𝑄 = √(√6)
2
+ (√2)
2
𝑃𝑄 = √6 + 2
𝑃𝑄 = √8 = 2√2
39. Look at the following picture!
60ᵒ
30ᵒ 45ᵒ
The value from sin D is
°
=
∠ 105
C
2
2
2
2
2
2
1
2
2
1
2
=
.
=
=
sin
=
sin
BC
BC
BC
A
BC
B
AC

26. 6
8
1
4
6
2
1
3
2
1
4
2
60
=
=
sin
=
sin
°
sin
=
sin
D
D
BD
D
BC
40. Given a triangle ABC with side length a = 3√2cm, b = 6cm, side A = 30𝑂
, the value
of tg side B is…
Solution :
𝑎
𝑠𝑖𝑛𝐴
=
𝑏
𝑠𝑖𝑛𝐵
3√2
𝑠𝑖𝑛30
=
6
𝑠𝑖𝑛𝐵
3√2
1
2
=
6
𝑠𝑖𝑛𝐵
3 = 3√2𝑠𝑖𝑛𝐵
𝑠𝑖𝑛𝐵 =
3
3√2
×
√2
√2
𝑠𝑖𝑛𝐵 =
3√2
6
=
1
2
√2
𝑎𝑟𝑐 𝑠𝑖𝑛 (
1
2
√2) = 45ᵒ =< 𝐵
So that value of tg <B = tg 45ᵒ = 1
41. A triangle PQR with side P = 9 cm, side 𝑃 = 120𝑂
, side 𝑄 = 45𝑂
, so that side
length q is…
𝑝
𝑠𝑖𝑛𝑃
=
𝑞
𝑠𝑖𝑛𝑄
9
𝑠𝑖𝑛120
=
𝑞
𝑠𝑖𝑛45

27. 9
1
2 √3
=
𝑞
1
2 √2
√3𝑞 = 9√2
𝑞 =
9√2
√3
×
√3
√3
𝑞 =
9√6
3
= 3√6
42. Given a triangle PQR where P = 10 cm, side 𝑃 = 45𝑂
, and side 𝑄 = 105𝑂
, side
length r is…
Solution :
< 𝑃+< 𝑄+< 𝑅 = 180
45 + 105+< 𝑅 = 180
150 + <R = 180
<R=30
𝑝
𝑠𝑖𝑛𝑃
=
𝑟
𝑠𝑖𝑛𝑅
10
𝑠𝑖𝑛45
=
𝑟
𝑠𝑖𝑛30
10
1
2 √2
=
𝑟
1
2
√2𝑟 = 10
𝑟 =
10
√2
×
√2
√2
𝑟 =
10√2
2
= 5√2
43. In triangle ABC it is known that side 𝐴 = 60𝑂
, side 𝐵 = 90𝑂
and AC = 12 cm,
length AC + BC is…
Solution :
𝑎
𝑠𝑖𝑛𝐴
=
𝑏
𝑠𝑖𝑛𝐵

28. 𝑎
𝑠𝑖𝑛60
=
12
𝑠𝑖𝑛90
𝑎
1
2 √3
=
12
1
𝑎 = 12.
1
2
√3
𝑎 = 6√3 equal side BC
So that length AC + BC = 12 + 6√3
44. Right triangle ABC at B, AB = 5 cm, BC
̅̅̅̅ = 12 cm. length of the circumcircle of
triangle ABC is…cm
Solution :
C
12
B 5 A
𝐴𝐶 = √122 + 52
𝐴𝐶 = √144 + 25
𝐴𝐶 = √169 = 13
The formula for finding the length of the outer radius of a triangle ABC is:
𝑟𝑙 =
𝑎𝑏𝑐
4√𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)
You have to find the value of s first, so that is obtained:
𝑠 =
1
2
(𝑎 + 𝑏 + 𝑐)
=
1
2
(12 + 13 + 5)
=
1
2
(30) = 15

29. So that the length of the outer radius of triangle ABC is:
𝑟𝑙 =
𝑎𝑏𝑐
4√𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)
=
12.13.5
4√15(15 − 12)(15 − 13)(15 − 5)
=
780
4√15(3)(2)(10)
=
780
4√900
=
780
4.30
=
780
120
= 6,5