Triginometry problem with solution

Last Name 1
Student’s Name
Professor’s Name
Subject
Date
TASK 3
1. In the triangle ABC , AB = 6 cm and BC = 9 cm. Given that the area of the triangle is 18 cm2 ,
find the possible values of the angle B .
Solution:
Given: AB = 6cm & BC = 9cm, Area of triangle is 18cm2
To find: Value of angle B
Area of triangle ABC = 0.5 x AB x BC x sinB
18 = 0.5 x 6 x 9 x sinB
18/27 = sinB
SinB = 0.66
B = 41⁰
2. Find the exact value of tan 30 (6sin135 4cos 210 ) , giving your answer in the form a b + √6.
Solution:
tan 30⁰ (6sin135⁰ 4cos 210⁰ )
tan 30⁰ = √3/3
sin135⁰ = √2/2
cos 210⁰ = -√3/2, substitute the values
= √3/3 (6 x √2/2 + 4 x (-√3/2))
Last Name 2
= √3/3 (3√2 – 2√3)
=√6–2
Since, a +b √ 6
= -2 + √ 6 ,where a = -2 & b = 1
3. Show that for a small angle  , measured in radians, sin 3 Ɵ (1 -cos 2 Ɵ) = 6  3
Solution:
1 – cos 2 Ɵ = 1 – ( 1 – 2sin2 Ɵ) = 2sin2 Ɵ
Base in the figure shown:
If the angle is very small, sin Ɵ = Ɵ ,
Therefore: sin3 Ɵ = 3 Ɵ
&
sin2 Ɵ = Ɵ2 & 1 – cos2 Ɵ = 2 Ɵ2
sin 3 Ɵ (1 – cos 2 Ɵ ) = 6 Ɵ3
3 Ɵ * 2 Ɵ2 = 6 Ɵ3
So the answer is sin 3 Ɵ (1 – cos 2 Ɵ ) = 6 Ɵ3
4. Prove that sin Ɵ (3sin2Ɵ + 4 cos2 Ɵ ) /tan Ɵ = 3 cos Ɵ + cos3 Ɵ
Solution:
tan Ɵ = sin Ɵ/cos Ɵ, substitute
= sin Ɵ (3sin2Ɵ + 4 cos2 Ɵ ) /tan Ɵ
= sin Ɵ (3sin2Ɵ + 4 cos2 Ɵ ) / sin Ɵ/cos Ɵ , cancel out sin Ɵ
= cos Ɵ (3sin2Ɵ + 4 cos2 Ɵ )
remember 1 = sin2Ɵ + cos2 Ɵ &
, substitute
= cos Ɵ (3(1 – cos2 Ɵ ) + 4 cos2 Ɵ )
sin2Ɵ = 1 – cos2 Ɵ
Last Name 3
= cos Ɵ (3 – 3cos2 Ɵ + 4 cos2 Ɵ )
= 3 cos Ɵ – 3cos3 Ɵ + 4 cos3 Ɵ
= 3 cos Ɵ + cos3 Ɵ
5. Simplify (36 + 4 . 27 ÷ 4 – 18) ÷ 5.
Solution: MDAS Rule
(36 + 4 * 27 ÷ 4 – 18) ÷ 5
(36 + 27 – 18) ÷ 5
(63 – 18) ÷ 5
45 ÷ 5 = 9
6. Proper aerobic exercise involves exercising at a person’s correct training heart rate. To find the
correct training heart rate the following formulas are used.
a. Training heart rate = (maximum heart rate – resting heart rate) × 0.65 + resting heart rate
b. maximum heart rate = 220 – person’s age Find the training heart rate for a 50-year-old with a
resting heart rate of 65 beats per minute
Solution:
Maximum heart rate = 220 -50 =170 beats per minute
Training heart rate = (maximum heart rate – resting heart rate) × 0.65 +
resting heart rate
Training heart rate = (170 -65) x 0.65 + 65
Training heart rate = 133.25 beats per minute
Last Name 4
7. Translate the following into an algebraic expression:
the sum of five times a number and three
Solution:
5X + 3
8. Solve 4(x + 1) = –15.
Solution:
4(x + 1) = -15
4x + 4 = -15
4x = -15 – 4
4x = -19
x = -19/4
9. Evaluate | y + 7 | if y = –2.
Solution:
| y + 7 | if y = –2.
| -2 + 7 |
| -5 | = 5
The solution is 5
Last Name 5
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