# 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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