High School Mathematics - 2 10.18 Trignometry Review

Function

0^{o}

30°

45°

60°

90°

sin

Ö0/2

Ö1/2

Ö2/2

Ö3/2

Ö4/2

cos

Ö4/2

Ö3/2

Ö2/2

Ö1/2

0

tan

0

Ö3/3

1

Ö3

undefined

sec

1

2Ö3/3

Ö2

2

undefined

csc

undefined

2

Ö2

2Ö3/3

1

cot

undefined

Ö3

1

Ö3/3

0

sin

cos

tan

0

0

1

0

90

1

0

infinity

180

0

-1

0

270

-1

0

infinity

Functions of angles in all quadrants in terms of those in quadrant 1

-A

90^{0} ± A P/2 ± A

180^{0} ± A P ± A

270^{0} ± A 3P/2 ± A

360^{0} ± A 2P ± A

sin

-sinA

cos A

±sin A

-cosA

±sin A

cos

cos A

±sin A

-cos A

±sin A

cos A

tan

-tan A

±cot A

± tan A

±cot A

±tan A

csc

-csc A

sec A

± csc A

-sec A

±csc A

sec

sec A

± csc A

-sec A

±csc A

sec A

cot

-cot A

± tan A

± cot A

±tan A

± cot A

Q 1: What is the angle of elevation of the sun when the length of the shadow of a pole is 3^{1/2} times the height of the pole? 60^{o} 30^{o} 90^{o}

Q 2: The angle of elevation of the top of a tower which is yet incomplete at a point 120 metres from its base is 45^{o}. How much higher should it be raised so that the elevation at the same point may become 60^{o}? 87.84 metres 100 metres 36.85 metres

Q 3: In triangle ABC b = 4, c = 6, B = 30^{o}, find sin C. 4/3 3/4 1/2

Q 4: The angles of a triangle ABC are in AP and b:c = 3^{1/2}:2^{1/2}. Find angle A. 45 degrees 60 degrees 75 degrees

Q 5: From the light house the angles of depression of two ships on opposite sides of the light house are observed to be 30^{o} and 45^{o}. If the height of the light house is 300 metres, find the distance between the ships if the line joining them passes through foot of the light house. 819.6 metres 809 metres 800 metres

Q 6: If tan C = 11, find sin C. cos C. Answer:

Q 7: Find the degree measure of 3/4. 45^{o} 42^{0}57^{l}18^{ll} 35^{o}

Q 8: The sines of the angles of a triangle are in the ratio of 4:5:6, find the ratio of the cosines of the angles. 4:5:2 12:9:2 2:3:4

Question 9: This question is available to subscribers only!

Question 10: This question is available to subscribers only!