In right triangle Δ ABC, ∠ B = 90o, tan C = 1/2. If AC = 6 cm, then AB =
6/√5 cm
2 cm
3/√5 cm
√5 cm
If a sin θ = 1 and b tan θ = 1, then the relation between a and b is
a2 = b2
a2 - b2 = 1
a2 + b2 = 1
a2 + b2 = -1
If cos θ = 3/4, then sin θ =
3/4
4/3
√7/4
√2/3
In the figure find sin (90o - θ )
4/5
3/5
5/3
5/4
If cosec θ = √5, then cot θ - cos θ =
1/5
1/3
If 2 sin θ = √3, then cos θ =
√3/2
2/√3
1/2
2
If sec θ = 2 then sin θ =
3/2
1/√3
In Δ PQR,∠ Q = 90o, PQ = 4 cm and PR = 9 cm, then the value of sinP is
10/9
√95/9
√65/9
√17/4
If tan θ = 5/12, then sec θ =
10/24
25/144
13/12
15/14
