Advertisements
Advertisements
Question
From the trigonometric table, write the values of cos 23°17'.
Solution
From the tables of natural cosines, we have cos 23°12' = 0.9191
Mean difference for 5' = 0.0006 (to be subtracted)
cos 23°17' = 0.9185.
APPEARS IN
RELATED QUESTIONS
In the below given figure, a tower AB is 20 m high and BC, its shadow on the ground, is 20√3 m long. Find the sun’s altitude.
Without using tables evaluate: 3cos 80°. cosec 10° + 2sin 59° sec 31°
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`((1+tan^2A)/(1+cot^2A))=((1-tanA)/(1-cotA))^2=tan^2A`
Without using trigonometric tables, evaluate :
`sin 16^circ/cos 74^circ`
Without using trigonometric tables, prove that:
(sin72° + cos18°)(sin72° − cos18°) = 0
Prove that:
cos1° cos2° cos3° ... cos180° = 0
If sec2A = cosec(A - 42°), where 2A is an acute angle, then find the value of A.
If 5 tan θ = 4, find the value of `(5 sin θ + 3 cos θ)/(5 sin θ + 2 cos θ)`
From trigonometric table, write the values of sin 37°19'.
Solve the following equation: `(cos^2θ - 3 cosθ + 2)/sin^2θ` = 1.
