Advertisements
Advertisements
प्रश्न
If tan 2 A = cot (A − 12°), where 2 A is an acute angle, find the value of A.
उत्तर
\[\begin{array}{l}sin3A = \cos(A - {26}^\circ ) \\ \end{array}\]
\[\begin{array}{l}\Rightarrow cos( {90}^\circ - 3A) = \cos(A - {26}^\circ )[ \because \sin\theta = \cos( {90}^\circ- \theta)] \\ \end{array}\]
\[\begin{array}{l}\Rightarrow {90}^\circ- 3A = A - {26}^\circ \\ \end{array}\]
\[\begin{array}{l}\Rightarrow {116}^\circ = 4A \\ \end{array}\]
\[ \Rightarrow A = \frac{{116}^\circ}{4} = {29}^\circ \]
APPEARS IN
संबंधित प्रश्न
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 trigonometric tables, evaluate
`sin^2 34^@ + sin^2 56^@ + 2tan 18^@ tan 72^@ - cot^2 30^@`
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 :
`cos 35^circ/sin 55^circ`
Without using trigonometric tables, prove that:
tan266° − cot224° = 0
Without using trigonometric tables, prove that:
(sin 65° + cos 25°)(sin 65° − cos 25°) = 0
Without using trigonometric tables, prove that:
sin35° sin55° − cos35° cos55° = 0
Without using trigonometric tables, prove that:
(sin72° + cos18°)(sin72° − cos18°) = 0
Prove that:
`cos 80^circ/(sin 10^circ) + cos 59^circ "cosec" 31^circ = 2`
Prove that:
sin θ cos (90° - θ ) + sin (90° - θ) cos θ = 1
Prove that:
\[cot\theta \tan\left( 90° - \theta \right) - \sec\left( 90° - \theta \right)cosec\theta + \sqrt{3}\tan12° \tan60° \tan78° = 2\]
Prove that:
cos1° cos2° cos3° ... cos180° = 0
If sin 3 A = cos (A − 26°), where 3 A is an acute angle, find the value of A.
Prove the following:
`1/(1+sin^2theta) + 1/(1+cos^2theta) + 1/(1+sec^2theta) + 1/(1+cosec^2theta) = 2`
Without using tables evaluate: `(2tan 53°)/(cot 37°) - (cot 80°)/(tan 10°)`.
Solve : Sin2θ - 3sin θ + 2 = 0 .
Using trigonometric table evaluate the following:
tan 25°45' + cot 45°25'.
`(sin 40° + cos 50°)/(tan 38°20')`
The length of a shadow of a tower standing on a level plane is found to be 2y meters longer when the seen's altitude is 30° than when it was 45° prove that the height of the tower is y ( √3 + 1 ) meter.
