Advertisements
Advertisements
प्रश्न
If 8 tan A = 15, find sin A – cos A.
उत्तर
8 tan A = 15 find. Sin A – cos A
`tan A = 15/8`
tan A = `"Opposite side"/"adjacent side"`
Let x be the hypotenuse By applying theorem
𝐴𝐶2 = 𝐴𝐵2 + 𝐵𝐶2
𝑥2 = 152 + 82
𝑥2 = 225 + 64
𝑥2 = 289 ⇒ x = 17
`sin A = (AB)/(AC) = 15/17`
`sin A - cos A = 15/17 - 8/17`
`= 7/17`
APPEARS IN
संबंधित प्रश्न
In rectangle ABCD AB = 20cm ∠BAC = 60° BC, calculate side BC and diagonals AC and BD.
In a ΔABC , ∠B = 90° , AB = 12 cm and BC = 5 cm Find
(i) cos A (ii) cosec A (iii) cos C (iv) cosec C
Evaluate:
`(sin^2 30^0 + 4 cot^2 45^0-sec^2 60^0)(cosec^2 45^0 sec^2 30^0)`
Given: sec A = `( 29 )/(21), "evaluate : sin A" - 1/tan "A"`
In a right-angled triangle, it is given that A is an acute angle and tan A = `(5) /(12)`.
find the value of :
(i) cos A
(ii) sin A
(iii) ` (cosA+sinA)/(cosA– sin A)`
In the following figure:
AD ⊥ BC, AC = 26 CD = 10, BC = 42, ∠DAC = x and ∠B = y.
Find the value of :
(i) cot x
(ii) `1/sin^2 y – 1/tan^2 y`
(iii) `6/cos x – 5/cos y + 8 tan y`.
If 5 cos = 6 sin ; evaluate:
(i) tan θ
(ii) `(12 sin θ – 3 cos θ)/(12 sin θ + 3 cos θ)`
In ΔABC, ∠B = 90°. If AB = 12units and BC = 5units, find: cos C
In the given figure, PQR is a triangle, in which QS ⊥ PR, QS = 3 cm, PS = 4 cm and QR = 12 cm, find the value of: cot2P - cosec2P
If sin A = `(7)/(25)`, find the value of : cot2A - cosec2A