हिंदी

In ΔABC, if a = 18, b = 24, c = 30 then find the values of sinA. - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

In ΔABC, if a = 18, b = 24, c = 30 then find the values of sinA.

योग
Advertisements

उत्तर

Given: a = 18, b = 24 and c = 30

`cos A = (b^2+c^2-a^2)/(2bc)`

`cos A = (24^2+30^2-18^2)/(2xx24xx30)`

= `(576+900-324)/1440`

= `1152/1440`

= `4/5`

`sin^2 A = 1-cos^2 A`

= `1 - (4/5)^2`

= `1- 16/25 = 9/25`

= `sinA = sqrt(9/25)`

= `3/5`

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Trigonometric Functions - Exercise 3.2 [पृष्ठ ८८]

APPEARS IN

बालभारती Mathematics and Statistics 1 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 3 Trigonometric Functions
Exercise 3.2 | Q 10.6 | पृष्ठ ८८

वीडियो ट्यूटोरियलVIEW ALL [3]

संबंधित प्रश्न

If sin−1 x = y, then ______.


Evaluate the following:

`tan^-1(tan  (5pi)/6)+cos^-1{cos((13pi)/6)}`


Evaluate the following:

`cot^-1{2cos(sin^-1  sqrt3/2)}`


Evaluate the following:

`\text(cosec)^-1(-2/sqrt3)+2cot^-1(-1)`


Evaluate: tan `[ 2 tan^-1  (1)/(2) – cot^-1 3]`


In ΔABC, if a = 18, b = 24, c = 30 then find the values of tan `A/2`


Find the principal value of the following: `sin^-1 (1/2)`


Evaluate the following:

`cos^-1(1/2) + 2sin^-1(1/2)`


Evaluate the following:

`"cosec"^-1(-sqrt(2)) + cot^-1(sqrt(3))`


Prove the following:

`cos^-1(3/5) + cos^-1(4/5) = pi/(2)`


Prove the following:

`tan^-1["cosθ + sinθ"/"cosθ - sinθ"] = pi/(4) + θ, if θ ∈ (- pi/4, pi/4)`


In ΔABC, prove the following:

`(cos A)/a + (cos B)/b + (cos C)/c = (a^2 + b^2 + c^2)/(2abc)`


Find the principal solutions of the following equation:

sin 2θ = `− 1/(sqrt2)`


Find the principal solutions of the following equation:

cot 2θ = 0.


sin−1x − cos−1x = `pi/6`, then x = ______


Show that `sin^-1 (- 3/5) - sin^-1 (- 8/17) = cos^-1 (84/85)`


Express `tan^-1 ((cos x - sin x)/(cos x + sin x))`, 0 < x < π in the simplest form.


Find the principal value of `sec^-1 (- sqrt(2))`


Find the principal value of `tan^-1 (sqrt(3))`


A man standing directly opposite to one side of a road of width x meter views a circular shaped traffic green signal of diameter ‘a’ meter on the other side of the road. The bottom of the green signal Is ‘b’ meter height from the horizontal level of viewer’s eye. If ‘a’ denotes the angle subtended by the diameter of the green signal at the viewer’s eye, then prove that α = `tan^-1 (("a" + "b")/x) - tan^-1 ("b"/x)`


The value of cot `(tan^-1 2x + cot^-1 2x)` is ______ 


In ΔABC, tan`A/2 = 5/6` and tan`C/2 = 2/5`, then ______


If sin `(sin^-1  1/3 + cos^-1 x) = 1`, then the value of x is ______.


`cos(2sin^-1  3/4+cos^-1  3/4)=` ______.


The value of `sin^-1(cos  (53pi)/5)` is ______ 


The domain of the function y = sin–1 (– x2) is ______.


If 2 tan–1(cos θ) = tan–1(2 cosec θ), then show that θ = π 4, where n is any integer.


When `"x" = "x"/2`, then tan x is ____________.


If `"x + y" = "x"/4` then (1+ tanx)(1 + tany) is equal to ____________.


If 6sin-1 (x2 – 6x + 8.5) = `pi`, then x is equal to ____________.


`2"tan"^-1 ("cos x") = "tan"^-1 (2 "cosec x")`


If a = `(2sin theta)/(1 + costheta + sintheta)`, then `(1 + sintheta - costheta)/(1 + sintheta)` is 


If A = `[(cosx, sinx),(-sinx, cosx)]`, then A1 A–1 is 


sin 6θ + sin 4θ + sin 2θ = 0, then θ =


What is the value of `sin^-1(sin  (3pi)/4)`?


Values of tan–1 – sec–1(–2) is equal to


`tan^-1  (1 - x)/(1 + x) = 1/2tan^-1x, (x > 0)`, x then will be equal to.


`2tan^-1 (cos x) = tan^-1 (2"cosec"  x)`, then 'x' will be equal to


Consider f(x) = sin–1[2x] + cos–1([x] – 1) (where [.] denotes greatest integer function.) If domain of f(x) is [a, b) and the range of f(x) is {c, d} then `a + b + (2d)/c` is equal to ______. (where c < d) 


If y = `tan^-1  (sqrt(1 + x^2) - sqrt(1 - x^2))/(sqrt(1 + x^2) + sqrt(1 - x^2))`, then `dy/dx` is equal to ______.


Find the value of `cos(x/2)`, if tan x = `5/12` and x lies in third quadrant.


If tan 4θ = `tan(2/θ)`, then the general value of θ is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×