Maine kr liya but 2 pages ka solution aaya hai.. Soln) ( to find cos theta) As per the question, Sin theta = (17-15costheta)/8...eq1 Also sin²theta= 1-cos² theta (identity) And if we square on both sides in equation no.1 We get two values of sin²theta , hence making it easy to solve the question.. Now as per eq 1 Sin² theta= (17-15costheta)²/64 So, 1-cos²theta=[289+225cos²theta-2(17)(15costheta)]/64 Further, 1-cos²theta=(289+225cos²theta-510costheta)/64 By cross multiplying, 64-64cos²theta=289+225cos²theta-510costheta 64-289=225cos²theta+64cos²theta-510costheta -225=225cos²theta+64 cos²theta - 510costheta 0=289cos²theta-510costheta+225 Now , the equation formed is quadratic and is containing 2-3 digit no.s making it tougher and more complex to use mid term spitting, so we shall use quadratic formula , which is x=(-b +- _/b²-4ac)/2a According to the equation, a=289 b=-510 c=225 Now, putting the values , Costheta= [-(-510) +- _/(-510)²-4(289)(225)]/2(289) After solving we get, Cos theta= (510+-_/260100-260100)/578 Cos theta= (510+0)/578, (510-0)/578 Cos theta =510/578,510/578 Cos theta = 15/17 answer.... Guys ek like toh banta hai yrr boht time laga type krne me
A trigonometric equation! We can solve this equation using the Pythagorean identity: sin²(θ) + cos²(θ) = 1 First, let's rewrite the equation: 8sin(θ) + 15cos(θ) = 17 Now, divide both sides by 17: (8/17)sin(θ) + (15/17)cos(θ) = 1 Let's introduce new variables: a = 8/17 b = 15/17 So, the equation becomes: a sin(θ) + b cos(θ) = 1 Now, we can use the Pythagorean identity: a² + b² = 1 (8/17)² + (15/17)² = 1 Simplifying, we get: 64/289 + 225/289 = 1 289/289 = 1 Which is true! Now, to find the value of θ, we can use the inverse trig functions: θ = arctan(b/a) = arctan(15/8) ≈ 1.07 radians ≈ 61.44° So, the solution is θ ≈ 61.44° or approximately 1.07 radians.
Maine kr liya but 2 pages ka solution aaya hai..
Soln) ( to find cos theta)
As per the question,
Sin theta = (17-15costheta)/8...eq1
Also sin²theta= 1-cos² theta (identity)
And if we square on both sides in equation no.1
We get two values of sin²theta , hence making it easy to solve the question..
Now as per eq 1
Sin² theta= (17-15costheta)²/64
So,
1-cos²theta=[289+225cos²theta-2(17)(15costheta)]/64
Further,
1-cos²theta=(289+225cos²theta-510costheta)/64
By cross multiplying,
64-64cos²theta=289+225cos²theta-510costheta
64-289=225cos²theta+64cos²theta-510costheta
-225=225cos²theta+64 cos²theta - 510costheta
0=289cos²theta-510costheta+225
Now , the equation formed is quadratic and is containing 2-3 digit no.s making it tougher and more complex to use mid term spitting, so we shall use quadratic formula , which is
x=(-b +- _/b²-4ac)/2a
According to the equation,
a=289
b=-510
c=225
Now, putting the values ,
Costheta= [-(-510) +- _/(-510)²-4(289)(225)]/2(289)
After solving we get,
Cos theta= (510+-_/260100-260100)/578
Cos theta= (510+0)/578, (510-0)/578
Cos theta =510/578,510/578
Cos theta = 15/17 answer....
Guys ek like toh banta hai yrr boht time laga type krne me
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Sin30= 2sin15.cos15
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Sin - लंब (8)
Cos आधार (15)
17 - कण
_________________
Sin 8/17
Cos 15/17
Ten 8/ 15. ✅
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A trigonometric equation!
We can solve this equation using the Pythagorean identity: sin²(θ) + cos²(θ) = 1
First, let's rewrite the equation:
8sin(θ) + 15cos(θ) = 17
Now, divide both sides by 17:
(8/17)sin(θ) + (15/17)cos(θ) = 1
Let's introduce new variables:
a = 8/17
b = 15/17
So, the equation becomes:
a sin(θ) + b cos(θ) = 1
Now, we can use the Pythagorean identity:
a² + b² = 1
(8/17)² + (15/17)² = 1
Simplifying, we get:
64/289 + 225/289 = 1
289/289 = 1
Which is true!
Now, to find the value of θ, we can use the inverse trig functions:
θ = arctan(b/a)
= arctan(15/8)
≈ 1.07 radians
≈ 61.44°
So, the solution is θ ≈ 61.44° or approximately 1.07 radians.
15²+8²=17²
See my solution
I am currently in class 10th and Arctans etc are deleted
I have done easily by making quadratic equation
You shall find my comment in this video only.
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Bhai mera solution dekho
Maine bas cos theta ki value nikali hai lekin uske thriugh through sin theta aur baki ki value bhi nikal loge
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right
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No of elements 18
The possible orders are
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No of elements 6
The possible orders are
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sin 30 =2 sin 15 . cos 15
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8/15