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Yogita Ingle 5 years, 1 month ago
We know that in a triangle, sum of the angles = 180°
A+B+C = 180 → (1)
sin 30
cos45
So,
sin (A+B-C) = sin 30
A+B-C = 30 → (2)
And
cos (B+C-A) = cos 45
B+C-A = 45 → (3)
On solving equation (1) and (2), we get,
A+B+C-A-B+C = 180-30 = 150
2C = 150
C = 75°
Substituting C=75 in equation (2), we get,
A+B-75 = 30
A+B = 105 → (4)
Also, substituting in equation (3), we get,
B+75-A =45
A-B = 30 → (5)
Adding equations (4) and (5), we get,
2A = 135 → A = 67.5°
B = A-30 = 67.5 - 30 = 37.5°
Therefore, A=67.5°; B=37.5°; and C=75°
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Gaurav Seth 5 years, 1 month ago
The nature of the roots depends on the value of b2 – 4ac. bx2 – 4ac is called the discriminant of the quadratic equation ax2 + bx + c = 0 and is generally, denoted by D.
∴ D = b2 – 4ac
If D > 0, i..e., b2 – 4ac > 0, i.e., b2 – 4ac is positive; the roots are real and unequal. Also,
(i) If b2 – 4ac is a perfect square, the roots are rational and unequal.
(ii) If b2 – 4ac is positive but not perfect square, the roots are irrational and unequal.
If D = 0, i.e., b2 – 4ac = 0; the roots are real and equal.
If D < 0, i.e., b2 – 4ac < 0; i.e., b2 – 4ac is negative; the roots are not real, i.e., the roots are imaginary.
Posted by Sachin Raj 5 years, 1 month ago
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Gaurav Seth 5 years ago
Nature of Roots
A quadratic equation ax2 + bx + c = 0 has
(i) two distinct real roots, if b2 – 4ac > 0,
(ii) two equal real roots, if b2 – 4ac = 0,
(iii) no real roots, if b2 – 4ac < 0.
Since b2 – 4ac determines whether the quadratic equation ax2 + bx + c = 0 has real roots or not, b2 – 4ac is called the discriminant of this quadratic equation
Posted by S S 5 years, 1 month ago
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Gaurav Seth 5 years, 1 month ago
(i) (cosec θ – cot θ)2 = (1-cos θ)/(1+cos θ)
To prove this, first take the Left-Hand side (L.H.S) of the given equation, to prove the Right Hand Side (R.H.S)
L.H.S. = (cosec θ – cot θ)2
The above equation is in the form of (a-b)2, and expand it
Since (a-b)2 = a2 + b2 – 2ab
Here a = cosec θ and b = cot θ
= (cosec2θ + cot2θ – 2cosec θ cot θ)
Now, apply the corresponding inverse functions and equivalent ratios to simplify
= (1/sin2θ + cos2θ/sin2θ – 2cos θ/sin2θ)
= (1 + cos2θ – 2cos θ)/(1 – cos2θ)
= (1-cos θ)2/(1 – cosθ)(1+cos θ)
= (1-cos θ)/(1+cos θ) = R.H.S.
Therefore, (cosec θ – cot θ)2 = (1-cos θ)/(1+cos θ)
Hence proved.
(ii) (cos A/(1+sin A)) + ((1+sin A)/cos A) = 2 sec A
Now, take the L.H.S of the given equation.
L.H.S. = (cos A/(1+sin A)) + ((1+sin A)/cos A)
= [cos2A + (1+sin A)2]/(1+sin A)cos A
= (cos2A + sin2A + 1 + 2sin A)/(1+sin A) cos A
Since cos2A + sin2A = 1, we can write it as
= (1 + 1 + 2sin A)/(1+sin A) cos A
= (2+ 2sin A)/(1+sin A)cos A
= 2(1+sin A)/(1+sin A)cos A
= 2/cos A = 2 sec A = R.H.S.
L.H.S. = R.H.S.
(cos A/(1+sin A)) + ((1+sin A)/cos A) = 2 sec A
Hence proved.
For more click on the given link:
<a href="https://mycbseguide.com/blog/ncert-solutions-class-10-maths-exercise-8-4/" ping="/url?sa=t&source=web&rct=j&url=https://mycbseguide.com/blog/ncert-solutions-class-10-maths-exercise-8-4/&ved=2ahUKEwiAh6LGlLPsAhWhQ3wKHRY_BY0QFjADegQIAhAC" rel="noopener" target="_blank">NCERT Solutions for Class 10 Maths Exercise 8.4 ...</a>
Posted by Wanpynbiang Wansit 5 years, 1 month ago
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Yogita Ingle 5 years, 1 month ago
Length of the boundary of the semicircular park = 90 m
⇒ πr + 2r = 90
⇒ r(22/7 + 2) = 90
⇒ r = (90*7)/36
⇒ r = 17.5 m
Area of the semicircular park = 1/2πr²
= 1/2*22/7*17.5*17.5
= 481.25 sq m
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Navnoor Singh 5 years, 1 month ago
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Gaurav Seth 5 years, 1 month ago
Height of the well = 14 m
Diameter of the well = 3 m
So, Radius of the well = 3/2 m
Volume of the earth taken out of the well = πr²h
= 22/7*(3/2)²*14
= 99 cu m
Outer radius of the embankment = R = (3/2 + 4)m = 11/2 m
Area of embankment = outer area - inner area
⇒ = πR² - πr²
= 22/7*[(11/2)² - (3/2)²]
= 22/7*[(121/4) - (9/4)]
= 22/7 × 112/4
= 88 m²
Height of the embankment = Volume/Area
= 99/88
Height of the embankment = 1.125 m
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Navnoor Singh 5 years, 1 month ago
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Gaurav Seth 5 years ago
A n s w e r
Step-by-step explanation: We are given the following :
We are to prove the following relation :
We will be using the following trigonometric formula :
We have
Hence proved.
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Yogita Ingle 5 years, 1 month ago
Hence, solution of the system of equation is x= -5/4, y = -1/4
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Gaurav Seth 5 years, 1 month ago
Frequency distribution table of less than type is as follows:
|
Daily income (in Rs) (upper class limits) |
Cumulative frequency |
|
Less than 120 |
12 |
|
Less than 140 |
12 + 14 = 26 |
|
Less than 160 |
26 + 8 = 34 |
|
Less than 180 |
34 + 6 = 40 |
|
Less than 200 |
40 + 10 = 50 |
Now taking upper class limits of class intervals on x-axis and their respective frequencies on y-axis, we can draw its ogive as follows:
Here, N = 50
N/2 = 25
Now, mark the point on curve whose y-coordinate is 25, its corresponding x-coordinate is 138.5. So median of this data is 138.5 (approximately).
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