Trigonometry
In a right-angled ∆ ABC, angle B is a right angle.
If sin C = 4/5, then find the value of cos A.
A) 4/5
B) -4/5
C) 5/4
D) -5/4
Solution:
In triangle ABC, angle B = 90°
Therefore, angle C + angle A = 90°
Angle C = 90° – A
Sin C = 4/5
Sin (90° – A) = 4/5 (sum of angle A and angle C is 90°)
Cos A = 4/5 (trigonometry formula)
Correct Answer: A
2) Circle and Tangent
Circle with Tangent
2)A circle with center O (-4,6) and a point (-7,-7 )on the circle. A tangent through the point (-7, -7)
what is the slope of the tangent?
solution
The tangent is perpendicular to the radius of the circle (Theorem)
let N is the slope of the tangent
Let M is the slope of the radius
therefore M×N=-1 (properties of tangents of any two perpendicular lines)
slope of the radius M = y2 – y1/x 2-x1 (distance formula when the two coordinates of two points are given)
the slope of the radius M= -7- (-6)/-7-(-4) =-7+6/-7+4 =-1/-3=1/3
product of the slope =M×N=1/3N= -1
N/3=-1
N= -1×3slope of the tangent =-3
ANS -3
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3) Arc length and chord sum
Circle with Chord AB
In the above circle with center O and chord AB it subtanen angle 45 at rhe center of the circle drawn in XY- plane
what is the Arc AB length?
Option
A)πr/4
B)πr/5
C)3πr/4
D)πr/6
solution
chord length = angle subtended at the center ×2πradius/360=θ2πr/360
chord length AB =45×2πr/360=45×2×πr/360 =2 ×πr/8 (simplify)
=2πr/8=πr/4
correct ANS :A
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Circle with Radii
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equation of a circle
general equation of a circle with center with radius (h,K) and radius r
(x-h)2+(y-k) 2=r2
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1)The graph x2+x+ y2+y=31/2 in the XY -plane is a circle.
What is the length of the diameter?
Option:
A) 10
B) 6
C) 8
D)12
Solution:
The given equation x2+x+ y2+y=31/2
Write this equation in perfect squire
(x+1/2)2-1/4+(y+1/2)2-1/4=31/2
Add ¼+1/4 on both side
Therefore (x+1/2) 2-1/4+1/4+(y+1/2) 2-1/4+1/4=31/2+1/4+1/4
(x+1/2)2+(y+1/2)2=31/2+2×1/4
(x+1/2) 2+(y+1/2) 2= (31×2)/4+2×1/4=62/4+2/4 (by taking LCM)
(x+1/2)2+(y+1/2)2= (62+2)/4=64/4=16=42
Therefore radius =4 Units
The he diameter of the circle =2×r=2×4=8 unit
Correct ANS: C
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2)The equation of a circle in XY -plane is x2+5x+y2+3y=15/2
What is the center and radius of the above circle?
A)(-5/2,-3/2) and -4
B)(-5/2,-3/2) and 4
C)(5/2,-3/2) and 4
D)(-5/2,3/2) and 4
Solution:
The given equation = x2+5x+y2+3y=15/2
By perfect squaring method we write the above equation as
(x+5/2)2+(y+3/2)2-25/4-9/4=15/2
Adding both side by 25/4 and 9/4
We got (x+5/2)2-25/4+(y+3?2)2-9/4+25/4+9/4=15/2+25/4+9/4
Simply the Right-hand side by taking LCM
=(15×2+25+9)/4=(30+25+9)/4=(30+34)/4=64/4=16
The equation is (x+5/2)2+(y+3/2)2=16=42
Therefore by comparing with the standard form of equation (x-h) 2+(y-k) 2=r2
The center of the circle is (-5/2, -3/2) and the radius is 4
Correct ANS: B
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3) the function 3x2-6x+3y2+12y=12 in XY -plane is circle.
What is the radius of this circle.
Option:
A) (-2,1)
B) (2, -1)
C) (-1,2)
D (1, -2)
Solution:
The given of the circle is 3x2-6x+3y2+12y=12
By dividing by 3 on both side we got
X2-2x+y2+4y=4
By method of perfect square we got
(x-1)2+(y+2)2-1-4=4
By adding 1 and 4 on both side we got
(x-1)2+(y+2) 2-1-4+1+4=4+1+4=9
(x-1)2+(y+2)2=9=32
By comparing the standard form of the equation of the circle
(x-h)2+(y-K)2=r2
The center of the circle is (2)
Correct ANS : C
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4) (x+8)2+(y-6)2=16 is the equation of a circle A. another circle with the same center
and half of the radius of the circ A.
find the radius of the Circle B.
Solution:
The given circle is (x+8)2 + 9y-6)2=16
This can be written (x+8) 2+(y-6) 2=16=42
By comparing with the standard equation of the circle
(x-h)2+(y-6)2=r2
(x+8)2+(y-6) 2=42
Radius of the above circle is r=4
The radius of the circle B is = half of the radius of the circle A =1/2×4=2
The radius of the circle is 2.
ANS:2
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5) convert 1080 degrees in to Radians
A)5 π
B)3 π
C)7 π
D)6 π
Solution: radians
We know that 180 degree = π
Therefore 1080 degrees =π×1080/180=6π
Correct ANS:D
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Properties Of Triangle
!) In General Triangle has Three sides and three angles
.
sum of all three angles = 180 degrees,< p>
Sum of any two sides is greater than the third side
Example : In a triangle ABC
AB+BC>Ac
BC+AC>AB
AB+AC>BC
2) Area of a triangle = ½ ×base ×height
= 1/2×b×h squire units
3)Isosceles Triangle;
Any two sides are equal then it is called Isosceles Triangle
4) Equilateral Triangle
Let the side of the equilateral triangle is = a units
Then area of triangle = √3/4×a2 squire units
Side = a =√3/2×a units
4) In triangle any one of the angle is 90 degree then it is called Right angled Triangle
Exterior angle is equal to sum the interior opposite angles in the triangle
Right-Angled Triangle
hypotenuse2=Base2+height22
Hypotenuse =√〖base〗2 )+√(〖height〗2
Base =√〖hypotenuse]2 – √〖height〗2
Height =√〖hypotenuse〗2 – √〖base〗2
Area of this triangle=1/2×base×height squire units
5) Similar triangle:
If two triangles are similar then their ANGLES are EQUAL.
Their sides are in the same ratio
6)Congruent Triangle
if two triangle are congruent then their ANGLES are equal
and their sides are equal
6)Trigonometry ratios
Sinθ = opposite side/Hypotenuse, Cosθ= adjacent side /Hypotenuse ,
Tanθ=Opposite side / adjacent side
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Right-Angled Similar Triangles
1) triangle ABC and DEF are similar.
Angle A corresponds to angle D
Angle C corresponds to Angele F
Angle B corresponds to angle E
Then SinA=1/2 and DE =100.
Find the Value of DF
Solution:
∆ ABC and ∆DEV are similar
Therefore corresponding angles are equal
SinA =SinD=1/2
SinD=(Opposite side )/Hypotenuse = FE/ED=FE/100=1/2
FE /100×100=1/2×100 (multiply by 100 on both side)
FE=1/2×100 = 50
DE2= DF2
+ EF
2 ( Pythagoras theorem)
DF =√(DE2-EF2 ) = √(1002 -502)= 10√75=10×√(25×3)=10×5√3
Correct ANS:50√3
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3)
Right Angled Triangle ABC
in the above diagram SinB=3/5.if BC=25
Find the length of the sides of the triangle?
Option:
A)15,20,25
B)20,25,30
C)15,25,30
D)10,15,15
solution:
consider the∆ BDE
it is given SinB=3/5 , it means opposite side = DE=3 and Hypotenuse=BE=5
therefore BD =√(52-32)=√(25-9)=√(16)=4
the side of the ∆ BDE are 3,4and 5
the sides of the ∆ BAC are BA,AC and 25
sides of the ∆s are similar
the ratio of the sides are BA/BD=AC/DE=BC/BE
BA/4=Ac/3=25/5
BA/4=5
BA=5×4=20 (multiply by 5 on both side)
AC/3=5
AC=5×3=15 (Multiply by 5 on both side)
the three sides are 15,20,25
correct ANS:A
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3)sum with similar triangle
Right Angled Triangle with Perpendicular
In the above triangle ABC .
the line DE is parallel to side AC of the triangle
the following measurements are given for the diagram.
angle A is 90 degrees and side AC= 15 , DE =3 and DB=4 .
find the value of BC in the diagram?
Option:
A)35
B) 45
C) 25
D) 15
solution:
consider the Triangle DBE
it is a right angled triangle
in the triangle sides DB=4 and DE=3 the hypotenuse BE = √DB2+DE2=√42+32
That is BE = √16+9=√25=5
now consider ∆ABC and ∆ DBE
They are similar , therefore their sides are in the same ratio
AC/DE=BC/BE, 15/3=BC/5
BC/5=15/3
BC/5=5
BC=5×5 (multiply by 5 on both side)
BC= 25
Correct ANS:C
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Solution
4 In a right-angled triangle with two legs are 24 and 32 units what is the length of the hypotenuse ?
Option:
A)40
B)42
C)45
D)35
solution
By Pythagoras Theorem hypotenuse2=sum of the squires of the two sides
Hypotenuse2= 242+322
Hypotenuse=√(242+322 )
hypotenuse=√(576+1600) =40
Correct ANS:A
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5) Sum related to Angle property of a Triangle
Triangle ABC with Perpendicular AD