5. Find the nature of the roots of the quadratic equation 3x²+ 2x + 2 =0
6. Solve 2x² + x - 6=0 by using formula,
7. If Tan2A = cot(1 - 18) where 2A is an acute angle. Find the value of A.
18. Prove that
1+tan²A/1+cot²A=tan²A
9. The angle of elevation of the top of a tower from a point on the ground which is 40m away from the foot of the tower is 30°.Find the height of the tower.
Answers
5.The nature of quadratic equation is determined by discriminant. 2) then roots are real and unequal. 3) then roots are unequal and not real i.e. imaginary.
6.Given Quadratic equation
2x² + x - 6 = 0
Splitting the middle term , we get
2x² + 4x -3x - 6 = 0
=> 2x( x + 2 ) - 3( x + 2 ) = 0
=> ( x + 2 )( 2x - 3 ) = 0
Therefore ,
x + 2 = 0 or 2x - 3 = 0
x = -2 or x = 3/2
7.This is related to Trigonometric
Ratios of Complementary angles ,
Two angles are said to be Complementary , if their sum is equal
to 90°.
As you know that ,
Cot ( 90 - x ) = tan x
According to the problem given,
Tan 2A = cot ( A - 18 )
Cot ( 90 - 2A ) = cot ( A - 18 )
Remove the cot both sides
we get ,
90 - 2A = A - 18
-2A - A = - 18 - 90
- 3A = - 108
A = ( - 108 ) / ( - 3 )
A = 36°
18.(1+tan2A)/(1+cot2A) =tan2A
L.H.S.
(1+tan2A) = sec2A = (1/cos2A)
(1+cot2A) = cosec2A = (1/sin2A)
(1/cos2A) / (1/sin2A) =1/cos2A×sin2A =sin2A/cos2A =tan2A
L.H.S. =R.H.S
( NOTE : 2 MEANS SQUARE OR POWER 2)
19.let h m be the height of tower and m be the depth of tank (as figure).
Now, in ΔABC:
tan30°=BC/AB;
1/√3=h/40;
h=40/√3 m (1)
In ΔABD:
tan45°=DC+BC/AB;
1=x+h/40;
x+h=40;
using (1):
x+40/√3=40;
x=40(1-1/√3) m
Then height of the tower = 40/√3 m,
and depth up tank :
40(1-1/√3)m
In ΔABD:
L.H.S. =R.H.S
Math
( NOTE : 2 MEANS SQUARE OR POWER 2)
5 points
19.let h m be the height of tower and m be the depth of tank (as figure).
Now, in ΔABC:
tan30°=BC/AB;
1/√3=h/40;
h=40/√3 m (1)
tan45°=DC+BC/AB;
1=x+h/40;
x+h=40;
using (1):
x+40/√3=40;
x=40(1-1/√3) m
Then height of the tower = 40/√3 m,
and depth up tank :