1.Two sides of a triangle are 6 cm and 8 cm. If height of the triangle corresponding to 6 cm side is 4 cm ; find :
(i) area of the triangle
(ii) height of the triangle corresponding to 8 cm side.
5. For any angle θ, state the value of:
Sin2 θ + cos2 θ.
Sin2 ? + cos2 ? = Sin2 ? + 1 – sin2 ?.
= 1
3.If sin x = cos y; write the relation between x and y, if both the angles x and y are acute.
sin x = cos y = sin (90o – y)
If x and y are acute angles
x = 90o – y
which implies,
x + y = 90o
hence x and y are complementary angles.
2. If sin x = cos x and x is acute, state the value of x.
The angle, x is acute and hence we have, 0 < x
We know that
Cos2x + sin2 x = 1
Since cos x = sin x
Above equation will become
2 sin2 x = 1
Sin x = 1/v2
Therefore, x = 45o
Prove that:
3 cosec2 60o – 2 cot2 30o + sec2 45o = 0.
Given 3 cosec2 60o – 2 cot2 30o + sec2 45o = 0.
LHS = 3 cosec2 60o – 2 cot2 30o + sec2 45o = 0.
= 3 (2/v3)2 – 2 (v3)2 + (v2)2
= 4 – 6 + 2
= 0
= RHS
what is even number?
Factorize:
5 - (3a2 - 2a) (6 - 3a2 + 2a)
3. Factorize:
(3x - 2y)2 + 3 (3x - 2y) - 10
2. Factorize by taking out the common factors:
2 (2x - 5y) (3x + 4y) - 6 (2x - 5y) (x - y)
Taking (2x - 5y) common from both terms
= (2x - 5y)[2(3x + 4y) - 6(x - y)]
=(2x - 5y)(6x + 8y - 6x + 6y)
=(2x - 5y)(8y + 6y)
=(2x - 5y)(14y)
=(2x - 5y)14y
1.In triangle ABC, M is mid-point of AB and a straight line through M and parallel to BC cuts AC in N. Find the lengths of AN and MN if BC = 7 cm and AC = 5 cm.
The triangle is shown below,
Since M is the midpoint of AB and MN||BC hence N is the midpoint of AC.Therefore
And 
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