Question

3. Fill in the blanks. I. Find θ if value of sin3θ =Cos(θ-6)______________. II. Mode = 3__________ −2__________. III. The point P(3,2) is at distance 3 units from y–axis and __________units from x-axis. IV. Sum of zeroes of cubic polynomials is _____________. V. (Cotθ+tan)(−)=_____________.

Answer 1

The x co-ordinate of A(−3, 2) is negative and its y coordinate is positive. Therefore, point A(−3, 2) is in the second quadrant.

The x co-ordinate of B(−5, −2) is negative and its y coordinate is negative. Therefore, point B(−5, −2) is in the third quadrant.

The x co-ordinate of K(3.5, 1.5) is positive and its y coordinate is positive. Therefore, point K(3.5, 1.5) is in the first quadrant.

The x co-ordinate of D(2, 10) is positive and its y coordinate is positive. Therefore, point D(2, 10) is in the first quadrant.

The x co-ordinate of E(37, 35) is positive and its y coordinate is positive. Therefore, point E(37, 35) is in the first quadrant.

The x co-ordinate of F(15, −18) is positive and its y coordinate is negative. Therefore, point F(15, −18) is in the fourth quadrant.

The x co-ordinate of G(3, −7) is positive and its y coordinate is negative. Therefore, point G(3, −7) is in the fourth quadrant.

The x co-ordinate of H(0, −5) is zero. Therefore, point H(0, −5) is on the Y-axis.

The y co-ordinate of M(12, 0) is zero. Therefore, point M(12, 0) is on the X-axis.

The x co-ordinate of N(0, 9) is zero. Therefore, point N(0, 9) is on the Y-axis.

The x co-ordinate of P(0, 2.5) is zero. Therefore, point P(0, 2.5) is on the Y-axis.

The x co-ordinate of Q(−7, −3) is negative and its y coordinate is negative. Therefore, point Q(−7, −3) is in the third quadrant.

Answer 2

Solution 1 ...

sin3θ = cos(θ - 6)

sin3θ = sin{90 - (θ - 6)}

sin3θ = sin (90 - θ + 6)

sin3θ = sin(96 - θ)

3θ = 96 - θ

4θ = 96

$\boxed{θ = 24°}$

Solution 2 ...

$\boxed{Mode = 3(median) - 2(mean)}$

Solution 3 ...

The distance of the point P(2,3) from the x-axis is 3 unit.

Solution 4 ...

★ The general form of a cubic equation is ax3 + bx2 + cx + d = 0 where a, b, c and d are constants and a ≠ 0.

$\boxed{Sum\: of\: zeroes = \frac{-b}{a}}$

Solution 5 ...

The general solution of cotθ+ tanθ=2. A ... θ=nπ+(−1)nα.