Q03 Test 38 English

Test No. 38 – Complete Solutions

151

Prove: \(\sin\theta \cdot \csc\theta = 1\)

🔑 Key Concept – Reciprocal Identity: By definition, cosecant is the reciprocal of sine: \[ \csc\theta = \frac{1}{\sin\theta} \qquad (\sin\theta \neq 0) \] Multiplying a number by its reciprocal always gives 1.

Step 1 – Start with the Left Hand Side (LHS): \[ \text{LHS} = \sin\theta \cdot \csc\theta \]

Step 2 – Replace \(\csc\theta\) with its reciprocal definition: \[ = \sin\theta \cdot \frac{1}{\sin\theta} \]

Step 3 – Cancel \(\sin\theta\): \[ = \frac{\sin\theta}{\sin\theta} = 1 = \text{RHS} \]

\( \sin\theta \cdot \csc\theta = \sin\theta \cdot \dfrac{1}{\sin\theta} = 1 \qquad \textbf{Proved} \)

152

Find slope and inclination of the line formed by points \((8,\,4)\) and \((4,\,6)\).

🔑 Key Concept – Slope and Inclination: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \qquad\qquad \alpha = \tan^{-1}(m),\quad 0° \leq \alpha < 180° \] For a negative slope, add 180° to the reference angle to get the inclination.

Given: \((x_1, y_1) = (8, 4)\),  \((x_2, y_2) = (4, 6)\)

Step 1 – Calculate the slope: \[ m = \frac{6 - 4}{4 - 8} = \frac{2}{-4} = -\frac{1}{2} \]

Step 2 – Find the inclination: Reference angle: \(\tan^{-1}\!\left(\dfrac{1}{2}\right) \approx 26.57°\)
Since slope is negative, inclination is in second quadrant: \[ \alpha = 180° - 26.57° \approx 153.43° \]

\( \boxed{m = -\dfrac{1}{2} \qquad \alpha \approx 153.43°} \)

153

Find the value of \(\sin 60°\cos 30° + \cos 60°\sin 30°\).

🔑 Key Concept – Sine Addition Formula: \[ \sin(A + B) = \sin A\cos B + \cos A\sin B \] So \(\sin 60°\cos 30° + \cos 60°\sin 30° = \sin(60° + 30°) = \sin 90°\).
Standard values: \[ \sin 60° = \frac{\sqrt{3}}{2},\quad \cos 30° = \frac{\sqrt{3}}{2},\quad \cos 60° = \frac{1}{2},\quad \sin 30° = \frac{1}{2} \]

Method 1 – Using the sine addition formula: \[ \sin 60°\cos 30° + \cos 60°\sin 30° = \sin(60° + 30°) = \sin 90° = 1 \]

Method 2 – Direct substitution: \[ = \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} + \frac{1}{2} \times \frac{1}{2} \] \[ = \frac{3}{4} + \frac{1}{4} = \frac{4}{4} = 1 \]

\( \boxed{\sin 60°\cos 30° + \cos 60°\sin 30° = 1} \)

154

Two similar triangles have areas \(16\ \text{cm}^2\) and \(25\ \text{cm}^2\). Find the ratio of their corresponding sides.

🔑 Key Concept – Ratio of Sides from Ratio of Areas: For two similar figures: \[ \frac{A_1}{A_2} = \left(\frac{l_1}{l_2}\right)^2 \] Therefore the ratio of corresponding sides is: \[ \frac{l_1}{l_2} = \sqrt{\frac{A_1}{A_2}} \]

Given: \(A_1 = 16\ \text{cm}^2\),  \(A_2 = 25\ \text{cm}^2\)

Step 1 – Apply the formula: \[ \frac{l_1}{l_2} = \sqrt{\frac{A_1}{A_2}} = \sqrt{\frac{16}{25}} \]

Step 2 – Simplify: \[ \frac{l_1}{l_2} = \frac{\sqrt{16}}{\sqrt{25}} = \frac{4}{5} \]

\( \boxed{\dfrac{l_1}{l_2} = \dfrac{4}{5} \quad \text{i.e. } 4:5} \)

155

Find equation: Line passing through \((-3,\,8)\) with slope \(0\).

🔑 Key Concept – Line with Zero Slope: A slope of zero means the line is horizontal. A horizontal line has the same y-coordinate for every point on it: \[ y = c \] Use the point-slope form \(y - y_1 = m(x - x_1)\) with \(m = 0\).

Given: Point \((-3,\ 8)\),  slope \(m = 0\)

Step 1 – Apply point-slope form: \[ y - 8 = 0\,(x - (-3)) \] \[ y - 8 = 0 \]

Step 2 – Simplify: \[ y = 8 \]

\( \boxed{y = 8} \)