Q03 Test 36 English

Test No. 36 – Complete Solutions

143

Find the value of \(2\sin 45° + 2\cos 45°\).

🔑 Key Concept – Standard Trigonometric Values: \[ \sin 45° = \cos 45° = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \] Since both values are equal, the expression simplifies quickly.

Step 1 – Substitute standard values: \[ 2\sin 45° + 2\cos 45° = 2 \times \frac{\sqrt{2}}{2} + 2 \times \frac{\sqrt{2}}{2} \]

Step 2 – Simplify each term: \[ = \sqrt{2} + \sqrt{2} \]

Step 3 – Add: \[ = 2\sqrt{2} \approx 2 \times 1.4142 \approx 2.828 \]

\( \boxed{2\sin 45° + 2\cos 45° = 2\sqrt{2} \approx 2.828} \)

144

Find the slope and inclination of the line formed by points \((5,\,11)\) and \((-2,\,4)\).

🔑 Key Concept – Slope and Inclination: The slope of a line through two points is: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \] The inclination \(\alpha\) is the angle the line makes with the positive x-axis: \[ \alpha = \tan^{-1}(m) \qquad 0° \leq \alpha < 180° \]

Given: \((x_1, y_1) = (5, 11)\),  \((x_2, y_2) = (-2, 4)\)

Step 1 – Calculate the slope: \[ m = \frac{4 - 11}{-2 - 5} = \frac{-7}{-7} = 1 \]

Step 2 – Find the inclination: \[ \alpha = \tan^{-1}(1) = 45° \] Since \(\tan 45° = 1\), the inclination is exactly \(45°\).

\( \boxed{m = 1 \qquad \alpha = 45°} \)

145

Find equation: Vertical line passing through \((-5,\,3)\).

🔑 Key Concept – Vertical Line: A vertical line has an undefined slope and every point on it shares the same x-coordinate. Its equation is: \[ x = c \] where \(c\) is the x-coordinate of any point on the line.

Given: Point \((-5,\ 3)\) lies on the line.

Step 1 – Identify the x-coordinate: The x-coordinate of the given point is \(-5\).

Step 2 – Write the equation: Since the line is vertical, every point has \(x = -5\): \[ x = -5 \]

\( \boxed{x = -5} \)

146

Find the unknown quantities in the following figures (similar triangles).

🔑 Key Concept – Ratio of Areas of Similar Figures: \[ \frac{A_1}{A_2} = \left(\frac{l_1}{l_2}\right)^2 \] Rearranging to find the unknown area: \[ A_2 = A_1 \times \left(\frac{l_2}{l_1}\right)^2 \]

Given: Triangle 1: \(l_1 = 15\ \text{cm},\quad A_1 = 60\ \text{cm}^2\)
Triangle 2: \(l_2 = 20\ \text{cm},\quad A_2 = ?\)

Step 1 – Write the formula: \[ \frac{A_1}{A_2} = \left(\frac{l_1}{l_2}\right)^2 \]

Step 2 – Substitute known values: \[ \frac{60}{A_2} = \left(\frac{15}{20}\right)^2 \]

Step 3 – Simplify the ratio and square it: \[ \frac{15}{20} = \frac{3}{4} \qquad \left(\frac{3}{4}\right)^2 = \frac{9}{16} \]

Step 4 – Solve for \(A_2\): \[ \frac{60}{A_2} = \frac{9}{16} \] \[ A_2 = \frac{60 \times 16}{9} = \frac{960}{9} = \frac{320}{3} \approx 106.67\ \text{cm}^2 \]

\( \boxed{A_2 = \dfrac{320}{3} \approx 106.67\ \text{cm}^2} \)