Live practice

The tangent at \(A\) is perpendicular to radius \(OA\), and the tangent at \(B\) is perpendicular to radius \(OB\). Since \(OA\) and \(OB\) lie on the same straight line \(AB\), both tangents are perpendicular to \(AB\).

Two lines perpendicular to the same line are parallel. Hence the tangents at \(A\) and \(B\) are parallel.

Formula or theorem used: Draw and label a right triangle for each line of sight. Use \(\tan\theta=\frac{\text{vertical height}}{\text{horizontal distance}}\).

Working: Define the unknown height or distance, write a tangent equation for each observation, solve the equations, and include the observer's height when the question requires it.

After substitution and simplification:

( )0.8 3 \(1\,\mathrm{m}\)+

Formula or theorem used: In a right triangle use \(\sin\theta=\frac{P}{H}\), \(\cos\theta=\frac{B}{H}\), and \(\tan\theta=\frac{P}{B}\), together with the standard-angle values and identities.

Working: Write the relevant ratio or identity, substitute the known value, simplify one equality at a time, and state restrictions when a denominator could be zero.

After substitution and simplification:

Yes

Formula or theorem used: Use \(PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\) or the section formula \(\left(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n}\right)\).

Working: Substitute the coordinates before simplifying. For a shape, compare the required side lengths or diagonals and state the conclusion.

After substitution and simplification:

– 9, 3

In \(\triangle POQ\), \(DE\parallel OQ\). By the Basic Proportionality Theorem,

In \(\triangle POR\), \(DF\parallel OR\). Hence,

From \(1\) and \(2\),

Thus \(E\) and \(F\) divide \(PQ\) and \(PR\) in the same ratio. By the converse of the Basic Proportionality Theorem, \(EF\parallel QR\).

Formula or theorem used: For an AP use \(a_n=a+(n-1)d\) and \(S_n=\frac n2[2a+(n-1)d]\).

Working: Identify \(a\), \(d\), and the required term or sum; substitute them in the appropriate formula and simplify line by line.

After substitution and simplification:

16th term

Formula or theorem used: Write the equation in the form \(ax^2+bx+c=0\). Factorise where possible; otherwise use \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\).

Working: Display the standard form, the factorisation or discriminant calculation, both roots, and reject any value that is impossible in the stated context.

After substitution and simplification:

Number of articles = 6, Cost of each article = ₹ 15

Formula or theorem used: Translate the information into two linear equations. Use the method named in the exercise, keeping equivalent equations aligned.

Working: Eliminate or substitute one variable, obtain the other variable, substitute back, and verify the ordered pair in both original equations.

After substitution and simplification:

(i) x = 9, y = 5 (ii) s = 9, t = 6 (iii) y = 3x – 3,
where x can take any real value, i.e., infinitely many solutions.
(iv) x = 2, y = 3 (v) x = 0, y = 0 (vi) x = 2, y = 3

Total marbles (=5+8+4=17).

Formula or theorem used: For grouped data use class marks and frequencies systematically. Apply the direct, assumed-mean, or step-deviation formula and show the working table totals.

Working: Write the relevant mean, median, or mode formula, identify the modal/median class, substitute the table totals, and interpret the numerical result in the question's context.

After substitution and simplification:

69.43 %