11 Heights and Distances
A car is moving away from the base of a $30 \text{ m}$ high tower. The angle of elevation of the top of the tower from the car at an instant, when the car is $10\sqrt{3} \text{ m}$ away from the base of the tower, is:
Concept:
1
Trigonometric Ratios: In a right-angled triangle, the tangent of the angle of elevation connects the Perpendicular (height of the tower) to the Base (distance from the tower). The formula is $\tan \theta = \frac{\text{Perpendicular}}{\text{Base}}$.
Hint:
1
Set up the equation: $\tan \theta = \frac{30}{10\sqrt{3}}$. Simplify the fraction by canceling the 10, then rationalize the denominator. Which angle gives you that tangent value?
Mistakes to Avoid:
!
Failing to Rationalize: Getting stuck at $\tan \theta = \frac{3}{\sqrt{3}}$ is a common error. Remember that $3$ is the same as $\sqrt{3} \times \sqrt{3}$. This allows you to easily cancel out the bottom root!
👉 Scroll horizontally to read the full handwritten solution.
Let the height of the tower be the perpendicular, $P = 30 \text{ m}$.
Let the distance of the car from the base be the base, $B = 10\sqrt{3} \text{ m}$.
Let the angle of elevation be $\theta$.
Using the tangent trigonometric ratio:
$\tan \theta = \frac{\text{Perpendicular}}{\text{Base}}$
$\tan \theta = \frac{30}{10\sqrt{3}}$ [½ M]
Simplify the fraction by dividing by 10:
$\tan \theta = \frac{3}{\sqrt{3}}$
Rewrite $3$ as $\sqrt{3} \times \sqrt{3}$:
$\tan \theta = \frac{\sqrt{3} \times \sqrt{3}}{\sqrt{3}}$
$\tan \theta = \sqrt{3}$
From standard trigonometric values, we know that $\tan 60^\circ = \sqrt{3}$.
Therefore, $\theta = 60^\circ$. [½ M]
Correct Option: (d) $60^\circ$
12 Circles
If TP and TQ are two tangents to a circle with centre O from an external point T so that $\angle POQ = 120^\circ$, then $\angle PTQ$ is equal to:
Concept:
1
Tangent to a Circle Theorem: A tangent drawn from an external point to a circle is strictly perpendicular ($90^\circ$) to the radius at the point of contact. Furthermore, the sum of all interior angles in any quadrilateral is $360^\circ$.
Hint:
1
Look at the quadrilateral $OPTQ$. You know the central angle is $120^\circ$. The two angles where the tangents touch the radii are both $90^\circ$. Add them up and subtract from $360^\circ$ to find the final angle at $T$.
Mistakes to Avoid:
!
Wrong Angle Sum: A quick trap is assuming the angle at the center and the angle between the tangents must add up to $90^\circ$. They actually add up to $180^\circ$ because the other two angles in the quadrilateral consume $180^\circ$ themselves ($90 + 90$).
👉 Scroll horizontally to read the full handwritten solution.
$TP$ and $TQ$ are tangents drawn from point $T$ to the circle at points $P$ and $Q$.
A fundamental property states that a tangent is perpendicular
to the radius passing through the point of contact.
Therefore: $\angle OPT = 90^\circ$ and $\angle OQT = 90^\circ$. [½ M]
Consider the quadrilateral $OPTQ$.
The sum of the interior angles of a quadrilateral is always $360^\circ$.
$\angle PTQ + \angle OPT + \angle POQ + \angle OQT = 360^\circ$
Substitute the known values into the equation:
$\angle PTQ + 90^\circ + 120^\circ + 90^\circ = 360^\circ$
$\angle PTQ + 300^\circ = 360^\circ$
$\angle PTQ = 360^\circ - 300^\circ = 60^\circ$ [½ M]
Correct Option: (a) $60^\circ$
13 Circles
In the given figure, PA is a tangent from an external point P to a circle with centre O. If $\angle AOB = 125^\circ$, then $\angle APO$ is equal to:
Concept:
1
Linear Pairs and Tangents: A straight line creates a $180^\circ$ angle (linear pair). Also, a radius drawn to the point of tangency is always perpendicular ($90^\circ$) to the tangent line, creating a right-angled triangle.
Hint:
1
First, use the linear pair property on the straight line $POB$ to find $\angle AOP$ (subtract $125^\circ$ from $180^\circ$). Then, look at the right-angled triangle $\Delta AOP$. You know two angles now ($90^\circ$ and $\angle AOP$). Find the third!
Mistakes to Avoid:
!
Bisector Assumption: Don't assume that the line $PO$ exactly bisects an angle here just because it looks like it does in other tangent theorems. Always rely on the linear pair ($180^\circ$) rule when dealing with a straight line passing through the center!
👉 Scroll horizontally to read the full handwritten solution.
Since points $P$, $O$, and $B$ are collinear (on a straight line),
$\angle AOP$ and $\angle AOB$ form a linear pair.
$\angle AOP + \angle AOB = 180^\circ$
$\angle AOP + 125^\circ = 180^\circ$
Therefore, $\angle AOP = 55^\circ$. [½ M]
$PA$ is a tangent at point $A$. The radius $OA$ is perpendicular
to the tangent, so $\angle OAP = 90^\circ$.
In right-angled $\Delta AOP$, the interior angles sum to $180^\circ$:
$\angle APO + 90^\circ + 55^\circ = 180^\circ$
$\angle APO + 145^\circ = 180^\circ$
$\angle APO = 35^\circ$. [½ M]
Correct Option: (d) $35^\circ$
14 Areas Related to Circles
The length of the arc of the sector of a circle with radius 21 cm and of central angle $60^\circ$, is:
Concept:
1
Arc Length Formula: An arc is a fraction of the total circumference of a circle. The length of an arc of a sector with central angle $\theta$ and radius $r$ is given by $l = \frac{\theta}{360^\circ} \times 2\pi r$.
Hint:
1
Plug $\theta = 60^\circ$ and $r = 21$ into the formula. Since 21 is a multiple of 7, use $\pi = \frac{22}{7}$ for easy cancellation!
Mistakes to Avoid:
!
Area vs Arc Length: A very common mistake is accidentally calculating the Area of the sector using $\pi r^2$ instead of the length of the arc using $2\pi r$. Always double-check what the question is asking for!
👉 Scroll horizontally to read the full handwritten solution.
Given parameters for the circle sector:
Radius ($r$) = $21 \text{ cm}$
Central angle ($\theta$) = $60^\circ$
The formula for the length of an arc of a sector is:
$\text{Length of arc } (l) = \frac{\theta}{360^\circ} \times 2\pi r$
Substitute the given values (using $\pi = \frac{22}{7}$):
$l = \frac{60^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 21$ [½ M]
Simplify the fraction $\frac{60}{360}$ to $\frac{1}{6}$:
$l = \frac{1}{6} \times 2 \times 22 \times 3$
$l = \frac{1}{6} \times 132$
$l = 22 \text{ cm}$ [½ M]
Correct Option: (a) 22 cm
15 Areas Related to Circles
The hour hand of a clock is 7 cm long. The angle swept by it between 7:00 a.m. and 8:10 a.m. is:
Concept:
1
Clock Math (Hour Hand Velocity): The hour hand of a clock completes one full circle ($360^\circ$) in 12 hours. Since 12 hours is 720 minutes, the hour hand moves at a speed of $\frac{360^\circ}{720} = 0.5^\circ$ per minute.
Hint:
1
First, calculate the total time elapsed in minutes between 7:00 a.m. and 8:10 a.m. Then multiply that total time by $0.5^\circ$.
Mistakes to Avoid:
!
Using the Minute Hand Speed: The minute hand moves much faster ($6^\circ$ per minute). If you use $6^\circ$, you will get a wildly incorrect answer. Also, the "7 cm long" measurement is a distractor; you don't need it to find the angle!
👉 Scroll horizontally to read the full handwritten solution.
Let's find the speed of the hour hand of a clock.
The hour hand completes one full circle ($360^\circ$) in 12 hours.
Total minutes in 12 hours = $12 \times 60 = 720 \text{ minutes}$.
Angle swept by the hour hand in 1 minute:
$= \frac{360^\circ}{720} = 0.5^\circ \text{ per minute} = \left(\frac{1}{2}\right)^\circ \text{ per minute}$
Next, calculate the total time elapsed between 7:00 a.m. and 8:10 a.m.:
From 7:00 a.m. to 8:00 a.m. = 60 minutes
From 8:00 a.m. to 8:10 a.m. = 10 minutes
Total time elapsed = $60 + 10 = 70 \text{ minutes}$. [½ M]
Finally, calculate the angle swept in 70 minutes:
$\text{Angle} = 70 \text{ minutes} \times 0.5^\circ \text{ per minute}$
$\text{Angle} = \frac{70}{2} = 35^\circ$ [½ M]
Correct Option: (c) $35^\circ$
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