31 Trigonometry
In a right triangle ABC, right-angled at A, if $\cos B = \frac{1}{4}$, then the value of $\cot B$ is:
Concept:
1
Right Triangle Trigonometry (SOH CAH TOA): We know $\cos B = \frac{\text{Base}}{\text{Hypotenuse}}$. We also know $\cot B = \frac{\text{Base}}{\text{Perpendicular}}$. Use the Pythagoras theorem to find the missing Perpendicular side.
Hint:
1
If Base = 1 and Hypotenuse = 4, find the Perpendicular using $1^2 + P^2 = 4^2$. Then plug the Base and Perpendicular into the $\cot B$ formula!
Mistakes to Avoid:
!
Mixing up Trig Ratios: A common mistake is answering with $\frac{\sqrt{15}}{1}$ (which is $\tan B$) or $\frac{4}{\sqrt{15}}$ (which is $\csc B$). Always double-check what sides the ratio specifically asks for.
👉 Scroll horizontally to read the full handwritten solution.
In a right-angled triangle $\Delta ABC$, right-angled at $A$, the angle is $B$.
By definition of cosine:
$\cos B = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{AB}{BC}$
Given $\cos B = \frac{1}{4}$, let Base $AB = 1$ and Hypotenuse $BC = 4$.
Use Pythagoras' theorem to find the perpendicular $AC$:
$AB^2 + AC^2 = BC^2$
$1^2 + AC^2 = 4^2$
$1 + AC^2 = 16$
$AC^2 = 15 \implies AC = \sqrt{15}$ [½ M]
We need to find the value of $\cot B$. By definition:
$\cot B = \frac{\text{Base}}{\text{Perpendicular}} = \frac{AB}{AC}$
$\cot B = \frac{1}{\sqrt{15}}$ [½ M]
Correct Option: (B) $\frac{1}{\sqrt{15}}$
32 Heights and Distances
The length of the shadow on the ground of a pole of height $18 \text{ m}$, when angle of elevation $\theta$ of the Sun is such that $\tan \theta = \frac{6}{7}$, is:
Concept:
1
Heights & Distances: The tangent of the angle of elevation relates the height of an object (Perpendicular) to the length of its shadow on the ground (Base). Thus, $\tan \theta = \frac{\text{Perpendicular (Height)}}{\text{Base (Shadow)}}$.
Hint:
1
Set up the equation: $\frac{6}{7} = \frac{18}{s}$. Cross-multiply to solve for the shadow length $s$.
Mistakes to Avoid:
!
Inverting the Ratio: Mixing up Height and Shadow in the numerator/denominator is a frequent error. If you write $\frac{6}{7} = \frac{s}{18}$, you will get the wrong answer. Remember: $\tan \theta$ is always Height/Shadow.
👉 Scroll horizontally to read the full handwritten solution.
Let the height of the pole be $h = 18 \text{ m}$ and the shadow length be $s$.
The pole, its shadow, and the line of sight form a right-angled triangle.
From the properties of right-angled triangles:
$\tan \theta = \frac{\text{Perpendicular (Height)}}{\text{Base (Shadow)}} = \frac{h}{s}$
We are given $\tan \theta = \frac{6}{7}$. Substituting the known values:
$\frac{6}{7} = \frac{18}{s}$ [½ M]
Cross-multiply to solve for $s$:
$6 \times s = 18 \times 7$
$s = \frac{18 \times 7}{6}$
$s = 3 \times 7 = 21 \text{ m}$ [½ M]
Correct Option: (D) 21 m
33 Circles
In the given figure, tangents PA and PB to the circle centred at O, from an external point P are perpendicular to each other. If $AB = 5\sqrt{2} \text{ cm}$, then the length of PA is equal to:
Concept:
1
Tangent Properties: Tangents drawn from an external point to a circle are exactly equal in length ($PA = PB$). If the angle between them is $90^\circ$, the triangle formed by the two tangents and the chord ($\Delta PAB$) is a right-angled isosceles triangle!
Hint:
1
Apply Pythagoras theorem on $\Delta PAB$: $PA^2 + PB^2 = AB^2$. Since $PA = PB$, this becomes $2PA^2 = (5\sqrt{2})^2$. Solve for $PA$.
Mistakes to Avoid:
!
Squaring Roots Incorrectly: A very common calculation error is squaring $5\sqrt{2}$ and getting $10$ instead of $50$. Remember: $(5\sqrt{2})^2 = 5^2 \times (\sqrt{2})^2 = 25 \times 2 = 50$.
👉 Scroll horizontally to read the full handwritten solution.
$PA$ and $PB$ are tangents drawn from the same external point $P$.
Therefore, tangent lengths are equal: $PA = PB$. [½ M]
Also, $\angle APB = 90^\circ$ because the two tangents are perpendicular.
Thus, $\Delta APB$ is a right-angled isosceles triangle.
By Pythagoras theorem: $PA^2 + PB^2 = AB^2$
Since $PA = PB$, we can write:
$2PA^2 = AB^2$ [½ M]
Given $AB = 5\sqrt{2} \text{ cm}$:
$2PA^2 = (5\sqrt{2})^2 = 25 \times 2 = 50$
$PA^2 = 25 \implies PA = 5 \text{ cm}$
Correct Option: (A) 5 cm
34 Circles
In the given figure, AT is a tangent to a circle centred at O. If $\angle ABC = 50^\circ$, then $\angle CAT$ is equal to:
Concept:
1
Circle Theorems Combined:
1. Angle in a semicircle: An angle inscribed in a semicircle (subtended by a diameter) is always $90^\circ$.
2. Tangent-Radius Property: A tangent is perpendicular ($90^\circ$) to the radius/diameter at the point of contact.
1. Angle in a semicircle: An angle inscribed in a semicircle (subtended by a diameter) is always $90^\circ$.
2. Tangent-Radius Property: A tangent is perpendicular ($90^\circ$) to the radius/diameter at the point of contact.
Hint:
1
AB is a diameter, so $\angle ACB = 90^\circ$. Use the angle sum property in $\Delta ABC$ to find the inner angle $\angle BAC$. Then, subtract $\angle BAC$ from the full $90^\circ$ tangent angle ($\angle BAT$) to find $\angle CAT$.
Mistakes to Avoid:
!
Assuming it's $90^\circ$: Students sometimes see the tangent and immediately assume $\angle CAT = 90^\circ$. Only the FULL angle $\angle BAT$ (between the diameter and the tangent) is $90^\circ$. $\angle CAT$ is just a piece of it!
👉 Scroll horizontally to read the full handwritten solution.
$AB$ passes through the centre $O$, so $AB$ is a diameter.
Therefore, by the angle in a semicircle theorem, $\angle ACB = 90^\circ$. [½ M]
Given $\angle ABC = 50^\circ$. In $\Delta ABC$:
$\angle BAC = 180^\circ - 90^\circ - 50^\circ = 40^\circ$.
$AT$ is tangent at $A$, so $OA \perp AT$.
Since $AB$ is along $OA$, the full angle $\angle BAT = 90^\circ$. [½ M]
From the figure, $\angle BAT = \angle BAC + \angle CAT$.
$90^\circ = 40^\circ + \angle CAT$
$\angle CAT = 50^\circ$.
Correct Option: (B) $50^\circ$
35 Areas Related to Circles
A sector of a circle with area $\frac{264}{7} \text{ cm}^2$ is cut from a circle of radius 6 cm. The central angle of the sector is:
Concept:
1
Area of a Sector: A sector's area is a proportional fraction of the entire circle's area. The formula is $\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2$. We can reverse-engineer this formula to find the missing central angle $\theta$.
Hint:
1
Plug in the knowns: $\frac{264}{7} = \frac{\theta}{360} \times \frac{22}{7} \times 6^2$. Notice how perfectly the $7$ in the denominators will cancel out on both sides!
Mistakes to Avoid:
!
Using Arc Length Formula: Do not accidentally use $2\pi r$ (the circumference/arc length formula) instead of $\pi r^2$. The question specifically gives you the Area in $\text{cm}^2$, not the length.
👉 Scroll horizontally to read the full handwritten solution.
Given:
Radius ($r$) = $6 \text{ cm}$
Area of the sector = $\frac{264}{7} \text{ cm}^2$
Let the central angle of the sector be $\theta$.
The formula for the area of a sector is: $\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2$
Substitute the known values (using $\pi = \frac{22}{7}$):
$\frac{264}{7} = \frac{\theta}{360^\circ} \times \frac{22}{7} \times (6)^2$
$\frac{264}{7} = \frac{\theta}{360^\circ} \times \frac{22}{7} \times 36$
Multiply both sides by 7 to clear the denominators:
$264 = \frac{\theta}{360^\circ} \times 22 \times 36$ [½ M]
Simplify the fraction $\frac{36}{360} = \frac{1}{10}$:
$264 = \theta \times \frac{22}{10} = \theta \times 2.2$
$\theta = \frac{264}{2.2} = \frac{2640}{22} = 120^\circ$ [½ M]
Correct Option: (D) $120^\circ$
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