NCERT Exemplar Solutions - Class 10 - Mathematics - Unit 8: Introduction to Trignometry & its Equation

Exercise 8.4

Step 1: We are given that:

\[ \csc\theta + \cot\theta = p \]

Here, \(\csc\theta = \dfrac{1}{\sin\theta}\) and \(\cot\theta = \dfrac{\cos\theta}{\sin\theta}\).

Step 2: Multiply the given expression by its conjugate \((\csc\theta - \cot\theta)\):

\[(\csc\theta + \cot\theta)(\csc\theta - \cot\theta) = \csc^2\theta - \cot^2\theta.\]

Step 3: Recall the identity:

\[ \csc^2\theta - \cot^2\theta = 1. \]

So,

\[ (p)(\csc\theta - \cot\theta) = 1. \]

Therefore,

\[ \csc\theta - \cot\theta = \dfrac{1}{p}. \]

Step 4: Now we have two equations:

\[ \csc\theta + \cot\theta = p \]

\[ \csc\theta - \cot\theta = \dfrac{1}{p} \]

Step 5: Add these two equations:

\[ (\csc\theta + \cot\theta) + (\csc\theta - \cot\theta) = p + \dfrac{1}{p} \]

\[ 2\csc\theta = p + \dfrac{1}{p} \]

\[ \csc\theta = \dfrac{p + \tfrac{1}{p}}{2} \]

Step 6: Subtract the two equations:

\[ (\csc\theta + \cot\theta) - (\csc\theta - \cot\theta) = p - \dfrac{1}{p} \]

\[ 2\cot\theta = p - \dfrac{1}{p} \]

\[ \cot\theta = \dfrac{p - \tfrac{1}{p}}{2} \]

Step 7: Use the relation:

\[ \cos\theta = \dfrac{\cot\theta}{\csc\theta}. \]

Step 8: Substitute the values of \(\cot\theta\) and \(\csc\theta\):

\[ \cos\theta = \dfrac{\dfrac{p - \tfrac{1}{p}}{2}}{\dfrac{p + \tfrac{1}{p}}{2}} \]

The 2’s cancel out, so:

\[ \cos\theta = \dfrac{p - \tfrac{1}{p}}{p + \tfrac{1}{p}} \]

Step 9: Simplify the fraction by multiplying numerator and denominator by \(p\):

\[ \cos\theta = \dfrac{p^2 - 1}{p^2 + 1} \]

Final Answer:

\[ \cos\theta = \dfrac{p^2 - 1}{p^2 + 1} \]

Step 1: We need to prove:

\[\sqrt{\sec^2\theta+\csc^2\theta} = \tan\theta + \cot\theta.\]

Step 2: To make things easier, let us square both sides. (Because if two positive numbers are equal, their squares are also equal.)

So, we want to check if:

\[(\tan\theta + \cot\theta)^2 = \sec^2\theta + \csc^2\theta.\]

Step 3: Expand the square on the left-hand side:

\[(\tan\theta + \cot\theta)^2 = \tan^2\theta + \cot^2\theta + 2.\]

Step 4: Recall the trigonometric identities:

• \(\tan^2\theta = \sec^2\theta - 1\)
• \(\cot^2\theta = \csc^2\theta - 1\)

Step 5: Substitute these into the equation:

\[(\tan\theta + \cot\theta)^2 = (\sec^2\theta - 1) + (\csc^2\theta - 1) + 2.\]

Step 6: Simplify:

\[(\tan\theta + \cot\theta)^2 = \sec^2\theta + \csc^2\theta.\]

Step 7: Now take square root on both sides:

\[\tan\theta + \cot\theta = \sqrt{\sec^2\theta + \csc^2\theta}.\]

Step 8: Since \(\theta\) is acute (between 0° and 90°), both \(\tan\theta\) and \(\cot\theta\) are positive. So the equality holds without any sign issues.

Final Result:

\[\sqrt{\sec^2\theta + \csc^2\theta} = \tan\theta + \cot\theta.\]

Step 1: Assume variables.

Let the height of the tower be \(h\,\text{m}\).

Let the initial distance of the observer from the tower be \(x\,\text{m}\).

Step 2: Use the first situation (angle = \(30^\circ\)).

From the definition of tangent,

\(\tan\theta = \dfrac{\text{opposite side}}{\text{adjacent side}}\).

Here, opposite side = tower height = \(h\), adjacent side = distance = \(x\).

So, \(\tan30^\circ = \dfrac{h}{x}\).

We know \(\tan30^\circ = \dfrac{1}{\sqrt{3}}\).

Therefore, \(\dfrac{1}{\sqrt{3}} = \dfrac{h}{x} \Rightarrow h = \dfrac{x}{\sqrt{3}}.\)

Step 3: Use the second situation (angle = \(45^\circ\)).

Now the observer moves \(20\,\text{m}\) closer. So, new distance = \((x - 20)\,\text{m}\).

Again using tangent,

\(\tan45^\circ = \dfrac{h}{x - 20}\).

We know \(\tan45^\circ = 1\).

So, \(1 = \dfrac{h}{x - 20} \Rightarrow h = x - 20.\)

Step 4: Combine the two equations for \(h\).

From first situation: \(h = \dfrac{x}{\sqrt{3}}\).

From second situation: \(h = x - 20\).

Therefore, \(\dfrac{x}{\sqrt{3}} = x - 20.\)

Step 5: Solve for \(x\).

Multiply both sides by \(\sqrt{3}\):

\(x = (x - 20)\sqrt{3}\).

Expand: \(x = x\sqrt{3} - 20\sqrt{3}\).

Bring terms with \(x\) together: \(x\sqrt{3} - x = 20\sqrt{3}\).

Factor \(x\): \(x(\sqrt{3} - 1) = 20\sqrt{3}\).

So, \(x = \dfrac{20\sqrt{3}}{\sqrt{3} - 1}.\)

Simplify by multiplying numerator and denominator by \((\sqrt{3}+1)\):

\(x = 10(3 + \sqrt{3})\,\text{m}.\)

Step 6: Find height \(h\).

From \(h = x - 20\):

\(h = 10(3 + \sqrt{3}) - 20 = 10(\sqrt{3} + 1)\,\text{m}.\)

Final Answer: The height of the tower is \(10(\sqrt{3}+1)\,\text{m}\).

Step 1: We are given:

\[1 + \sin^2\theta = 3\sin\theta \cos\theta.\]

Step 2: To solve this, let us write everything in terms of \(\tan\theta\). Recall the identity: \(\tan\theta = \dfrac{\sin\theta}{\cos\theta}.\)

Step 3: Put \(\tan\theta = t\). Then \(\sin\theta = t \cos\theta.\)

Step 4: Substitute this into the given equation:

\[1 + (t\cos\theta)^2 = 3(t\cos\theta)(\cos\theta).\]

Step 5: Simplify both sides:

\[1 + t^2 \cos^2\theta = 3t \cos^2\theta.\]

Step 6: Divide the whole equation by \(\cos^2\theta\) (valid if \(\cos\theta \neq 0\)):

\[\dfrac{1}{\cos^2\theta} + t^2 = 3t.\]

Step 7: Recall that \(\dfrac{1}{\cos^2\theta} = \sec^2\theta.\) Also, \(\sec^2\theta = 1 + \tan^2\theta = 1 + t^2.\)

Step 8: Substitute this:

\[(1 + t^2) + t^2 = 3t.\]

Step 9: Simplify:

\[1 + 2t^2 = 3t.\]

Step 10: Rearrange into standard quadratic form:

\[2t^2 - 3t + 1 = 0.\]

Step 11: Factorize the quadratic:

\[(2t - 1)(t - 1) = 0.\]

Step 12: Solve for \(t\):

\[t = 1 \quad \text{or} \quad t = \dfrac{1}{2}.\]

Final Answer: Since \(t = \tan\theta\),

\[\tan\theta = 1 \quad \text{or} \quad \tan\theta = \dfrac{1}{2}.\]

Step 1: Recall a useful identity:

For any real numbers \(a, b, c, d\):

\((a\sin\theta + b\cos\theta)^2 + (c\sin\theta + d\cos\theta)^2 = (a^2+c^2)\sin^2\theta + (b^2+d^2)\cos^2\theta + 2(ab+cd)\sin\theta\cos\theta.\)

This is often used to combine expressions involving sine and cosine.

Step 2: Consider the two expressions in the question:

• First expression: \(\sin\theta + 2\cos\theta\)
• Second expression: \(2\sin\theta - \cos\theta\)

Step 3: Square and add them:

\[(2\sin\theta - \cos\theta)^2 + (\sin\theta + 2\cos\theta)^2.\]

Step 4: Expand each square:

\((2\sin\theta - \cos\theta)^2 = 4\sin^2\theta - 4\sin\theta\cos\theta + \cos^2\theta\)

\((\sin\theta + 2\cos\theta)^2 = \sin^2\theta + 4\sin\theta\cos\theta + 4\cos^2\theta\)

Step 5: Add them together:

\(4\sin^2\theta + \sin^2\theta = 5\sin^2\theta\)

\(\cos^2\theta + 4\cos^2\theta = 5\cos^2\theta\)

The cross terms: \(-4\sin\theta\cos\theta + 4\sin\theta\cos\theta = 0\)

So the sum is:

\(5\sin^2\theta + 5\cos^2\theta = 5(\sin^2\theta + \cos^2\theta).\)

Step 6: Recall the fundamental trigonometric identity:

\(\sin^2\theta + \cos^2\theta = 1\)

So, the sum becomes:

\(5(1) = 5\).

Step 7: We now know:

\((2\sin\theta - \cos\theta)^2 + (\sin\theta + 2\cos\theta)^2 = 5.\)

Step 8: From the question, we are given:

\(\sin\theta + 2\cos\theta = 1.\)

Therefore:

\((\sin\theta + 2\cos\theta)^2 = 1^2 = 1.\)

Step 9: Substitute this into the earlier equation:

\((2\sin\theta - \cos\theta)^2 + 1 = 5\)

Step 10: Simplify:

\((2\sin\theta - \cos\theta)^2 = 4\)

Step 11: Take square root:

\(|2\sin\theta - \cos\theta| = \sqrt{4} = 2.\)

Final Answer:

Hence, \(|2\sin\theta - \cos\theta| = 2.\)

(The absolute value is used because the expression could be +2 or -2 depending on the quadrant of \(\theta\).)

Step 1: Understand the situation.

We have a vertical tower of height \(h\,\text{metres}\). Two people stand on the ground, at distances \(s\,\text{metres}\) and \(t\,\text{metres}\) away from the foot of the tower. They look up to the top of the tower, and their angles of elevation are complementary. This means that if one angle is \(\alpha\), the other is \(90^\circ - \alpha\).

Step 2: Write trigonometric relations.

For the person at distance \(s\):

\[ \tan(\alpha) = \frac{\text{opposite side}}{\text{adjacent side}} = \frac{h}{s}. \]

For the person at distance \(t\):

The angle here is \(90^\circ - \alpha\). But we know that: \[ \tan(90^\circ - \alpha) = \cot(\alpha). \]

So, \[ \tan(90^\circ - \alpha) = \frac{h}{t} = \cot(\alpha). \]

Step 3: Relating the two equations.

From the first equation: \[ \tan(\alpha) = \frac{h}{s}. \]

From the second equation: \[ \cot(\alpha) = \frac{h}{t}. \]

We know that: \[ \tan(\alpha) \times \cot(\alpha) = 1. \]

Substitute the values: \[ \left( \frac{h}{s} \right) \times \left( \frac{h}{t} \right) = 1. \]

Step 4: Simplify the equation.

\[ \frac{h^2}{st} = 1. \]

Multiply both sides by \(st\):

\[ h^2 = st. \]

Step 5: Take square root.

\[ h = \sqrt{st}. \]

So, the height of the tower is \(\sqrt{st}\,\text{metres}.\)

Step 1: Let the height of the tower be \(h\,\text{m}\).

Step 2: When the Sun's angle of elevation is \(\theta\), the relation between height and shadow is:

\(\tan \theta = \dfrac{\text{opposite side}}{\text{adjacent side}} = \dfrac{h}{\text{shadow length}}\)

So, \(\text{shadow length} = \dfrac{h}{\tan \theta}\).

Step 3: For \(30^\circ\):

\(L_{30} = \dfrac{h}{\tan 30^\circ}\).

Since \(\tan 30^\circ = \dfrac{1}{\sqrt{3}}\), we get:

\(L_{30} = \dfrac{h}{1/\sqrt{3}} = h\sqrt{3}\).

Step 4: For \(60^\circ\):

\(L_{60} = \dfrac{h}{\tan 60^\circ}\).

Since \(\tan 60^\circ = \sqrt{3}\), we get:

\(L_{60} = \dfrac{h}{\sqrt{3}}\).

Step 5: According to the question, the shadow at \(30^\circ\) is 50 m longer than the shadow at \(60^\circ\):

\(L_{30} - L_{60} = 50\).

Substitute the values:

\(h\sqrt{3} - \dfrac{h}{\sqrt{3}} = 50\).

Step 6: Take LHS to a single fraction:

\(\dfrac{3h - h}{\sqrt{3}} = 50\).

\(\dfrac{2h}{\sqrt{3}} = 50\).

Step 7: Multiply both sides by \(\sqrt{3}\):

\(2h = 50\sqrt{3}\).

Step 8: Divide both sides by 2:

\(h = 25\sqrt{3}\,\text{m}\).

Final Answer: The height of the tower is \(25\sqrt{3}\,\text{m}\).

Step 1: Let the height of the tower be \(H\), and the horizontal distance of the observation point from the tower be \(x\).

Step 2: From the point of observation, the angle of elevation to the top of the tower only is \(\alpha\). By definition of tangent in a right-angled triangle: \[ \tan\alpha = \frac{\text{opposite side}}{\text{adjacent side}} = \frac{H}{x} \] Therefore, \[ H = x \tan\alpha. \]

Step 3: The top of the flag staff is at height \(H + h\) (tower height + flag staff height). The angle of elevation to the top of the flag staff is \(\beta\). Again, using tangent: \[ \tan\beta = \frac{H + h}{x} \] So, \[ H + h = x \tan\beta. \]

Step 4: We now have two equations: (i) \(H = x \tan\alpha\) (ii) \(H + h = x \tan\beta\)

Step 5: Eliminate \(x\). From (i): \[ x = \frac{H}{\tan\alpha}. \] Substitute this into (ii): \[ H + h = \frac{H}{\tan\alpha} \cdot \tan\beta. \]

Step 6: Simplify: \[ H + h = H \cdot \frac{\tan\beta}{\tan\alpha}. \]

Step 7: Multiply both sides by \(\tan\alpha\): \[ (H + h) \tan\alpha = H \tan\beta. \]

Step 8: Rearranging: \[ H \tan\beta - H \tan\alpha = h \tan\alpha. \] \[ H(\tan\beta - \tan\alpha) = h \tan\alpha. \]

Step 9: Finally, divide both sides by \((\tan\beta - \tan\alpha)\): \[ H = \frac{h \tan\alpha}{\tan\beta - \tan\alpha}. \]

Therefore, the height of the tower is proved as required.

Step 1: We are given that \(\tan\theta + \sec\theta = \ell\).

Step 2: Recall the identity: \((\sec\theta + \tan\theta)(\sec\theta - \tan\theta) = 1\). This means that if we know \(\sec\theta + \tan\theta\), then we can also find \(\sec\theta - \tan\theta\).

Step 3: Since \(\sec\theta + \tan\theta = \ell\), substitute this into the identity:

\(\ell(\sec\theta - \tan\theta) = 1\)

So, \(\sec\theta - \tan\theta = \dfrac{1}{\ell}\).

Step 4: Now we have two equations:

• \(\sec\theta + \tan\theta = \ell\)
• \(\sec\theta - \tan\theta = \dfrac{1}{\ell}\)

Step 5: Add these two equations to remove \(\tan\theta\):

\((\sec\theta + \tan\theta) + (\sec\theta - \tan\theta) = \ell + \dfrac{1}{\ell}\)

\(2\sec\theta = \ell + \dfrac{1}{\ell}\)

Step 6: Divide both sides by 2 to isolate \(\sec\theta\):

\(\sec\theta = \dfrac{\ell + \dfrac{1}{\ell}}{2}\)

Step 7: Simplify the fraction:

\(\sec\theta = \dfrac{\dfrac{\ell^2 + 1}{\ell}}{2}\)

\(\sec\theta = \dfrac{\ell^2 + 1}{2\ell}\)

Final Answer: \(\sec\theta = \dfrac{\ell^2 + 1}{2\ell}\).

Step 1: We are given two equations:

• \(\sin\theta + \cos\theta = p\)
• \(\sec\theta + \csc\theta = q\)

Step 2: First, let us square the first equation:

\[(\sin\theta + \cos\theta)^2 = p^2\]

Expanding: \(\sin^2\theta + \cos^2\theta + 2\sin\theta\cos\theta = p^2\).

Step 3: Recall the identity: \(\sin^2\theta + \cos^2\theta = 1\).

So: \(1 + 2\sin\theta\cos\theta = p^2\).

Step 4: Rearranging, we get:

\(p^2 - 1 = 2\sin\theta\cos\theta\).   (Equation A)

Step 5: Now, consider the second given expression:

\(q = \sec\theta + \csc\theta\).

Write \(\sec\theta = \dfrac{1}{\cos\theta}\), and \(\csc\theta = \dfrac{1}{\sin\theta}\).

So: \(q = \dfrac{1}{\cos\theta} + \dfrac{1}{\sin\theta}\).

Step 6: Take LCM:

\(q = \dfrac{\sin\theta + \cos\theta}{\sin\theta \cos\theta}\).

Step 7: But from Step 1, \(\sin\theta + \cos\theta = p\).

So: \(q = \dfrac{p}{\sin\theta \cos\theta}\).   (Equation B)

Step 8: Now, multiply Equation (A) and (B):

\[ q(p^2 - 1) = \dfrac{p}{\sin\theta\cos\theta} \times (2\sin\theta\cos\theta) \]

Step 9: The \(\sin\theta\cos\theta\) cancels out:

\[ q(p^2 - 1) = 2p \]

Final Answer: Hence proved.

Step 1: We are given the relation:

\[ a\sin\theta + b\cos\theta = c. \]

Step 2: Let us consider the expression we want to prove:

\[ a\cos\theta - b\sin\theta. \]

Our goal is to show it equals \( \pm \sqrt{a^2 + b^2 - c^2} \).

Step 3: To connect the two expressions, square and add them:

\[(a\sin\theta + b\cos\theta)^2 + (a\cos\theta - b\sin\theta)^2.\]

Step 4: Expand each square.

First part: \((a\sin\theta + b\cos\theta)^2 = a^2\sin^2\theta + 2ab\sin\theta\cos\theta + b^2\cos^2\theta.\)

Second part: \((a\cos\theta - b\sin\theta)^2 = a^2\cos^2\theta - 2ab\sin\theta\cos\theta + b^2\sin^2\theta.\)

Step 5: Add both results together:

\[ (a^2\sin^2\theta + 2ab\sin\theta\cos\theta + b^2\cos^2\theta) \\ + (a^2\cos^2\theta - 2ab\sin\theta\cos\theta + b^2\sin^2\theta). \]

Step 6: Simplify the terms.

• \(+2ab\sin\theta\cos\theta\) and \(-2ab\sin\theta\cos\theta\) cancel out.
• Combine like terms: \(a^2(\sin^2\theta + \cos^2\theta) + b^2(\sin^2\theta + \cos^2\theta).\)

Step 7: Use the identity \(\sin^2\theta + \cos^2\theta = 1\).

So the sum becomes: \(a^2 + b^2.\)

Step 8: Now replace the first square with the given condition:

Since \(a\sin\theta + b\cos\theta = c\), we have \((a\sin\theta + b\cos\theta)^2 = c^2.\)

Step 9: Therefore,

\[ c^2 + (a\cos\theta - b\sin\theta)^2 = a^2 + b^2. \]

Step 10: Rearranging,

\[ (a\cos\theta - b\sin\theta)^2 = a^2 + b^2 - c^2. \]

Step 11: Taking square roots on both sides,

\[ a\cos\theta - b\sin\theta = \pm \sqrt{a^2 + b^2 - c^2}. \]

Hence Proved.

Step 1: We are asked to prove:

\[ \frac{1 + \sec\theta - \tan\theta}{1 + \sec\theta + \tan\theta} = \frac{1 - \sin\theta}{\cos\theta}. \]

Step 2: To simplify the left-hand side (LHS), multiply numerator and denominator by the conjugate of the denominator. The denominator is \(1 + \sec\theta + \tan\theta\). Its conjugate is \(1 + \sec\theta - \tan\theta\).

So,

\[ \text{LHS} = \frac{1 + \sec\theta - \tan\theta}{1 + \sec\theta + \tan\theta} \times \frac{1 + \sec\theta - \tan\theta}{1 + \sec\theta - \tan\theta}. \]

Step 3: Now numerator becomes:

\[(1 + \sec\theta - \tan\theta)^2.\]

The denominator becomes a difference of squares:

\[(1 + \sec\theta)^2 - (\tan\theta)^2.\]

Step 4: Expand the denominator.

\[(1 + \sec\theta)^2 - \tan^2\theta = 1 + 2\sec\theta + \sec^2\theta - \tan^2\theta.\]

Step 5: Use the Pythagoras identity:

\[1 + \tan^2\theta = \sec^2\theta.\]

This means \(\sec^2\theta - \tan^2\theta = 1\).

So denominator = \(1 + 2\sec\theta + 1 = 2(1 + \sec\theta).\)

Step 6: Numerator is \((1 + \sec\theta - \tan\theta)^2\). Take square root idea to simplify later: keep it as is for now.

Step 7: So overall we have:

\[ \text{LHS} = \frac{(1 + \sec\theta - \tan\theta)^2}{2(1 + \sec\theta)}. \]

Step 8: Expand numerator:

\[(1 + \sec\theta - \tan\theta)^2 = (1 + \sec\theta)^2 + \tan^2\theta - 2\tan\theta(1 + \sec\theta).\]

Step 9: Again, replace \(\tan^2\theta = \sec^2\theta - 1.\)

So numerator = \[(1 + 2\sec\theta + \sec^2\theta) + (\sec^2\theta - 1) - 2\tan\theta(1 + \sec\theta).\]

Simplify: numerator = \[2\sec^2\theta + 2\sec\theta - 2\tan\theta(1 + \sec\theta).\]

Step 10: Take common factor 2 from numerator:

Numerator = \[2(\sec^2\theta + \sec\theta - \tan\theta(1 + \sec\theta)).\]

So,

\[ \text{LHS} = \frac{2(\sec^2\theta + \sec\theta - \tan\theta(1 + \sec\theta))}{2(1 + \sec\theta)}. \]

Cancel 2:

\[ = \frac{\sec^2\theta + \sec\theta - \tan\theta(1 + \sec\theta)}{1 + \sec\theta}. \]

Step 11: Split numerator:

\[= \frac{\sec^2\theta + \sec\theta}{1 + \sec\theta} - \frac{\tan\theta(1 + \sec\theta)}{1 + \sec\theta}.\]

Cancel terms:

\[= \sec\theta - \tan\theta.\]

Step 12: Write in terms of sine and cosine:

\[\sec\theta - \tan\theta = \frac{1}{\cos\theta} - \frac{\sin\theta}{\cos\theta}.\]

Combine: \[= \frac{1 - \sin\theta}{\cos\theta}.\]

Step 13: This is exactly the Right-Hand Side (RHS).

Therefore, identity is proved.

Step 1: Understand the situation

There are two towers standing on the same horizontal ground.

• Height of the first tower = \(30\,\text{m}\)
• Let the height of the second tower be \(H\,\text{m}\) (unknown)
• Let the horizontal distance between the towers be \(D\,\text{m}\)

Step 2: Use the first angle of elevation

From the foot of the second tower, the top of the first tower is seen at an angle of \(60^\circ\).

So, in the right triangle formed:

\[ \tan(60^\circ) = \frac{\text{opposite side}}{\text{adjacent side}} = \frac{30}{D} \]

We know \(\tan(60^\circ) = \sqrt{3}\).

So, \(\sqrt{3} = \frac{30}{D}\).

Cross multiply: \(D = \frac{30}{\sqrt{3}}\).

Simplify: \(D = 10\sqrt{3}\,\text{m}\).

Step 3: Use the second angle of elevation

From the foot of the first tower, the top of the second tower is seen at an angle of \(30^\circ\).

So, in the right triangle formed:

\[ \tan(30^\circ) = \frac{\text{opposite side}}{\text{adjacent side}} = \frac{H}{D} \]

We know \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\).

So, \(\frac{1}{\sqrt{3}} = \frac{H}{10\sqrt{3}}\).

Cross multiply: \(H = \frac{10\sqrt{3}}{\sqrt{3}}\).

Simplify: \(H = 10\,\text{m}\).

Final Answer:

• Distance between the two towers = \(10\sqrt{3}\,\text{m}\)
• Height of the second tower = \(10\,\text{m}\)

Step 1: Draw a vertical tower of height \(h\). Let the top of the tower be point A and the bottom (foot) be point O.

Step 2: Two objects are lying on the ground in a straight line from O. Let the nearer object be at point P and the farther one at point Q.

Step 3: From the top of the tower (point A), the line of sight to P makes angle of depression \(\beta\) and the line of sight to Q makes angle of depression \(\alpha\).

Step 4: In right-angled triangle AOP:

\(\tan\beta = \dfrac{\text{opposite side}}{\text{adjacent side}} = \dfrac{AO}{OP} = \dfrac{h}{x}\).

So, \(x = h \cot\beta\).

Step 5: In right-angled triangle AOQ:

\(\tan\alpha = \dfrac{AO}{OQ} = \dfrac{h}{y}\).

So, \(y = h \cot\alpha\).

Step 6: The distance between the two objects is:

\(PQ = y - x = h \cot\alpha - h \cot\beta\).

Final Answer: Distance \(= h(\cot\alpha - \cot\beta)\).

Step 1: Understand the setup

A ladder of fixed length \(L\) metres is leaning against a vertical wall. The bottom of the ladder (its foot) touches the ground, and the top touches the wall.

The angle between the ladder and the ground is initially \(\alpha\). After moving the foot of the ladder outward, the new angle with the ground is \(\beta\).

Step 2: Express the initial position

From right-angled triangle trigonometry:

• Base (horizontal distance from wall) = \(L \cos \alpha\) metres
• Height (vertical distance up the wall) = \(L \sin \alpha\) metres

Step 3: Express the final position

When the angle changes to \(\beta\):

• Base = \(L \cos \beta\) metres
• Height = \(L \sin \beta\) metres

Step 4: Calculate the change in base (\(p\))

The base has increased from \(L \cos \alpha\) to \(L \cos \beta\).

So, the change in base =

\[p = (L \cos \beta) - (L \cos \alpha) = L(\cos \beta - \cos \alpha)\]

Step 5: Calculate the change in height (\(q\))

The height has decreased from \(L \sin \alpha\) to \(L \sin \beta\).

So, the change in height =

\[q = (L \sin \alpha) - (L \sin \beta) = L(\sin \alpha - \sin \beta)\]

Step 6: Form the required ratio

We need to calculate \(\dfrac{p}{q}\).

Substitute the values of \(p\) and \(q\):

\[\dfrac{p}{q} = \dfrac{L(\cos \beta - \cos \alpha)}{L(\sin \alpha - \sin \beta)}\]

Step 7: Simplify

The ladder length \(L\) cancels out:

\[\dfrac{p}{q} = \dfrac{\cos \beta - \cos \alpha}{\sin \alpha - \sin \beta}\]

Final Result: The relation is proved.

Step 1: Draw the situation

Let the height of the tower be \(H\) (in metres). Let the horizontal distance from the tower to the point on the ground be \(x\) (in metres).

Step 2: Use the first observation point (on the ground)

From the ground, the angle of elevation is \(60^\circ\). By definition of tangent,

\[ \tan 60^\circ = \frac{\text{opposite side}}{\text{adjacent side}} = \frac{H}{x} \]

We know \(\tan 60^\circ = \sqrt{3}\). So,

\[ H = \sqrt{3}x \]

Step 3: Use the second observation point (10 m above the ground)

The new point is 10 m higher. So, the vertical distance from this point to the top of the tower is \(H - 10\).

The angle of elevation is given as \(45^\circ\). Again using tangent,

\[ \tan 45^\circ = \frac{H - 10}{x} \]

But \(\tan 45^\circ = 1\). So,

\[ \frac{H - 10}{x} = 1 \quad \Rightarrow \quad x = H - 10 \]

Step 4: Substitute value of \(x\)

From Step 2: \(H = \sqrt{3}x\). Replace \(x\) by \(H - 10\):

\[ H = \sqrt{3}(H - 10) \]

Step 5: Solve the equation

\[ H = \sqrt{3}H - 10\sqrt{3} \]

Bring terms together:

\[ H - \sqrt{3}H = -10\sqrt{3} \]

\[ H(1 - \sqrt{3}) = -10\sqrt{3} \]

Divide both sides by \((1 - \sqrt{3})\):

\[ H = \frac{-10\sqrt{3}}{1 - \sqrt{3}} \]

Step 6: Simplify the fraction

Multiply numerator and denominator by \((1 + \sqrt{3})\) to remove the surd from the denominator:

\[ H = \frac{-10\sqrt{3}(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})} \]

\[ H = \frac{-10\sqrt{3}(1 + \sqrt{3})}{1 - (\sqrt{3})^2} \]

\[ H = \frac{-10\sqrt{3}(1 + \sqrt{3})}{1 - 3} = \frac{-10\sqrt{3}(1 + \sqrt{3})}{-2} \]

\[ H = 5\sqrt{3}(1 + \sqrt{3}) \]

\[ H = 5\sqrt{3} + 15 \]

Final Answer: The height of the tower is \(H = 15 + 5\sqrt{3}\,\text{m}\).

Step 1: Draw the figure.

Imagine a window at height \(h\) from the ground. From this window:

• You look up at the top of the other house. This is the angle of elevation \(\alpha\).
• You look down at the bottom of the other house. This is the angle of depression \(\beta\).

Let the height of the other house be \(H\) and the horizontal distance between the two houses be \(d\).

Step 2: Use right-angled triangles and trigonometry.

• From the triangle formed at the top of the house:
  \(\tan\alpha = \dfrac{\text{vertical difference}}{\text{horizontal distance}} = \dfrac{H - h}{d}.\)
• From the triangle formed at the bottom of the house:
  \(\tan\beta = \dfrac{\text{vertical difference}}{\text{horizontal distance}} = \dfrac{h}{d}.\)

Step 3: Express \(d\) in terms of \(h\) and \(\beta\).

From \(\tan\beta = h/d\), rearrange to get:
\(d = \dfrac{h}{\tan\beta}.\)

Step 4: Substitute \(d\) into the first equation.

From \(\tan\alpha = (H - h)/d\):
\(H - h = d \cdot \tan\alpha.\)

Substitute \(d = h/\tan\beta\):
\(H - h = \dfrac{h}{\tan\beta} \cdot \tan\alpha.\)

Step 5: Simplify.

\(H - h = h \cdot \tan\alpha \cdot \cot\beta.\)

So, \(H = h + h \tan\alpha \cot\beta.\)

Final Result:
\(H = h \big(1 + \tan\alpha\,\cot\beta\big).\)

Step 1: Understand the problem

The house has two windows. The lower window is 2 m above the ground. The upper window is 4 m higher than this, so it is at \(2 + 4 = 6\,\text{m}\) above the ground.

From the lower window, the angle of elevation to the balloon is \(60^\circ\). From the upper window, the angle of elevation is \(30^\circ\).

We need to find the total height of the balloon from the ground (let us call this \(H\)).

Step 2: Draw right triangles

Let the horizontal distance between the balloon and the wall of the house be \(x\) metres. This forms two right-angled triangles with vertical sides \((H - 2)\) and \((H - 6)\).

Step 3: Write equation using tan for the lower window

For the lower window (at 2 m):

\(\tan 60^\circ = \dfrac{H - 2}{x}\)

We know \(\tan 60^\circ = \sqrt{3}\). So:

\(\sqrt{3} = \dfrac{H - 2}{x} \;\; \Rightarrow \; H - 2 = \sqrt{3}\,x.\)   (Equation 1)

Step 4: Write equation using tan for the upper window

For the upper window (at 6 m):

\(\tan 30^\circ = \dfrac{H - 6}{x}\)

We know \(\tan 30^\circ = \dfrac{1}{\sqrt{3}}\). So:

\(\dfrac{1}{\sqrt{3}} = \dfrac{H - 6}{x} \;\; \Rightarrow \; H - 6 = \dfrac{x}{\sqrt{3}}.\)   (Equation 2)

Step 5: Eliminate x

From Equation (1): \(x = \dfrac{H - 2}{\sqrt{3}}\).

Substitute this into Equation (2):

\(H - 6 = \dfrac{1}{\sqrt{3}} \times \dfrac{H - 2}{\sqrt{3}} = \dfrac{H - 2}{3}.\)

Step 6: Solve for H

Multiply both sides by 3:

\(3(H - 6) = H - 2\)

\(3H - 18 = H - 2\)

\(3H - H = -2 + 18\)

\(2H = 16\)

\(H = 8\,\text{m}\)

Final Answer: The height of the balloon above the ground is 8 metres.