CBSE Class 10 Mathematics Trigonometry Assignment Set 21

Question 18. To find \(AB = AD + DB\) where \(CD = h\) is the height of a tower, and angles of elevation from points \(A\) and \(B\) on the ground are \(\alpha\) and \(\beta\) respectively. Prove that \(AB = \frac{h(\tan\alpha + \tan\beta)}{\tan\alpha \tan\beta}\).
Answer: In right-angled triangle \(\triangle ADC\):
\(\tan\alpha = \frac{CD}{AD} = \frac{h}{AD}\)
\(AD = \frac{h}{\tan\alpha} \quad \text{--- (1)}\)
In right-angled triangle \(\triangle BDC\):
\(\tan\beta = \frac{CD}{BD} = \frac{h}{BD}\)
\(BD = \frac{h}{\tan\beta} \quad \text{--- (2)}\)
Total distance \(AB = AD + DB\)
Substituting values from (1) and (2):
\(AB = \frac{h}{\tan\alpha} + \frac{h}{\tan\beta}\)
Taking LCM and simplifying:
\(AB = h \left( \frac{\tan\beta + \tan\alpha}{\tan\alpha \tan\beta} \right)\)
\(AB = \frac{h(\tan\alpha + \tan\beta)}{\tan\alpha \tan\beta}\)
In simple words: We find the two base segments using the trigonometric ratio of tangent and add them together to get the total length.

Exam Tip: Always draw separate right triangles when dealing with multi-triangle height and distance problems to avoid trigonometric confusion.

Question 19. If \(AB = x\) is the height of a tower, \(BC = a\) and \(DB = b\) are distances from the base along a straight line, and the angles of elevation at the two points are complementary such that \(\angle ACB = \theta\) and \(\angle ADB = 90^\circ - \theta\), prove that \(x = \sqrt{ab}\).
Answer: In right-angled triangle \(\triangle ABC\):
\(\tan\theta = \frac{AB}{BC} = \frac{x}{a}\)
\(x = a \tan\theta \quad \text{--- (1)}\)
In right-angled triangle \(\triangle ABD\):
\(\tan(90^\circ - \theta) = \frac{AB}{DB} = \frac{x}{b}\)
Since \(\tan(90^\circ - \theta) = \cot\theta\):
\(\cot\theta = \frac{x}{b} \implies x = b \cot\theta \quad \text{--- (2)}\)
Multiplying equation (1) and equation (2):
\(x \cdot x = (a \tan\theta) \cdot (b \cot\theta)\)
\(x^2 = ab (\tan\theta \cdot \frac{1}{\tan\theta})\)
\(x^2 = ab\)
\(x = \sqrt{ab}\)
In simple words: Using complementary angle relations transforms tangent into cotangent, which cancels out when multiplied to leave a simple square root product.

Exam Tip: Remember the standard identity \(\tan(90^\circ - \theta) = \cot\theta\) as it frequently simplifies height and distance proofs involving collinear points.

Question 20(a). Let \(AB = x\) be the height of a tower and \(DC = a\) be the distance between two observation points on the ground where the angles of elevation are \(\beta\) and \(\alpha\) respectively. Prove that \(x = \frac{a \tan\alpha \tan\beta}{\tan\alpha - \tan\beta}\).
Answer: In right-angled triangle \(\triangle ABC\):
\(\tan\alpha = \frac{AB}{BC} = \frac{x}{BC}\)
\(BC = \frac{x}{\tan\alpha} \quad \text{--- (1)}\)
In right-angled triangle \(\triangle ABD\):
\(\tan\beta = \frac{AB}{BD} = \frac{x}{BD}\)
\(BD = \frac{x}{\tan\beta} \quad \text{--- (2)}\)
From the geometry of the figure, \(DB - CB = DC = a\):
Substitute (1) and (2) into the length relation:
\(\frac{x}{\tan\beta} - \frac{x}{\tan\alpha} = a\)
Factor out \(x\):
\(x \left( \frac{1}{\tan\beta} - \frac{1}{\tan\alpha} \right) = a\)
\(x \left( \frac{\tan\alpha - \tan\beta}{\tan\alpha \tan\beta} \right) = a\)
Isolate \(x\):
\(x = \frac{a \tan\alpha \tan\beta}{\tan\alpha - \tan\beta}\)
In simple words: By expressing both larger and smaller base distances in terms of height, subtracting them gives the given distance segment \(a\).

Exam Tip: Pay close attention to subtraction order in denominators when dealing with points lying on the same side of a tower.

Question 20(b). Let \(AB = h\) be the height of an observer above a lake, and point \(C\) be the position of a cloud at height \(x\) above the lake. If the angle of elevation of the cloud is \(\alpha\) and the angle of depression of its reflection in the lake is \(\beta\), prove that \(x = \frac{h(\tan\beta + \tan\alpha)}{\tan\beta - \tan\alpha}\).
Answer: Let height of cloud above the observer level be \(CE = x - h\), and depth of reflection below observer level be \(EF = x + h\). Let the horizontal distance be \(DE\).
From right-angled triangle \(\triangle CDE\):
\(\tan\alpha = \frac{CE}{DE} = \frac{x - h}{DE}\)
\(DE = \frac{x - h}{\tan\alpha} \quad \text{--- (1)}\)
From right-angled triangle \(\triangle DEF\):
\(\tan\beta = \frac{EF}{DE} = \frac{x + h}{DE}\)
\(DE = \frac{x + h}{\tan\beta} \quad \text{--- (2)}\)
Equating equation (1) and equation (2) since horizontal distance \(DE\) is constant:
\(\frac{x - h}{\tan\alpha} = \frac{x + h}{\tan\beta}\)
Cross-multiplying:
\((x - h) \tan\beta = (x + h) \tan\alpha\)
\(x \tan\beta - h \tan\beta = x \tan\alpha + h \tan\alpha\)
Grouping terms of \(x\) on one side:
\(x \tan\beta - x \tan\alpha = h \tan\beta + h \tan\alpha\)
\(x (\tan\beta - \tan\alpha) = h (\tan\beta + \tan\alpha)\)
\(x = \frac{h(\tan\beta + \tan\alpha)}{\tan\beta - \tan\alpha}\)
In simple words: The height of a cloud above water equals the depth of its reflection below water. Equating the shared horizontal distance from two triangles solves for height.

Exam Tip: For reflection problems over water, remember that the height of the object above the surface equals the depth of the image below the surface.