CBSE Solutions for Class 10 Maths Chapter 9: Some Applications of Trigonometry || CBSE Class 10 Maths Chapter 9 Some Applications of Trigonometry solutions

Let AD be the height of the statue which is 1.6 m , And DB be the height of the pedestal.
And C be the point where the observer is present.

To find: height of pedestal, BD
Now,
In \( \Delta \mathrm{ABC} \),
\( \tan \theta=\frac{p}{b} \)
where \( \mathrm{p}= \) perpendicular and \( \mathrm{b}= \) base
\( \tan 60^{\circ}=\frac{p}{b} \)
\( \sqrt{3}=\frac{D B+1.6}{b} \)
\( b=\frac{D B+1.6}{\sqrt{3}} \ldots\) (i)
In \( \triangle D B C \),
\( \tan \theta=\frac{p}{b} \)
\( \tan 45^{\circ}=\frac{p}{b} \)
\( 1=\frac{D B}{b} \)
\( b=\mathrm{DB} \ldots\) (ii)
On comparing (i) and (ii),
\(D B=\frac{D B+1.6}{\sqrt{3}}\)
\(
D B \sqrt{3}=D B+1.6\)
\(D B(\sqrt{3}-1)=1.6\)
\(D B=\frac{1.6}{\sqrt{3}-1}\)
\(D B=\frac{1.6}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}\)
\(D B=\frac{1.6 \times(\sqrt{3}+1)}{3-1}\)
\(D B=0.8(\sqrt{3}+1)
\)
Hence, height of pedestal is \( 0.8(\sqrt{3}+1) \mathrm{m} \).

Let us take AB as building and Let us take DC as tower \( =50 \mathrm{~m} \), \( \angle \mathrm{ACB}=30^{\circ} \)
And,
\(
\angle \mathrm{DBC}=60^{\circ},
\)
\( \mathrm{AB}= \) ?
In \( \Delta \) DCB;
\( \tan \theta=\frac{p}{b} \)
where, \( \mathrm{p}= \) perpendicular and \( \mathrm{b}= \) base
\( \tan 60^{\circ}=\frac{50}{b} \)
\( \sqrt{3}=\frac{50}{b} \)
\( b=\frac{50}{\sqrt{3}} \)
In \( \Delta \mathrm{ACB} \)
\( \tan \theta=\frac{p}{b} \)
where \( \mathrm{p}= \) perpendicular and \( \mathrm{b}= \) base
\( \tan 30^{\circ}=\frac{A B}{b} \)
\(
\frac{1}{\sqrt{3}}=\frac{A B}{\frac{50}{\sqrt{3}}}\)
\(A B=\frac{50}{\sqrt{3} \times \sqrt{3}}\)
\(\mathrm{AB}=16.67 \mathrm{~m}
\)

Let AB and DE be the two poles, and C be the point of observation. Given, width of road, \( \mathrm{BD}=80 \mathrm{~m} \) Angle of elevation to \( \mathrm{AB}, \angle \mathrm{ACB}= \) \( 30^{\circ} \) Angle of elevation to \( \mathrm{DE}, \angle \mathrm{ECD}=60^{\circ} \) To find: Height of buildings AB and DE .
In \( \Delta \mathrm{ACB} \)
\( \tan \theta=\frac{p}{b} \)
\( \tan 30^{\circ}=\frac{A B}{B C} \)
\( \frac{1}{\sqrt{3}}=\frac{A B}{B C} \)
\( A B=\frac{B C}{\sqrt{3}} \ldots \) (i)
In \( \triangle E D C \)
\( \tan \theta=\frac{p}{b} \)
\( \tan 60^{\circ}=\frac{E D}{C D} \)
\( \sqrt{3}=\frac{E D}{80-B C} \)
\( E D=\sqrt{3}(80-B C) \ldots\) (ii)
We know that \( \mathrm{AB}=\mathrm{ED} \) as the poles are of same height.
Hence, from (i) and (ii),
\(
\frac{B C}{\sqrt{3}}=\sqrt{3}(80-B C)
\)
cross multiplying, we get
\(
\mathrm{BC}=3(80-\mathrm{BC})\)
\(\mathrm{BC}=240-3 \mathrm{BC}\)
\(4 \mathrm{BC}=240\)
\(\mathrm{BC}=60 \mathrm{~m}
\)
Now using the value of BC in (i),
\(
\mathrm{AB}=20 \sqrt{3} \mathrm{~m}\)
\(\tan 60^{\circ}=\frac{E D}{C D}
\)
Now from triangle EDC, \( \sqrt{3}=\frac{20 \sqrt{3}}{C D} \)
\(
\mathrm{CD}=20 \mathrm{~m}
\)
Therefore, Height of Pole \( =20 {\sqrt{3}} \mathrm{~m} \). Distances of poles from observing point \( =60 \mathrm{~m} \) and \( 20\) m