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Height and Distance Class 10 Optional Math Exercise 7.6
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1. Heights and Distances: Class 10 Optional Mathematics Trigonometry Overview and Key Formulas

Unit 7: Trigonometry (Optional Mathematics, Grade 10, SEE, Nepal) • Exercise 7.6 Heights and Distances

Height and distance problems use the angle of elevation or the angle of depression with the tangent ratio \(\tan\theta=\dfrac{\text{height}}{\text{distance}}\). This page gives the height and distance formulas, height and distance examples, the angle formula for angles of elevation and depression and step by step solutions to every question of Exercise 7.6 (Class 10 Optional Mathematics, SEE, Nepal), with the Nepali question in brackets.

This is part of the unit prescribed by the Curriculum Development Centre (CDC), Nepal. Multiple choice solutions also have Nepali in brackets, and the other solutions are in English because the steps and symbols are the same in both languages.

  • Heights and distances (Exercise 7.6): draw a right-angled triangle for each observation, mark the angle of elevation or depression from the horizontal, and use \(\tan\theta=\dfrac{\text{height}}{\text{distance}}\). With two observations, write both distances in terms of the height and use the difference or sum.

Key formulas to remember

Heights and distances: angle of elevation (object above eye level) and angle of depression (object below eye level) are both measured from the horizontal. \(\tan\theta=\dfrac{\text{perpendicular}}{\text{base}}\); with \(\theta=45^\circ\) height and distance are equal; \(\tan30^\circ=\dfrac1{\sqrt3}\), \(\tan60^\circ=\sqrt3\).
Two-point method: for angles of elevation \(\alpha>\beta\) observed from two points on the same side, \(d_1=\dfrac h{\tan\alpha}\), \(d_2=\dfrac h{\tan\beta}\) and the gap between the points is \(d_2-d_1\). To find an angle from known sides use \(\theta=\tan^{-1}\dfrac{\text{height}}{\text{distance}}\).

For more chapters, visit the complete SEE Class 10 Optional Mathematics notes. You can also read about trigonometric identities for extra understanding.

2. Exercise 7.6 Heights and Distances: Class 10 Optional Mathematics Trigonometry

Exercise 7.6 Heights and Distances: Multiple Choice Questions (बहुवैकल्पिक प्रश्न)

Q.1 Tick (✓) the correct option for the given questions:

(दिइएका प्रश्नहरूको उपयुक्त विकल्पमा ठिक चिह्न (✓) लगाउनुहोस् : )

A. What is the angle called which is formed by the line of sight with the horizontal when looking at an object below the eye level? (तलतिर रहेको वस्तु हेर्दा दृष्टि रेखा र क्षितिज रेखाबिच बन्ने कोणलाई के भनिन्छ ?)

  • a) Angle of depression
  • b) Acute angle
  • c) Angle of elevation
  • d) Right angle

Correct Answer: a) Angle of depression

Solution:

  • The angle between the line of sight and the horizontal, when the object is below eye level, is the angle of depression.

(आँखाको स्तरभन्दा तलको वस्तु हेर्दा बन्ने कोणलाई अवनति कोण भनिन्छ ।)

B. What is the angle called that is formed by the line of sight with the horizontal when looking at an object above the eye level? (माथितिर रहेको वस्तु हेर्दा दृष्टि रेखा र क्षितिज रेखाबिच बन्ने कोणलाई के भनिन्छ ?)

  • a) Angle of depression
  • b) Acute angle
  • c) Angle of elevation
  • d) Right angle

Correct Answer: c) Angle of elevation

Solution:

  • The angle between the line of sight and the horizontal, when the object is above eye level, is the angle of elevation.

(आँखाको स्तरभन्दा माथिको वस्तु हेर्दा बन्ने कोणलाई उन्नतांश कोण भनिन्छ ।)

C. If the angle of elevation of an object observed from a place is \(0^\circ\), where is that object? (कुनै स्थानबाट एउटा वस्तुको अवलोकन गर्दा बन्ने उन्नतांश कोण \(0^\circ\) भयो भने सो वस्तु कहाँ हुन्छ ?)

  • a) Above
  • b) Below
  • c) At eye level
  • d) Not visible

Correct Answer: c) At eye level

Solution:

  • The angle of elevation is measured upward from the horizontal through the eye. If it is \(0^\circ\), the line of sight is the horizontal itself, so the object is at eye level.

(उन्नतांश कोण \(0^\circ\) हुँदा दृष्टि रेखा क्षितिज रेखासँग मिल्छ, त्यसैले वस्तु आँखाकै स्तरमा हुन्छ ।)

D. When the angle of elevation of the top of a pole observed from a place is \(45^\circ\), what is the relationship between the height of the pole and the distance from that place to the pole? (कुनै स्थानबाट खम्बाको टुप्पाको उन्नतांश कोण \(45^\circ\) हुँदा खम्बाको उचाइ र सो स्थानबाट खम्बासम्मको दुरीको सम्बन्ध के हुन्छ ?)

  • a) Height > Distance
  • b) Height < Distance
  • c) Equal
  • d) No relation

Correct Answer: c) Equal

Solution:

  • Let the height be \(h\) and the distance \(d\).
\[\tan45^\circ=\frac hd \implies 1=\frac hd \implies h=d\]

(\(\tan45^\circ=\frac hd=1\) भएकाले \(h=d\) हुन्छ ।)

E. As the angle of elevation of the top of a tower observed from a place increases, what happens to the distance? (कुनै स्थानबाट टावरको टुप्पाको उन्नतांश कोण बढ्दै जाँदा दुरी कस्तो हुन्छ ?)

  • a) Increases
  • b) Decreases
  • c) Remains the same
  • d) No relation

Correct Answer: b) Decreases

Solution:

  • The height \(h\) of the tower is fixed, and \(d=\dfrac{h}{\tan\theta}\).
  • As \(\theta\) increases from \(0^\circ\) to \(90^\circ\), \(\tan\theta\) increases, so \(d\) decreases.

(\(\theta\) बढ्दा \(\tan\theta\) बढ्छ र \(d=\frac h{\tan\theta}\) घट्छ ।)

F. What is the angle of elevation when observing the top of a tree of height \(10\sqrt3\) m from a point \(10\) m away? (कुनै स्थानबाट \(10\) m टाढा रहेको \(10\sqrt3\) m अग्लो रुखको टुप्पामा हेर्दा कति डिग्रीको उन्नतांश कोण बन्छ ?)

  • a) \(30^\circ\)
  • b) \(45^\circ\)
  • c) \(60^\circ\)
  • d) \(120^\circ\)

Correct Answer: c) \(60^\circ\)

Solution:

  • Here, perpendicular \(=10\sqrt3\) m and base \(=10\) m.
\[\tan\theta=\frac{10\sqrt3}{10}=\sqrt3=\tan60^\circ \implies \theta=60^\circ\]

(\(\tan\theta=\sqrt3\) भएकाले \(\theta=60^\circ\) हुन्छ ।)

Exercise 7.6 Heights and Distances: Written Questions (लेख्ने प्रश्न)

Q.2 Find the values of \(x\), \(y\) and \(\theta\) in the given figures:

  • a. Figure (a)
  • b. Figure (b)
  • c. Figure (c)
  • d. Figure (d)
  • e. Figure (e)
  • f. Figure (f)
Class 10 Optional Maths Exercise 7.6 Question 2: Heights and Distances
Figure: Triangles for Exercise 7.6, Question 2 (a) to (f).

(दिइएका चित्रहरूबाट \(x\), \(y\) र \(\theta\) का मान पत्ता लगाउनुहोस् : )

a. Figure (a)

The data given for the figure are \(AB=20\) m, \(\angle B=90^\circ\), \(\angle C=45^\circ\) and \(AC=y\). The side \(x\) is not described, so we take \(x=BC\).

\[\begin{aligned}\sin45^\circ&=\frac{AB}{AC}=\frac{20}{y} \\ \frac1{\sqrt2}&=\frac{20}{y} \\ y&=20\sqrt2\approx28.28\text{ m}\end{aligned}\]\[\begin{aligned}\tan45^\circ&=\frac{AB}{BC}=\frac{20}{x} \\ x&=20\text{ m}\end{aligned}\]
\(y=20\sqrt2\text{ m}\approx28.28\text{ m}\) (and \(x=BC=20\) m)
b. Figure (b)
\[\begin{aligned}\tan30^\circ&=\frac{PQ}{QR} \\ \frac1{\sqrt3}&=\frac{x}{10} \\ x&=\frac{10}{\sqrt3}\approx5.77\text{ m}\end{aligned}\]
\(x=\dfrac{10}{\sqrt3}\text{ m}\approx5.77\text{ m}\)
c. Figure (c)

In \(\triangle ABD\):

\[\begin{aligned}\tan30^\circ&=\frac{AB}{BD} \\ \frac1{\sqrt3}&=\frac{x}{30} \\ x&=\frac{30}{\sqrt3}=10\sqrt3\approx17.32\text{ m}\end{aligned}\]

In \(\triangle ABC\):

\[\begin{aligned}\tan\theta&=\frac{AB}{BC}=\frac{10\sqrt3}{10}=\sqrt3 \\ \theta&=60^\circ\end{aligned}\]
\(x=10\sqrt3\text{ m}\approx17.32\text{ m},\ \theta=60^\circ\)
d. Figure (d)

Here \(N,P,B\) lie on a line, so \(NB=NP+PB=45+y\). In \(\triangle LPB\):

\[\begin{aligned}\tan45^\circ&=\frac{LB}{PB}=\frac xy \\ x&=y\end{aligned}\]

In \(\triangle LNB\):

\[\begin{aligned}\tan30^\circ&=\frac{LB}{NB}=\frac{x}{45+y}=\frac{x}{45+x} \\ 45+x&=\sqrt3x \\ x(\sqrt3-1)&=45 \\ x&=\frac{45}{\sqrt3-1}=\frac{45(\sqrt3+1)}{2}\approx61.47\text{ m}\end{aligned}\]

Since \(y=x\), \(y\approx61.47\) m.

\(x\approx61.47\text{ m},\ y\approx61.47\text{ m}\)
e. Figure (e)

\(AB\) is the vertical height \(50\) m, \(E\) is on the horizontal through \(A\), and \(D\), \(C\) are points seen from \(A\) at depressions \(30^\circ\) and \(45^\circ\). Draw \(DF\perp AB\), so \(DF=BC=y\) and \(BF=DC=x\). By alternate angles, \(\angle ACB=45^\circ\) and \(\angle ADF=30^\circ\).

In \(\triangle ABC\):

\[\begin{aligned}\tan45^\circ&=\frac{AB}{BC}=\frac{50}{y} \\ y&=50\text{ m}\end{aligned}\]

In \(\triangle ADF\):

\[\begin{aligned}\tan30^\circ&=\frac{AF}{DF}=\frac{AF}{50} \\ AF&=\frac{50}{\sqrt3}\approx28.87\text{ m}\end{aligned}\]\[\begin{aligned}x&=BF=AB-AF=50-28.87 \\ &=21.13\text{ m}\end{aligned}\]
\(x\approx21.13\text{ m},\ y=50\text{ m}\)
f. Figure (f)

In \(\triangle WVT\):

\[\begin{aligned}\tan45^\circ&=\frac{WT}{VT}=\frac{40}{x} \\ x&=40\text{ m}\end{aligned}\]

In \(\triangle WUT\):

\[\begin{aligned}\tan30^\circ&=\frac{WT}{UT}=\frac{40}{y} \\ y&=40\sqrt3\approx69.28\text{ m}\end{aligned}\]
\(x=40\text{ m},\ y=40\sqrt3\text{ m}\approx69.28\text{ m}\)
Q.3
a. A point on the roof of a house is observed from a point on the ground and the angle of elevation is found to be \(60^\circ\). On moving \(50\) m away from the house, the angle of elevation is \(30^\circ\). Find the height of the house. (एउटा घरको छतको एउटा विन्दुलाई जमिनको एउटा विन्दुबाट अवलोकन गर्दा उन्नतांश कोण \(60^\circ\) पाइयो । सो विन्दुबाट \(50\) m टाढा गएर हेर्दा उन्नतांश कोण \(30^\circ\) पाइयो भने सो घरको उचाइ पत्ता लगाउनुहोस् ।)

Let \(AB=h\) be the height of the house and \(C,D\) the two observation points, with \(CD=50\) m, \(\angle ACB=60^\circ\), \(\angle ADB=30^\circ\).

\[\begin{aligned}\tan60^\circ&=\frac{h}{BC}\implies BC=\frac h{\sqrt3} \\ \tan30^\circ&=\frac{h}{BD}=\frac{h}{BC+50} \\ \frac1{\sqrt3}&=\frac{h}{\frac h{\sqrt3}+50} \\ \frac h{\sqrt3}+50&=\sqrt3h \\ 50&=h\left(\sqrt3-\frac1{\sqrt3}\right)=\frac{2h}{\sqrt3} \\ h&=25\sqrt3\approx43.30\text{ m}\end{aligned}\]
Height of the house \(=25\sqrt3\text{ m}\approx43.30\text{ m}\)
b. The angles of elevation of the top of a \(30\) ft tall pole, observed from two points on the same side of it, are \(60^\circ\) and \(45^\circ\). Find the distance between the two points. (एउटा \(30\) ft अग्लो खम्बाको एकैतिर समतल सतहमा रहेका दुई विन्दुबाट खम्बाको टुप्पोको उन्नतांश कोणहरू क्रमशः \(60^\circ\) र \(45^\circ\) पाइयो भने ती दुई विन्दुबिचको दुरी पत्ता लगाउनुहोस् ।)

Let \(AB=30\) ft be the pole, with \(\angle ACB=60^\circ\) and \(\angle ADB=45^\circ\).

\[\begin{aligned}BC&=\frac{30}{\tan60^\circ}=\frac{30}{\sqrt3}=10\sqrt3\approx17.32\text{ ft} \\ BD&=\frac{30}{\tan45^\circ}=30\text{ ft} \\ CD&=BD-BC=30-17.32=12.68\text{ ft}\end{aligned}\]
Distance between the points \(\approx12.68\) ft
Q.4
a. From the top of a \(500\) m high hill, two points on the ground on the same side are observed at angles of depression \(60^\circ\) and \(30^\circ\). Find the distance between the two points. (\(500\) m अग्लो पहाडको टुप्पोबाट जमिनमा एकैतिर रहेका दुई विन्दुहरूको अवनति कोणहरू क्रमशः \(60^\circ\) र \(30^\circ\) पाइयो । ती दुई विन्दुबिचको दुरी पत्ता लगाउनुहोस् ।)

Let \(AB=500\) m be the hill. By alternate angles, \(\angle ACB=60^\circ\) and \(\angle ADB=30^\circ\).

\[\begin{aligned}BC&=\frac{500}{\tan60^\circ}=\frac{500}{\sqrt3}\approx288.68\text{ m} \\ BD&=\frac{500}{\tan30^\circ}=500\sqrt3\approx866.03\text{ m} \\ CD&=BD-BC=866.03-288.68=577.35\text{ m}\end{aligned}\]
Distance between the two points \(\approx577.35\) m
b. From the roof of a house, two people sitting on the ground in a line, \(10\) ft apart, are observed at angles of depression \(60^\circ\) and \(45^\circ\). Find the height of the house. (एउटा घरको छतबाट चउरमा एकैतिर एउटै रेखामा \(10\) ft को दुरीमा बसेका दुई मानिसको अवनति कोणहरू क्रमशः \(60^\circ\) र \(45^\circ\) पाइयो भने त्यो घरको उचाइ पत्ता लगाउनुहोस् ।)

Let \(AB=h\) be the house, \(CD=10\) ft, \(\angle ACB=60^\circ\) and \(\angle ADB=45^\circ\).

\[\begin{aligned}BC&=\frac h{\sqrt3},\qquad BD=\frac h{\tan45^\circ}=h \\ BD-BC&=10 \\ h-\frac h{\sqrt3}&=10 \\ h&=\frac{10\sqrt3}{\sqrt3-1}=\frac{10\sqrt3(\sqrt3+1)}{2}=5(3+\sqrt3) \\ &\approx23.66\text{ ft}\end{aligned}\]
Height of the house \(\approx23.66\) ft
Q.5
a. The angles of elevation of the top of a pole, observed from two points \(40\) m and \(36\) m from its base on the same side, are supplementary. Find the height of the pole. (एउटा खम्बाको फेदबाट खम्बाको एकैतिर \(40\) m र \(36\) m दुरीमा रहेका दुई विन्दुबाट खम्बाको टुप्पो अवलोकन गर्दा बनेका कोणहरू एकअर्काका समपूरक पाइयो भने सो खम्बाको उचाइ पत्ता लगाउनुहोस् ।)

Note on the wording

Two angles of elevation of the same pole are both acute, so they cannot add up to \(180^\circ\). The intended meaning is that they add up to \(90^\circ\) (complementary), which is how the solution below proceeds.

Let \(AB=h\) be the pole, \(BC=36\) m, \(BD=40\) m, \(\angle ACB=\theta\) and \(\angle ADB=90^\circ-\theta\).

\[\begin{aligned}\tan\theta&=\frac h{36} \\ \tan(90^\circ-\theta)&=\cot\theta=\frac h{40}\implies\tan\theta=\frac{40}{h} \\ \frac h{36}&=\frac{40}{h} \\ h^2&=1440 \\ h&=\sqrt{1440}=12\sqrt{10}\approx37.95\text{ m}\end{aligned}\]
Height of the pole \(=12\sqrt{10}\text{ m}\approx37.95\text{ m}\)
b. Two poles are \(50\) m apart and one is twice as high as the other. From the midpoint between them, the angles of elevation of their tops are complementary (given as “supplementary”). Find the heights of the poles. (\(50\) m दुरीमा रहेका दुईओटा खम्बामा एउटाको उचाइ अर्कोको दोब्बर छ । ती खम्बाको बिचमा पर्ने एक विन्दुबाट दुबै खम्बाका टुप्पा अवलोकन गर्दा बनेका कोणहरू एकअर्काका समपूरक पाइयो भने ती खम्बाको उचाइ पत्ता लगाउनुहोस् ।)

As in Question 5(a), the two acute angles must add up to \(90^\circ\). The midpoint is \(25\) m from each pole. Let the shorter pole be \(h\) and the taller \(2h\), with angles \(\theta\) and \(90^\circ-\theta\).

\[\begin{aligned}\tan\theta&=\frac h{25} \\ \cot\theta&=\frac{2h}{25}\implies\tan\theta=\frac{25}{2h} \\ \frac h{25}&=\frac{25}{2h} \\ 2h^2&=625 \\ h&=\frac{25}{\sqrt2}\approx17.68\text{ m} \\ 2h&\approx35.36\text{ m}\end{aligned}\]
Heights of the poles \(\approx17.68\) m and \(35.36\) m
Q.6
a. The shadow of a vertical pole on level ground is \(50\) m longer when the sun\u2019s altitude is \(30^\circ\) than when it is \(60^\circ\). Find the height of the pole. (समतल सतहमा रहेको खम्बाको छाया सूर्यको उचाइ \(60^\circ\) भएको समयमा भन्दा \(30^\circ\) भएको समयमा \(50\) m लामो हुन्छ भने सो खम्बाको उचाइ पत्ता लगाउनुहोस् ।)

Let \(AB=h\) be the pole. The shadow is \(BC\) at \(60^\circ\) and \(BD\) at \(30^\circ\), with \(BD-BC=50\).

\[\begin{aligned}BC&=\frac h{\tan60^\circ}=\frac h{\sqrt3},\qquad BD=\frac h{\tan30^\circ}=\sqrt3h \\ \sqrt3h-\frac h{\sqrt3}&=50 \\ \frac{2h}{\sqrt3}&=50 \\ h&=25\sqrt3\approx43.30\text{ m}\end{aligned}\]
Height of the pole \(=25\sqrt3\text{ m}\approx43.30\text{ m}\)
b. When the sun\u2019s altitude is \(30^\circ\), the shadow of a \(35\) ft high pole is \(x\) ft longer than when the altitude is \(45^\circ\). Find \(x\). (समतल सतहमा रहेको \(35\) ft अग्लो खम्बाको छाया सूर्यको उचाइ \(45^\circ\) भएको समयमा भन्दा \(30^\circ\) भएको समयमा \(x\) ft लामो हुन्छ भने \(x\) को मान पत्ता लगाउनुहोस् ।)
\[\begin{aligned}\text{Shadow at }45^\circ&=\frac{35}{\tan45^\circ}=35\text{ ft} \\ \text{Shadow at }30^\circ&=\frac{35}{\tan30^\circ}=35\sqrt3\approx60.62\text{ ft} \\ x&=35\sqrt3-35=35(\sqrt3-1)\approx25.62\text{ ft}\end{aligned}\]
\(x\approx25.62\) ft
Q.7
a. From the roof of a \(30\) m tall building, the angles of depression of the top and the bottom of a column are \(45^\circ\) and \(60^\circ\). Find the height of the column. (\(30\) m अग्लो भवनको छतबाट एउटा स्तम्भको टुप्पो र फेदको अवनति कोणहरू क्रमशः \(45^\circ\) र \(60^\circ\) पाइयो भने सो स्तम्भको उचाइ पत्ता लगाउनुहोस् ।)

Let \(AB=30\) m be the building and \(CD\) the column (\(C\) its bottom, \(D\) its top). Draw \(DF\perp AB\), so \(DF=BC\) and \(CD=BF\). By alternate angles, \(\angle ACB=60^\circ\) and \(\angle ADF=45^\circ\).

\[\begin{aligned}BC&=\frac{30}{\tan60^\circ}=10\sqrt3\approx17.32\text{ m} \\ AF&=DF\tan45^\circ=17.32\text{ m} \\ CD&=BF=AB-AF=30-17.32=12.68\text{ m}\end{aligned}\]
Height of the column \(\approx12.68\) m
b. The pinnacle of a temple is observed from the roof and the foot of a building. The angle of depression from the roof is \(60^\circ\) and the angle of elevation from the foot is \(30^\circ\). If the building is \(20\) m high, find the height of the temple. (एउटा मन्दिरको गजुरलाई एउटा भवनको छत र फेदबाट अवलोकन गर्दा अवनति र उन्नतांश कोणहरू क्रमशः \(60^\circ\) र \(30^\circ\) पाइयो । यदि भवनको उचाइ \(20\) m भए मन्दिरको उचाइ पत्ता लगाउनुहोस् ।)

Let the building be \(AB=20\) m, the temple \(CD=h\), and the horizontal distance \(BC=d\). From \(B\) the pinnacle \(D\) is at elevation \(30^\circ\); from \(A\) it is at depression \(60^\circ\), so \(\angle ADF=60^\circ\) where \(F\) is the foot of the perpendicular from \(D\) to \(AB\).

\[\begin{aligned}\tan30^\circ&=\frac hd\implies d=\sqrt3h \\ \tan60^\circ&=\frac{AF}{DF}=\frac{20-h}{d} \\ \sqrt3&=\frac{20-h}{\sqrt3h} \\ 3h&=20-h \\ h&=5\text{ m}\end{aligned}\]
Height of the temple \(=5\) m
Q.8
a. A flag of Nepal is placed on top of a tower. From a point \(100\) m from the foot of the tower, the angles of elevation of the top of the tower and the top of the flag are \(45^\circ\) and \(60^\circ\). Find the height of the flag. (एउटा टावरमाथि नेपालको झन्डा खडा गरिएको छ । टावरको फेदबाट \(100\) m परेको विन्दुबाट टावरको टुप्पो र झन्डाको टुप्पोको उन्नतांश कोणहरू क्रमशः \(45^\circ\) र \(60^\circ\) बनेको छ भने सो झन्डाको उचाइ निकाल्नुहोस् ।)
\[\begin{aligned}\text{Tower}&=100\tan45^\circ=100\text{ m} \\ \text{Tower + flag}&=100\tan60^\circ=100\sqrt3\approx173.21\text{ m} \\ \text{Flag}&=173.21-100=73.21\text{ m}\end{aligned}\]
Height of the flag \(=100(\sqrt3-1)\text{ m}\approx73.21\text{ m}\)
b. A man \(1.8\) m tall observes the roof of a house and the top of a window at angles of elevation \(45^\circ\) and \(30^\circ\). If the top of the window is \(20\) m above the ground, find the height of the house. (\(1.8\) m अग्लो एक जना मानिसले एउटा घरको छत र झ्यालको माथिल्लो भागको अवलोकन गर्दा उन्नतांश कोणहरू क्रमशः \(45^\circ\) र \(30^\circ\) पाउँछ । यदि झ्यालको माथिल्लो भाग जमिनबाट \(20\) m को उचाइमा छ भने सो घरको उचाइ पत्ता लगाउनुहोस् ।)

The window top is \(20-1.8=18.2\) m above the man\u2019s eye level. Let \(d\) be his distance from the house.

\[\begin{aligned}\tan30^\circ&=\frac{18.2}{d}\implies d=18.2\sqrt3\approx31.52\text{ m} \\ \text{Roof above eye level}&=d\tan45^\circ=31.52\text{ m} \\ \text{Height of house}&=31.52+1.8=33.32\text{ m}\end{aligned}\]
Height of the house \(\approx33.32\) m
Q.9
a. From the foot and the roof of a \(15\) m tall house, the angles of elevation of the top of a column are \(60^\circ\) and \(45^\circ\). Find the height of the column. (\(15\) m अग्लो घरको फेद र छतबाट एउटा स्तम्भको टुप्पामा हेर्दा उन्नतांश कोणहरू क्रमशः \(60^\circ\) र \(45^\circ\) पाइयो भने स्तम्भको उचाइ पत्ता लगाउनुहोस् ।)

Let the column have height \(H\) and let \(d\) be its distance from the house.

\[\begin{aligned}\tan60^\circ&=\frac Hd\implies d=\frac H{\sqrt3} \\ \tan45^\circ&=\frac{H-15}{d}\implies H-15=d=\frac H{\sqrt3} \\ H\left(1-\frac1{\sqrt3}\right)&=15 \\ H&=\frac{15\sqrt3}{\sqrt3-1}=\frac{15(3+\sqrt3)}{2}\approx35.49\text{ m}\end{aligned}\]
Height of the column \(\approx35.49\) m
b. An aeroplane flying horizontally at a constant height of \(6000\) m is observed from a point on the ground at an angle of elevation of \(30^\circ\). After \(10\) seconds it is observed from the same point at an angle of elevation of \(45^\circ\). Find the speed of the aeroplane in km/hr. (जमिनबाट \(6000\) m को समान उचाइमा सिधा उडिरहेको एउटा हवाईजहाज जमिनको एउटा विन्दुबाट \(30^\circ\) को उन्नतांश कोणमा देखिन्छ । \(10\) सेकेन्डपछि सोही विन्दुबाट उन्नतांश कोण \(45^\circ\) मा देखिन्छ भने सो हवाईजहाजको गति कति km प्रतिघण्टा रहेछ, पत्ता लगाउनुहोस् ।)

We take the aeroplane to be flying towards the observer, so it covers the difference of the two horizontal distances.

\[\begin{aligned}d_1&=\frac{6000}{\tan30^\circ}=6000\sqrt3\approx10392.30\text{ m} \\ d_2&=\frac{6000}{\tan45^\circ}=6000\text{ m} \\ \text{Distance covered}&=d_1-d_2\approx4392.30\text{ m in }10\text{ s} \\ \text{Speed}&=\frac{4392.30}{10}=439.23\text{ m/s} \\ &=439.23\times\frac{18}{5}\approx1581.23\text{ km/hr}\end{aligned}\]
Speed of the aeroplane \(\approx1581.23\) km/hr
Q.10
a. Two poles of equal height stand on opposite sides of a \(30\) m wide road. From a point between them, the angles of elevation of their tops are \(60^\circ\) and \(30^\circ\). Find the height of the poles and the distances of the point from the poles. (\(30\) m फराकिलो बाटाको दुई किनारामा बराबर उचाइका दुईओटा खम्बा छन् । दुईओटा खम्बाका बिचबाट ती खम्बाका टुप्पामा हेर्दा क्रमशः \(60^\circ\) र \(30^\circ\) का उन्नतांश कोणहरू बन्छ भने ती खम्बाका उचाइ र खम्बादेखि अवलोकन विन्दुसम्मका दुरी पत्ता लगाउनुहोस् ।)

Let the point be \(x\) m from the first pole and \((30-x)\) m from the second, and let each pole have height \(h\).

\[\begin{aligned}\tan60^\circ&=\frac hx\implies h=\sqrt3x \\ \tan30^\circ&=\frac{h}{30-x}\implies h=\frac{30-x}{\sqrt3} \\ \sqrt3x&=\frac{30-x}{\sqrt3} \\ 3x&=30-x\implies x=7.5\text{ m} \\ 30-x&=22.5\text{ m} \\ h&=7.5\sqrt3\approx12.99\text{ m}\end{aligned}\]
Height of the poles \(\approx12.99\) m; distances \(7.5\) m and \(22.5\) m
b. A man \(1.6\) m tall standing midway between two poles observes their tops at angles of elevation \(60^\circ\) and \(45^\circ\). If the shorter pole is \(41.6\) m high, find the height of the taller pole and the distance between the poles. (दुईओटा खम्बाका बिचबाट एक जना \(1.6\) m अग्लो मानिसले दुबै खम्बाको टुप्पामा हेर्दा क्रमशः \(60^\circ\) र \(45^\circ\) का उन्नतांश कोणहरू पाउँछ । यदि होचो खम्बाको उचाइ \(41.6\) m छ भने अग्लो खम्बाको उचाइ र ती खम्बाहरूबिचको दुरी पत्ता लगाउनुहोस् ।)

Assumption

The data fix the answer only if the man is at the middle (\u201cबिचबाट\u201d), at equal distance \(d\) from both poles; we take that. With equal distances the larger angle belongs to the taller pole, so the \(45^\circ\) pole is the shorter one, which agrees with the given data.

\[\begin{aligned}\text{Shorter pole above eye level}&=41.6-1.6=40\text{ m} \\ \tan45^\circ&=\frac{40}{d}\implies d=40\text{ m} \\ \text{Taller pole above eye level}&=d\tan60^\circ=40\sqrt3\approx69.28\text{ m} \\ \text{Height of taller pole}&=69.28+1.6=70.88\text{ m} \\ \text{Distance between poles}&=2d=80\text{ m}\end{aligned}\]
Taller pole \(\approx70.88\) m; distance between the poles \(=80\) m
Q.11
a. A pole is divided by a point on it, from the ground, in the ratio \(4:5\). If both parts subtend equal angles at a point \(20\) m from the base, find the height of the pole. (एउटा खम्बालाई जमिनबाट एउटा विन्दुले \(4:5\) को अनुपातमा विभाजन गरेको छ । दुबै भागले फेदबाट \(20\) m टाढाको विन्दुमा बराबर कोण बनाउँछन् भने खम्बाको उचाइ पत्ता लगाउनुहोस् ।)

Let the lower part be \(4k\) and the upper part \(5k\), so the height is \(9k\). Each part subtends \(\theta\), so the whole pole subtends \(2\theta\).

\[\begin{aligned}\tan\theta&=\frac{4k}{20}=\frac k5 \\ \tan2\theta&=\frac{9k}{20} \\ \frac{2\cdot\frac k5}{1-\frac{k^2}{25}}&=\frac{9k}{20} \\ \frac{10k}{25-k^2}&=\frac{9k}{20} \\ 200&=9(25-k^2) \\ 9k^2&=25\implies k=\frac53 \\ \text{Height}&=9k=15\text{ m}\end{aligned}\]
Height of the pole \(=15\) m
b. A flag \(5\) m high is fixed on top of a tower. From a point on the ground, the tower and the flag subtend angles of \(45^\circ\) and \(15^\circ\) respectively. Find the height of the tower. (एउटा टावरमाथि \(5\) m अग्लो झन्डा ठडाइएको छ । जमिनको कुनै विन्दुमा टावर र झन्डाले क्रमशः \(45^\circ\) र \(15^\circ\) कोण बनाउँछन् भने सो टावरको उचाइ पत्ता लगाउनुहोस् ।)

Let the tower be \(h\) and the distance to the point \(d\). The tower top is at elevation \(45^\circ\), so the flag top is at \(45^\circ+15^\circ=60^\circ\).

\[\begin{aligned}\tan45^\circ&=\frac hd\implies d=h \\ \tan60^\circ&=\frac{h+5}{d}=\frac{h+5}{h} \\ \sqrt3h&=h+5 \\ h&=\frac5{\sqrt3-1}=\frac{5(\sqrt3+1)}{2}\approx6.83\text{ m}\end{aligned}\]
Height of the tower \(\approx6.83\) m

Q.12 Two poles of equal height are given. Standing at a point between the two poles, a person observes the top of each pole and finds the angle of elevation to be \(45^\circ\). Moving \(20\) m closer to one pole, the angle of elevation becomes \(60^\circ\).

  • a. Represent the given context in a diagram.
  • b. Find the height of the poles.
  • c. Find the distance between the two poles.
  • d. As the observer moves closer to the pole, the angle of elevation increases. Explain with reason.

(बराबर उचाइका दुईओटा खम्बाहरू छन् । दुवै खम्बाका बिचमा उभिएर एक जना मानिसले प्रत्येक खम्बाको टुप्पो अवलोकन गर्दा \(45^\circ\) को उन्नतांश कोण पाउँछ । एउटा खम्बातिर \(20\) m नजिक गएर फेरि अवलोकन गर्दा \(60^\circ\) को उन्नतांश कोण पाउँछ । (क) चित्र बनाउनुहोस् । (ख) खम्बाहरूको उचाइ पत्ता लगाउनुहोस् । (ग) खम्बाहरूबिचको दुरी पत्ता लगाउनुहोस् । (घ) खम्बाहरू नजिक गएर अवलोकन गर्दा उन्नतांश कोण किन बढ्दै जान्छ, कारणसहित पुष्टि गर्नुहोस् ।)

a. Diagram

Diagram: poles AP and BQ of equal height h; M is the midpoint (45° to both tops); N is 20 m closer to BQ (60°).

b. Height of the poles

Let each pole have height \(h\). At \(M\) the angle to the top of \(BQ\) is \(45^\circ\), so \(MQ=h\) (since \(\tan45^\circ=1\)). After moving \(20\) m towards \(Q\), \(NQ=h-20\).

\[\begin{aligned}\tan60^\circ&=\frac{h}{h-20} \\ \sqrt3(h-20)&=h \\ h(\sqrt3-1)&=20\sqrt3 \\ h&=\frac{20\sqrt3}{\sqrt3-1}=10(3+\sqrt3)\approx47.32\text{ m}\end{aligned}\]

c. Distance between the poles

From \(M\) both tops are seen at \(45^\circ\), so \(PM=MQ=h\). Hence

\[\begin{aligned}PQ&=2h\approx2\times47.32=94.64\text{ m}\end{aligned}\]

d. Why the angle increases

For a pole of fixed height \(h\) at distance \(d\), \(\tan\theta=\dfrac hd\). As the observer moves closer, \(d\) decreases while \(h\) stays the same, so \(\tan\theta\) increases. Since \(\tan\theta\) increases with \(\theta\) for \(0^\circ<\theta<90^\circ\), the angle of elevation \(\theta\) increases.

Height \(\approx47.32\) m; distance between the poles \(\approx94.64\) m

Q.13 Outside a building, a ladder \(10\) m long is placed so that its top rests at a point \(4\) m below the roof. From the foot of the ladder, the angle of elevation of the roof is \(60^\circ\).

  • a. Draw a diagram representing the given context.
  • b. Find the height of the building at which the top of the ladder touches.
  • c. Find the distance between the foot of the ladder and the building.
  • d. Find the angle of elevation, from the foot of the ladder, of a point \(20\) m below the roof.

(एउटा भवनको बाहिर भित्तामा भवनको छतबाट \(4\) m तल एउटा \(10\) m लामो भर्‍याङ तेर्साएर राखिएको छ । भर्‍याङको फेदबाट भवनको छतसम्मको उन्नतांश कोण \(60^\circ\) बनेको छ भने (क) चित्र निर्माण गर्नुहोस् । (ख) भर्‍याङको टुप्पाले घरको कुन उचाइमा छोएको छ, पत्ता लगाउनुहोस् । (ग) भर्‍याङको फेदबाट घर कति दुरीमा छ, पत्ता लगाउनुहोस् । (घ) भर्‍याङको फेदबाट भवनको छतबाट \(20\) m तलको विन्दुको उन्नतांश कोण पत्ता लगाउनुहोस् ।)

a. Diagram

Diagram: ladder FT of length 10 m rests on the wall at T, 4 m below the roof R; the angle of elevation of R from F is 60°.

b. Height at which the ladder touches the wall

Let \(BT=h\) and \(BF=d\). Then the roof is at height \(h+4\), and \(FT=10\) m.

\[\begin{aligned}\tan60^\circ&=\frac{h+4}{d}\implies d=\frac{h+4}{\sqrt3} \\ h^2+d^2&=10^2\ (\text{right triangle }FBT) \\ h^2+\frac{(h+4)^2}{3}&=100 \\ 3h^2+h^2+8h+16&=300 \\ h^2+2h-71&=0 \\ h&=\frac{-2+\sqrt{4+284}}{2}=-1+6\sqrt2\approx7.49\text{ m}\end{aligned}\]

c. Distance of the foot of the ladder from the building

\[\begin{aligned}d&=\frac{h+4}{\sqrt3}=\frac{11.49}{1.732}\approx6.63\text{ m}\end{aligned}\]

Check: \(h^2+d^2=7.49^2+6.63^2\approx100\).

d. Note on the printed question

The roof is only \(h+4\approx11.49\) m above the ground, so a point \(20\) m below the roof would lie below the ground. The question cannot be answered as printed (the number \(20\) is probably a misprint).

The method is: for a point \(k\) m below the roof (\(k<11.49\)), its height is \(11.49-k\) and \(\tan\theta=\dfrac{11.49-k}{6.63}\). For example, the point \(4\) m below the roof is the top of the ladder:

\[\begin{aligned}\tan\theta&=\frac{7.49}{6.63}\approx1.129 \\ \theta&\approx48.5^\circ\end{aligned}\]
Height touched \(\approx7.49\) m; distance \(\approx6.63\) m; part (d) cannot be answered as printed
Answer
1. A. a   B. c   C. c   D. c   E. b   F. c
2. a. \(y=20\sqrt2\) m   b. \(\frac{10}{\sqrt3}\) m   c. \(10\sqrt3\) m, \(60^\circ\)   d. \(61.47\) m, \(61.47\) m   e. \(21.13\) m, \(50\) m   f. \(40\) m, \(40\sqrt3\) m
3. a. \(43.30\) m   b. \(12.68\) ft
4. a. \(577.35\) m   b. \(23.66\) ft
5. a. \(37.95\) m   b. \(17.68\) m and \(35.36\) m
6. a. \(43.30\) m   b. \(25.62\) ft
7. a. \(12.68\) m   b. \(5\) m
8. a. \(73.21\) m   b. \(33.32\) m
9. a. \(35.49\) m   b. \(1581.23\) km/hr
10. a. \(12.99\) m; \(7.5\) m and \(22.5\) m   b. \(70.88\) m; \(80\) m
11. a. \(15\) m   b. \(6.83\) m
12. b. \(47.32\) m   c. \(94.64\) m
13. b. \(7.49\) m   c. \(6.63\) m   d. see the note (cannot be answered as printed)

3. Frequently Asked Questions on Height and Distance (Class 10 Optional Mathematics)

What is covered in Class 10 Optional Mathematics Trigonometry Exercise 7.6 Heights and Distances? (कक्षा १० ऐच्छिक गणित त्रिकोणमितिको अभ्यास 7.6 Heights and Distances मा के के समावेश छ ?)

This page solves all 32 questions of Exercise 7.6 Heights and Distances, in textbook order, with the Nepali question in brackets and an Answer box at the end. Height and distance formulas, angle of elevation examples and all Exercise 7.6 questions solved with the tangent ratio for Class 10 Optional Math SEE.

What is the angle of elevation and the angle of depression? (उन्नतांश कोण र अवनति कोणबिच के फरक छ ?)

Both are measured between the line of sight and the horizontal through the observer’s eye. If the object is above eye level the angle is the angle of elevation; if it is below eye level it is the angle of depression (Question 1). By alternate angles, the angle of depression from the top of a tower equals the angle of elevation of the top from the object.

What are the height and distance formulas? (उचाइ र दुरीका सूत्रहरू के के हुन् ?)
  • \(\tan\theta=\dfrac{\text{height}}{\text{distance}}\)
  • \(\sin\theta=\dfrac{\text{height}}{\text{line of sight}}\)
  • \(\cos\theta=\dfrac{\text{distance}}{\text{line of sight}}\)
  • \(\tan30^\circ=\dfrac1{\sqrt3}\), \(\tan45^\circ=1\), \(\tan60^\circ=\sqrt3\)
Can you give a height and distance example with the angle of elevation? (उन्नतांश कोणसहितको उचाइ र दुरीको उदाहरण दिनुहोस् ।)

A point on the roof is seen at \(60^\circ\) and, after moving \(50\) m away, at \(30^\circ\). Writing both distances in terms of the height \(h\) gives \(\sqrt3h-\dfrac h{\sqrt3}=50\), so \(h=25\sqrt3\approx43.30\) m (Question 3a).

How do I find the height of a building when the angle of elevation is observed from two points? (दुई विन्दुबाट उन्नतांश कोण दिइएको छ भने भवनको उचाइ कसरी निकाल्ने ?)

Let the height be \(h\). Write each distance from the foot as \(\dfrac h{\tan\theta}\), then use the given gap between the two points to form an equation in \(h\) (Question 3a and 4).

What if the angle of elevation is \(45^\circ\)? (उन्नतांश कोण \(45^\circ\) भए के हुन्छ ?)

Then \(\tan45^\circ=1\), so the height of the object equals its horizontal distance from the observer (Question 1D).

Is there a height and distance calculator? (के उचाइ र दुरी निकाल्ने क्याल्कुलेटर छ ?)

You can use a scientific calculator for the tangent values. Draw the right-angled triangle, write \(\tan\theta=\dfrac{\text{height}}{\text{distance}}\), and solve. To find an angle use \(\tan^{-1}\). Every solution on this page shows each step.

Is height and distance studied in Class 11 as well? Is there a PDF or math quiz? (के उचाइ र दुरी कक्षा ११ मा पनि पढाइन्छ ? PDF वा क्विज छ ?)

Height and distance problems also appear in higher classes (for example Class 11 trigonometry). For practice, start with the six multiple choice questions of Exercise 7.6 (Question 1), which work like a short math quiz, then solve the written questions. All of them have worked solutions and an Answer box.

Are there misprints in Exercise 7.6? (के अभ्यास ७.६ मा छापाइका गल्ती छन् ?)

Question 5 says the angles are “supplementary”, but two acute angles of elevation cannot add up to \(180^\circ\); the solutions take them as complementary. In Question 13(d) a point \(20\) m below the roof would lie below the ground, so the method is shown with another point.

Are these Exercise 7.6 questions useful for SEE preparation in Nepal? (के यी प्रश्नहरू SEE को तयारीका लागि उपयोगी छन् ?)

Yes. Exercise 7.6 covers one-point and two-point height and distance problems, which are common SEE-style word problems in Optional Mathematics.

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