NCERT Exemplar Solutions Class 10 Maths Introduction to Trigonometry and its Applications — Exercise 8.4

NCERT Exemplar Problems, Class 10 Mathematics — Introduction to Trigonometry and its Applications, Exercise 8.4. All 18 questions solved.

Eighteen questions, and the hardest in the unit. They split cleanly: Questions 1, 2, 4, 5, 9, 10, 11 and 12 are identities and equations, and the rest are heights and distances.

For the identities, one technique dominates. Three of them — Questions 1, 9 and 12 — are built on the same idea, and it is worth naming before you meet it:

$$\sec^2\theta – \tan^2\theta = 1 \implies (\sec\theta + \tan\theta)(\sec\theta – \tan\theta) = 1$$

So if you are told that $\sec\theta + \tan\theta = l$, you immediately know that $\sec\theta – \tan\theta = \dfrac{1}{l}$ — and adding or subtracting the two gives $\sec\theta$ and $\tan\theta$ separately. The same works for $\operatorname{cosec}$ and $\cot$. Whenever a question hands you a sum of two ratios whose squares differ by $1$, write down the reciprocal partner straight away.

For the heights and distances, one procedure dominates. Nearly every problem here has two observation points, and the recipe is always:

1. Draw the figure and label the unknown height $h$ and the unknown horizontal distance $d$.
2. Write one tangent equation per observation point — two equations, two unknowns.
3. Eliminate $d$, because it is almost never what is asked for.

Key insight. In a two-position problem the horizontal distance is a nuisance variable, not an answer. Write both equations as $d = h\cot\alpha$ and $d = h\cot\beta$ and subtract; the $d$ vanishes and what remains is the relation between the height and the known displacement. Ten of the eighteen questions here fall out of exactly that subtraction.

Question 1

If $\operatorname{cosec}\theta + \cot\theta = p$, then prove that $\cos\theta = \dfrac{p^2 – 1}{p^2 + 1}$.

Solution. Start from the identity $\operatorname{cosec}^2\theta – \cot^2\theta = 1$ and factorise it as a difference of squares:

$$\left(\operatorname{cosec}\theta + \cot\theta\right)\left(\operatorname{cosec}\theta – \cot\theta\right) = 1$$

The first bracket is $p$, so

$$\operatorname{cosec}\theta – \cot\theta = \frac{1}{p}$$

Two linear equations in $\operatorname{cosec}\theta$ and $\cot\theta$. Adding them:

$$2\operatorname{cosec}\theta = p + \frac1p = \frac{p^2+1}{p} \implies \operatorname{cosec}\theta = \frac{p^2+1}{2p}$$

Subtracting them:

$$2\cot\theta = p – \frac1p = \frac{p^2-1}{p} \implies \cot\theta = \frac{p^2-1}{2p}$$

Now $\cos\theta$ is the quotient of these two, because $\dfrac{\cot\theta}{\operatorname{cosec}\theta} = \dfrac{\cos\theta/\sin\theta}{1/\sin\theta} = \cos\theta$:

$$\cos\theta = \frac{\cot\theta}{\operatorname{cosec}\theta} = \frac{\frac{p^2-1}{2p}}{\frac{p^2+1}{2p}} = \frac{p^2 – 1}{p^2 + 1}$$

The $2p$ cancels, which is the tidy ending the question was built towards.

Question 2

Prove that $\sqrt{\sec^2\theta + \operatorname{cosec}^2\theta} = \tan\theta + \cot\theta$

Solution. Work inside the root. Write both terms over $\sin^2\theta\cos^2\theta$:

$$\sec^2\theta + \operatorname{cosec}^2\theta = \frac{1}{\cos^2\theta} + \frac{1}{\sin^2\theta} = \frac{\sin^2\theta + \cos^2\theta}{\sin^2\theta\cos^2\theta} = \frac{1}{\sin^2\theta\cos^2\theta}$$

Taking the square root:

$$\sqrt{\frac{1}{\sin^2\theta\cos^2\theta}} = \frac{1}{\sin\theta\cos\theta}$$

positive because $\theta$ is acute, so both $\sin\theta$ and $\cos\theta$ are positive.

And that is exactly the right-hand side, as Question 4 of Exercise 8.3 established:

$$\tan\theta + \cot\theta = \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta} = \frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} = \frac{1}{\sin\theta\cos\theta}$$

Both sides equal $\dfrac{1}{\sin\theta\cos\theta}$.

The acuteness matters, as it did in Question 4 of Exercise 8.2. For a general angle the left side, being a square root, is never negative, while $\tan\theta + \cot\theta$ can be; the identity as printed holds on the range Class 10 works in.

Question 3

The angle of elevation of the top of a tower from certain point is $30^\circ$. If the observer moves $20$ metres towards the tower, the angle of elevation of the top increases by $15^\circ$. Find the height of the tower.

Solution. “Increases by $15^\circ$” means the second angle is $30^\circ + 15^\circ = 45^\circ$ — a value that appears nowhere in the question and has to be computed.

Let the tower have height $h$, and let the first observation point be at horizontal distance $d_1$ from its foot, the second at $d_2$. Moving towards the tower shortens the distance, so

$$d_1 – d_2 = 20$$

Write a tangent equation at each point and solve each for the distance:

$$\tan 30^\circ = \frac{h}{d_1} \implies d_1 = \frac{h}{\tan 30^\circ} = h\sqrt3$$

$$\tan 45^\circ = \frac{h}{d_2} \implies d_2 = \frac{h}{\tan 45^\circ} = h$$

Subtracting, which eliminates nothing but gives the whole answer:

$$h\sqrt3 – h = 20 \implies h\left(\sqrt3 – 1\right) = 20 \implies h = \frac{20}{\sqrt3 – 1}$$

Rationalise by multiplying above and below by $\sqrt3 + 1$:

$$h = \frac{20\left(\sqrt3+1\right)}{\left(\sqrt3\right)^2 – 1^2} = \frac{20\left(\sqrt3+1\right)}{2} = 10\left(\sqrt3 + 1\right) \approx 27.32 \text{ m}$$

A note on the printed answer. The key for this question is set as “$10\ \sqrt3\ \ 1$ m” — with a gap where the operator should be. Examined at high magnification the character is simply absent: between the $\sqrt3$ and the $1$ there is only white space, with a Symbol-font space sitting inside it, so the sign was typeset in a symbol font and dropped when the file was made. The intended answer is $10(\sqrt3 + 1)$ m, which is what the derivation above gives. The nearby answers for Questions 7 and 16 print their operators correctly, so the fault is specific to this one entry.

Question 4

If $1 + \sin^2\theta = 3\sin\theta\cos\theta$, then prove that $\tan\theta = 1$ or $\dfrac12$.

Solution. The obstacle is the bare $1$, which mixes with nothing. Replace it using the Pythagorean identity:

$$\left(\sin^2\theta + \cos^2\theta\right) + \sin^2\theta = 3\sin\theta\cos\theta$$

$$2\sin^2\theta + \cos^2\theta = 3\sin\theta\cos\theta$$

Now every term is degree two in $\sin$ and $\cos$, so the equation is homogeneous and dividing throughout by $\cos^2\theta$ turns it entirely into tangents:

$$2\tan^2\theta + 1 = 3\tan\theta$$

$$2\tan^2\theta – 3\tan\theta + 1 = 0$$

A quadratic in $\tan\theta$. Factorise:

$$(2\tan\theta – 1)(\tan\theta – 1) = 0 \implies \tan\theta = \frac12 \ \text{ or } \ \tan\theta = 1$$

as required.

Replacing $1$ by $\sin^2\theta + \cos^2\theta$ is the move to remember. It is what makes an equation homogeneous, and homogeneity is what lets you divide by $\cos^2\theta$ and reduce two unknowns to one. Without it the equation cannot be turned into a polynomial at all.

Question 5

Given that $\sin\theta + 2\cos\theta = 1$, then prove that $2\sin\theta – \cos\theta = 2$.

Solution. There is a slick proof here, and it turns on noticing that the two expressions are companions.

Expand the sum of their squares:

$$\left(\sin\theta + 2\cos\theta\right)^2 + \left(2\sin\theta – \cos\theta\right)^2$$

$$= \left(\sin^2\theta + 4\sin\theta\cos\theta + 4\cos^2\theta\right) + \left(4\sin^2\theta – 4\sin\theta\cos\theta + \cos^2\theta\right)$$

The cross terms $+4\sin\theta\cos\theta$ and $-4\sin\theta\cos\theta$ cancel, leaving

$$= 5\sin^2\theta + 5\cos^2\theta = 5\left(\sin^2\theta + \cos^2\theta\right) = 5$$

So the sum of the squares is $5$ identically — for every angle, with no condition needed. Now impose the given condition, $\sin\theta + 2\cos\theta = 1$:

$$1^2 + \left(2\sin\theta – \cos\theta\right)^2 = 5 \implies \left(2\sin\theta – \cos\theta\right)^2 = 4$$

$$2\sin\theta – \cos\theta = \pm 2$$

And the sign has to be settled, which the question does not do for you. The square root leaves two possibilities, and only one of them is $+2$.

Solve the condition to see both. Squaring $\sin\theta = 1 – 2\cos\theta$ gives $1 – \cos^2\theta = 1 – 4\cos\theta + 4\cos^2\theta$, so $5\cos^2\theta = 4\cos\theta$ and hence $\cos\theta = 0$ or $\cos\theta = \frac45$.

• $\cos\theta = 0$ gives $\sin\theta = 1$, so $\theta = 90^\circ$, and $2(1) – 0 = 2$. ✓
• $\cos\theta = \frac45$ gives $\sin\theta = 1 – \frac85 = -\frac35$, and $2\left(-\frac35\right) – \frac45 = -2$. ✗

The second branch needs a negative sine, which is impossible for the acute angles Class 10 works with. So on the syllabus range the only solution is $\theta = 90^\circ$ and the result is $+2$, as claimed. The result as printed would be false if angles beyond $90^\circ$ were admitted.

Question 6

The angle of elevation of the top of a tower from two points distant $s$ and $t$ from its foot are complementary. Prove that the height of the tower is $\sqrt{st}$.

Solution. Complementary means the two angles sum to $90^\circ$, so call them $\alpha$ and $90^\circ – \alpha$. Let the tower have height $h$.

From the point at distance $s$:

$$\tan\alpha = \frac{h}{s} \implies h = s\tan\alpha$$

From the point at distance $t$, where the elevation is $90^\circ – \alpha$:

$$\tan(90^\circ – \alpha) = \frac{h}{t} \implies \cot\alpha = \frac{h}{t} \implies h = t\cot\alpha$$

Multiply the two expressions for $h$ rather than dividing or subtracting:

$$h \cdot h = \left(s\tan\alpha\right)\left(t\cot\alpha\right) = st \cdot \tan\alpha\cot\alpha = st$$

since $\tan\alpha\cot\alpha = 1$. Therefore

$$h^2 = st \implies h = \sqrt{st}$$

taking the positive root because $h$ is a length.

Multiplying is the whole idea, and it is worth seeing why it is the right operation. The two expressions involve $\tan\alpha$ and its reciprocal, so their product kills the unknown angle outright. Subtracting would leave $\tan\alpha – \cot\alpha$, which does not simplify.

The height is the geometric mean of the two distances, not the arithmetic mean. With $s = 4$ and $t = 9$ the tower is $6$ m, not $6.5$.

Question 7

The shadow of a tower standing on a level plane is found to be $50$ m longer when Sun’s elevation is $30^\circ$ than when it is $60^\circ$. Find the height of the tower.

Solution. Let the tower have height $h$, and let $s_{30}$ and $s_{60}$ be the shadow lengths at the two elevations. The lower Sun casts the longer shadow, so

$$s_{30} – s_{60} = 50$$

The shadow is the side adjacent to the elevation angle, so

$$\tan 30^\circ = \frac{h}{s_{30}} \implies s_{30} = h\sqrt3, \qquad \tan 60^\circ = \frac{h}{s_{60}} \implies s_{60} = \frac{h}{\sqrt3}$$

Subtract:

$$h\sqrt3 – \frac{h}{\sqrt3} = 50$$

Put the left side over $\sqrt3$:

$$\frac{3h – h}{\sqrt3} = 50 \implies \frac{2h}{\sqrt3} = 50 \implies h = 25\sqrt3 \approx 43.30 \text{ m}$$

A check on which shadow is longer, since getting it backwards gives a negative height: at $30^\circ$ the shadow is $25\sqrt3 \times \sqrt3 = 75$ m, and at $60^\circ$ it is $\frac{25\sqrt3}{\sqrt3} = 25$ m. The difference is $50$ m, as required, and the $30^\circ$ shadow is indeed the longer.

Question 8

A vertical tower stands on a horizontal plane and is surmounted by a vertical flag staff of height $h$. At a point on the plane, the angles of elevation of the bottom and the top of the flag staff are $\alpha$ and $\beta$, respectively. Prove that the height of the tower is $\left(\dfrac{h\tan\alpha}{\tan\beta – \tan\alpha}\right)$.

Solution. Let the tower have height $H$ — the quantity to be found — and let the observation point be at horizontal distance $d$ from its foot. The flagstaff sits on top, so the top of the flagstaff is at height $H + h$.

Two elevations, two equations:

$$\tan\alpha = \frac{H}{d} \qquad \text{(bottom of the flagstaff = top of the tower)}$$

$$\tan\beta = \frac{H + h}{d} \qquad \text{(top of the flagstaff)}$$

Eliminate $d$. From the first, $d = \dfrac{H}{\tan\alpha}$; substitute into the second:

$$\tan\beta = \frac{H+h}{\frac{H}{\tan\alpha}} = \frac{(H+h)\tan\alpha}{H}$$

Multiply across and expand:

$$H\tan\beta = H\tan\alpha + h\tan\alpha$$

Collect the $H$ terms on one side:

$$H\left(\tan\beta – \tan\alpha\right) = h\tan\alpha \implies H = \frac{h\tan\alpha}{\tan\beta – \tan\alpha}$$

as required.

The denominator is positive because $\beta > \alpha$ — the top of the flagstaff is higher than its bottom, so it subtends the larger angle from the same point. That is a useful sanity check on the direction of the subtraction: a denominator of $\tan\alpha – \tan\beta$ would give a negative height.

Question 9

If $\tan\theta + \sec\theta = l$, then prove that $\sec\theta = \dfrac{l^2 + 1}{2l}$.

Solution. The same reciprocal-partner trick as Question 1. From $\sec^2\theta – \tan^2\theta = 1$:

$$\left(\sec\theta + \tan\theta\right)\left(\sec\theta – \tan\theta\right) = 1$$

The first bracket is $l$, so

$$\sec\theta – \tan\theta = \frac{1}{l}$$

Add this to the given $\sec\theta + \tan\theta = l$; the tangents cancel:

$$2\sec\theta = l + \frac{1}{l} = \frac{l^2+1}{l} \implies \sec\theta = \frac{l^2+1}{2l}$$

Subtracting instead gives the companion result, $\tan\theta = \dfrac{l^2 – 1}{2l}$, which is worth writing down because questions often ask for it in the next part.

Compare with Question 1. There the pair was $\operatorname{cosec}\theta$ and $\cot\theta$; here it is $\sec\theta$ and $\tan\theta$. The two identities $\operatorname{cosec}^2 – \cot^2 = 1$ and $\sec^2 – \tan^2 = 1$ have the same shape, so every question built on one has a twin built on the other.

Question 10

If $\sin\theta + \cos\theta = p$ and $\sec\theta + \operatorname{cosec}\theta = q$, then prove that $q(p^2 – 1) = 2p$.

Solution. Handle the two given quantities separately and then combine.

From $p$. Squaring:

$$p^2 = \left(\sin\theta + \cos\theta\right)^2 = \sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta = 1 + 2\sin\theta\cos\theta$$

so

$$p^2 – 1 = 2\sin\theta\cos\theta$$

From $q$. Combine over a common denominator:

$$q = \frac{1}{\cos\theta} + \frac{1}{\sin\theta} = \frac{\sin\theta + \cos\theta}{\sin\theta\cos\theta} = \frac{p}{\sin\theta\cos\theta}$$

the numerator being $p$ by definition.

Now multiply:

$$q\left(p^2 – 1\right) = \frac{p}{\sin\theta\cos\theta} \times 2\sin\theta\cos\theta = 2p$$

The $\sin\theta\cos\theta$ cancels completely, which is why the result is as clean as it is.

Notice that $\sin\theta\cos\theta$ was never evaluated — it appeared in both pieces and cancelled. Identifying the quantity that will cancel, and leaving it alone, saves the whole calculation.

Question 11

If $a\sin\theta + b\cos\theta = c$, then prove that $a\cos\theta – b\sin\theta = \sqrt{a^2 + b^2 – c^2}$.

Solution. The two expressions are companions, exactly as in Question 5. Expand the sum of their squares:

$$\left(a\sin\theta + b\cos\theta\right)^2 + \left(a\cos\theta – b\sin\theta\right)^2$$

$$= \left(a^2\sin^2\theta + 2ab\sin\theta\cos\theta + b^2\cos^2\theta\right) + \left(a^2\cos^2\theta – 2ab\sin\theta\cos\theta + b^2\sin^2\theta\right)$$

The cross terms cancel. Grouping what is left by $a^2$ and $b^2$:

$$= a^2\left(\sin^2\theta + \cos^2\theta\right) + b^2\left(\cos^2\theta + \sin^2\theta\right) = a^2 + b^2$$

So the sum of the squares is $a^2 + b^2$ identically. Imposing the given condition $a\sin\theta + b\cos\theta = c$:

$$c^2 + \left(a\cos\theta – b\sin\theta\right)^2 = a^2 + b^2$$

$$\left(a\cos\theta – b\sin\theta\right)^2 = a^2 + b^2 – c^2$$

and taking square roots gives the stated result.

The question is slightly overstated, and it is worth knowing why. Taking the square root of both sides gives

$$\left|a\cos\theta – b\sin\theta\right| = \sqrt{a^2+b^2-c^2}$$

with a modulus, because a square root is non-negative by convention while $a\cos\theta – b\sin\theta$ need not be. It is easy to find cases where the left side is negative: with $a = 1$, $b = 3$ and $\theta = 60^\circ$, the quantity $a\cos\theta – b\sin\theta = 0.5 – 2.60 = -2.10$, while the square root is $+2.10$. Both magnitudes agree; the signs do not.

So the honest statement of the result is that $a\cos\theta – b\sin\theta = \pm\sqrt{a^2+b^2-c^2}$, and the question asserts the positive branch. In an examination, prove the squared relation — which is what is actually true — and note the sign; that is a complete answer and a more accurate one.

Question 12

Prove that $\dfrac{1 + \sec\theta – \tan\theta}{1 + \sec\theta + \tan\theta} = \dfrac{1 – \sin\theta}{\cos\theta}$

Solution. The trick is to replace the $1$ in the numerator only, using $1 = \sec^2\theta – \tan^2\theta$:

$$1 + \sec\theta – \tan\theta = \left(\sec^2\theta – \tan^2\theta\right) + \sec\theta – \tan\theta$$

Factorise the difference of squares, then take out the common factor $(\sec\theta – \tan\theta)$:

$$= \left(\sec\theta – \tan\theta\right)\left(\sec\theta + \tan\theta\right) + \left(\sec\theta – \tan\theta\right) = \left(\sec\theta – \tan\theta\right)\left(\sec\theta + \tan\theta + 1\right)$$

The second bracket is exactly the denominator of the original fraction. So

$$\frac{1 + \sec\theta – \tan\theta}{1 + \sec\theta + \tan\theta} = \frac{\left(\sec\theta – \tan\theta\right)\left(1 + \sec\theta + \tan\theta\right)}{1 + \sec\theta + \tan\theta} = \sec\theta – \tan\theta$$

Finally convert to sines and cosines:

$$\sec\theta – \tan\theta = \frac{1}{\cos\theta} – \frac{\sin\theta}{\cos\theta} = \frac{1 – \sin\theta}{\cos\theta}$$

which is the right-hand side.

Replacing the $1$ in only one place is the whole idea. Doing it in both numerator and denominator gets you nowhere; doing it in the numerator manufactures the denominator as a factor, and the cancellation is immediate.

Question 13

The angle of elevation of the top of a tower $30$ m high from the foot of another tower in the same plane is $60^\circ$ and the angle of elevation of the top of the second tower from the foot of the first tower is $30^\circ$. Find the distance between the two towers and also the height of the other tower.

Solution. Two towers, and each is observed from the foot of the other. Let the distance between them be $d$ and let the second tower have height $H$.

First observation. From the foot of the second tower, the top of the $30$ m tower has elevation $60^\circ$:

$$\tan 60^\circ = \frac{30}{d} \implies \sqrt3 = \frac{30}{d} \implies d = \frac{30}{\sqrt3} = 10\sqrt3 \approx 17.32 \text{ m}$$

rationalising $\frac{30}{\sqrt3} = \frac{30\sqrt3}{3} = 10\sqrt3$.

Second observation. From the foot of the first tower, the top of the second has elevation $30^\circ$, across the same distance $d$:

$$\tan 30^\circ = \frac{H}{d} \implies \frac{1}{\sqrt3} = \frac{H}{10\sqrt3} \implies H = \frac{10\sqrt3}{\sqrt3} = 10 \text{ m}$$

So the towers are $10\sqrt3$ m apart and the second is $10$ m high.

The shape of the answer is worth noting: the second tower is exactly one-third the height of the first, because $\tan 30^\circ = \frac13 \tan 60^\circ$ and both angles are measured across the same base. Shallower angle, shorter tower — as it must be.

Question 14

From the top of a tower $h$ m high, the angles of depression of two objects, which are in line with the foot of the tower are $\alpha$ and $\beta$ $(\beta > \alpha)$. Find the distance between the two objects.

Solution. Both objects lie on the ground in a straight line from the foot of the tower, and both are observed from the top.

The angle of depression from the top equals the angle of elevation from the object — they are alternate angles between the horizontal at the top and the horizontal ground. So for an object at horizontal distance $d$:

$$\tan(\text{depression}) = \frac{h}{d} \implies d = \frac{h}{\tan(\text{depression})} = h\cot(\text{depression})$$

Applying this to each object:

$$d_\alpha = h\cot\alpha, \qquad d_\beta = h\cot\beta$$

Which is nearer? Since $\beta > \alpha$ and the cotangent decreases as the angle grows, $\cot\beta < \cot\alpha$, so $d_\beta < d_\alpha$. The object seen at the steeper depression is the closer one — which matches intuition, since you look more steeply down at something near your feet.

Both objects are on the same side, so the distance between them is the difference:

$$d_\alpha – d_\beta = h\cot\alpha – h\cot\beta = h\left(\cot\alpha – \cot\beta\right)$$

The condition $\beta > \alpha$ is in the question precisely to fix the order of that subtraction and guarantee a positive answer.

Question 15

A ladder rests against a vertical wall at an inclination $\alpha$ to the horizontal. Its foot is pulled away from the wall through a distance $p$ so that its upper end slides a distance $q$ down the wall and then the ladder makes an angle $\beta$ to the horizontal. Show that $\dfrac{p}{q} = \dfrac{\cos\beta – \cos\alpha}{\sin\alpha – \sin\beta}$.

Solution. The one fact that makes this work: the ladder does not change length. Call it $L$.

First position, inclined at $\alpha$ to the horizontal:

• the foot is $L\cos\alpha$ from the wall
• the top is $L\sin\alpha$ up the wall

Second position, inclined at $\beta$:

• the foot is $L\cos\beta$ from the wall
• the top is $L\sin\beta$ up the wall

Now read off the two displacements the question names.

The foot is pulled away from the wall, so it moves from $L\cos\alpha$ out to $L\cos\beta$:

$$p = L\cos\beta – L\cos\alpha = L\left(\cos\beta – \cos\alpha\right)$$

The top slides down, so its height falls from $L\sin\alpha$ to $L\sin\beta$:

$$q = L\sin\alpha – L\sin\beta = L\left(\sin\alpha – \sin\beta\right)$$

Dividing, the unknown length cancels:

$$\frac{p}{q} = \frac{L\left(\cos\beta – \cos\alpha\right)}{L\left(\sin\alpha – \sin\beta\right)} = \frac{\cos\beta – \cos\alpha}{\sin\alpha – \sin\beta}$$

as required.

The two subtractions run in opposite orders, and that is not an oversight. Because the foot moves outward, $\beta < \alpha$, so $\cos\beta > \cos\alpha$ and $\sin\alpha > \sin\beta$. Writing $\cos\beta – \cos\alpha$ on top and $\sin\alpha – \sin\beta$ underneath makes both differences positive, which they must be since $p$ and $q$ are distances. Reversing either one would produce a negative ratio.

Question 16

The angle of elevation of the top of a vertical tower from a point on the ground is $60^\circ$. From another point $10$ m vertically above the first, its angle of elevation is $45^\circ$. Find the height of the tower.

Solution. The second observation point is directly above the first, so the horizontal distance to the tower is the same from both — that shared $d$ is what makes the problem solvable.

Let the tower have height $H$ and let both points be at horizontal distance $d$ from its foot.

From the ground point, the whole tower subtends $60^\circ$:

$$\tan 60^\circ = \frac{H}{d} \implies H = d\sqrt3$$

From the point $10$ m up, the observer’s own height is subtracted from the tower’s, and the elevation is $45^\circ$:

$$\tan 45^\circ = \frac{H – 10}{d} \implies H – 10 = d$$

Two equations in $H$ and $d$. Substituting the first into the second:

$$d\sqrt3 – 10 = d \implies d\left(\sqrt3 – 1\right) = 10 \implies d = \frac{10}{\sqrt3 – 1} = \frac{10\left(\sqrt3+1\right)}{2} = 5\left(\sqrt3+1\right)$$

Then

$$H = d\sqrt3 = 5\sqrt3\left(\sqrt3 + 1\right) = 5\left(3 + \sqrt3\right) = 5\left(\sqrt3 + 3\right) \approx 23.66 \text{ m}$$

A check: $d = 5(\sqrt3+1) \approx 13.66$ m, and $H – 10 \approx 13.66$ m — equal, as $\tan 45^\circ = 1$ requires.

Compare with Question 3. There the observer moved horizontally and the height stayed fixed; here the observer moves vertically and the distance stays fixed. Which quantity is shared between the two equations is the first thing to identify in any two-position problem.

Question 17

A window of a house is $h$ metres above the ground. From the window, the angles of elevation and depression of the top and the bottom of another house situated on the opposite side of the lane are found to be $\alpha$ and $\beta$, respectively. Prove that the height of the other house is $h\left(1 + \tan\alpha\cot\beta\right)$ metres.

Solution. Let the lane be $d$ metres wide and the other house $H$ metres tall. Both angles are measured from the window, along the horizontal through it.

The depression to the base. The base of the other house is at ground level, which is $h$ metres below the window:

$$\tan\beta = \frac{h}{d} \implies d = \frac{h}{\tan\beta} = h\cot\beta$$

That single equation gives the width of the lane — which is not asked for, but is needed for the next step.

The elevation to the top. The top of the other house is $H – h$ metres above the window:

$$\tan\alpha = \frac{H – h}{d} \implies H – h = d\tan\alpha$$

Substituting $d = h\cot\beta$:

$$H – h = h\cot\beta\tan\alpha = h\tan\alpha\cot\beta$$

$$H = h + h\tan\alpha\cot\beta = h\left(1 + \tan\alpha\cot\beta\right)$$

as required.

The structure of the answer says something useful: the $h$ outside the bracket is the part of the other house level with the window and below, and the $h\tan\alpha\cot\beta$ is the part above it. If the two houses were the same height, $\alpha$ would be zero and the formula would correctly return $H = h$.

Question 18

The lower window of a house is at a height of $2$ m above the ground and its upper window is $4$ m vertically above the lower window. At certain instant the angles of elevation of a balloon from these windows are observed to be $60^\circ$ and $30^\circ$, respectively. Find the height of the balloon above the ground.

Solution. First get the two window heights right. The lower is $2$ m above the ground; the upper is $4$ m above the lower, so it is $2 + 4 = 6$ m above the ground. Reading “$4$ m” as the upper window’s height above the ground is the commonest error in this question.

Let the balloon be $H$ metres above the ground and $d$ metres horizontally from the house. Both windows are on the same wall, so $d$ is the same for both.

From the lower window (height $2$ m), elevation $60^\circ$:

$$\tan 60^\circ = \frac{H – 2}{d} \implies H – 2 = d\sqrt3$$

From the upper window (height $6$ m), elevation $30^\circ$:

$$\tan 30^\circ = \frac{H – 6}{d} \implies H – 6 = \frac{d}{\sqrt3}$$

Subtract the second from the first, which eliminates $H$:

$$(H – 2) – (H – 6) = d\sqrt3 – \frac{d}{\sqrt3}$$

$$4 = \frac{3d – d}{\sqrt3} = \frac{2d}{\sqrt3} \implies d = 2\sqrt3$$

Then from the first equation:

$$H – 2 = 2\sqrt3 \cdot \sqrt3 = 6 \implies H = 8 \text{ m}$$

Check with the second: $H – 6 = 2$, and $\dfrac{d}{\sqrt3} = \dfrac{2\sqrt3}{\sqrt3} = 2$. Consistent.

The elevation falls as the observer rises, from $60^\circ$ at the lower window to $30^\circ$ at the upper. That is the expected direction: moving up towards the balloon’s level flattens the line of sight. If your two angles were increasing with height, the problem has been set up backwards.

Common mistakes

• Question 18 — mis-reading the upper window’s height. It is $4$ m above the lower window, so $6$ m above the ground. The question is worded to catch exactly this, and using $4$ gives a plausible but wrong answer.
• Question 3 — using $15^\circ$ as the second angle. The elevation increases by $15^\circ$, so the second angle is $30^\circ + 15^\circ = 45^\circ$. The number you need never appears in the question.
• Questions 5 and 11 — taking a square root without settling the sign. Both give $(\text{expression})^2 = \text{number}$, so the expression is $\pm$ the root. In Question 5 the acute-angle restriction selects the positive branch; in Question 11 the positive branch is simply asserted, and is often false. Prove the squared relation and then discuss the sign.
• Question 10 — computing $\sin\theta\cos\theta$. It appears in both halves and cancels. Spotting the quantity that will cancel, and refusing to evaluate it, is what keeps these proofs to five lines.
• Question 12 — substituting $1 = \sec^2\theta – \tan^2\theta$ in both numerator and denominator. Do it in the numerator only; that manufactures the denominator as a factor and the cancellation is immediate. Doing both leads nowhere.
• Questions 14 and 15 — subtracting in the wrong order. In Question 14 the condition $\beta > \alpha$ means $\cot\alpha > \cot\beta$, so the difference must be written $\cot\alpha – \cot\beta$; in Question 15 the two differences run in opposite orders for the same reason. Check that each distance comes out positive.
• Question 6 — dividing the two equations instead of multiplying. The complementary angles produce $\tan\alpha$ in one equation and $\cot\alpha$ in the other, so multiplying eliminates the angle and dividing does not.
• Questions 3 and 16 — assuming the same quantity is shared. In Question 3 the observer moves horizontally, so the height is shared; in Question 16 the observer moves vertically, so the distance is shared. Identify which before writing the equations, and the rest is algebra.

Practise next

• Exercise 9.1 — Circles, the next unit, which returns to pure geometry after this unit’s algebra. Tangent-and-radius arguments there use the same right-angled triangles.
• Exercise 8.3 — worth a second pass now, especially Questions 1, 2, 4 and 6, whose identities are the ingredients of Questions 1, 9, 10 and 12 here.
• Exercise 8.2 — its Questions 7 to 12 state as claims exactly the reasoning these eighteen problems require, and re-reading them after this exercise makes the justifications obvious.