NCERT Solutions Class 10 Maths Chapter 9 Some Applications of Trigonometry

How to choose the ratio for full marks. An examiner wants to see you name the triangle, mark the right angle, and justify the sine choice before any number appears.

• Read the sides: the rope length is the hypotenuse and the pole is what we want, which is the opposite side, so sine is the only ratio that uses just those two.
• Exact value: because sin 30^∘=12 is an exact value, the height comes out as a clean 10 m with no rounding, and that exactness tells the examiner the method is sound.
• Order of writing: put the ratio in words first, then the symbol form, then the substitution, and you secure every method mark even if a final arithmetic slip creeps in.
• Sanity check: the pole must be shorter than the slanting rope, and 10 m is comfortably less than the 20 m rope, so the answer sits in a sensible range.

Height of the pole =10 m, exactly half the rope length.

A faster single-formula route. Instead of finding the two parts separately, notice that the total height equals BC(tan 30^∘+sec 30^∘) and pack both pieces into one line.

• Shared base: reading the picture carefully is what makes this quick, since the horizontal ground distance between the foot and the touch point is shared by both right-triangle ratios, so the same eight metres feeds into both.
• Two pieces: the vertical standing piece is the base times the tangent of the angle, and the slanting broken piece is the base times the secant of the same angle, so the original height is just their sum.
• One step: the sum collapses to 8(13+23)=8·33=83, and the answer drops out in a single line once the common factor of eight is taken outside the bracket.
• Why the surd matters: a neat surd such as 83 signals to the examiner that the special-angle values were handled correctly, which earns more credit than a rounded decimal would.
• Rebuild the whole: always reconstruct the full original object, here the entire tree before it broke, rather than stopping at one of its parts, because the question asks for the original height and not just a piece.

83 m, obtained in one line as 8(tan 30^∘+sec 30^∘).

Keep the two cases clearly separated. The question asks for two lengths, so present them as two labelled mini-solutions, because mixing the numbers from the two slides is the easiest way to lose marks here.

• Same structure: the steps for both parts are identical, and it is worth pointing out to the examiner that each is height opposite the angle, slide as hypotenuse, sine chosen, then substitute.
• First slide: the exact value sin 30^∘=12 gives a tidy length of 3 m straight away.
• Second slide: sin 60^∘=32 leads to 63, which must be rationalised to 23 for full marks, since leaving a surd in the denominator is treated as an unfinished answer.
• Reasoning mark: a short sentence noting that the steeper slide turns out shorter shows physical understanding and is the kind of remark that earns the reasoning mark.

Younger children's slide =3 m; older children's slide =23 m.

Rationalise before you decimalise. Many students write 303 and then reach for a calculator, but the board scheme rewards converting to 103 first.

• Why tangent: once the right triangle is drawn, the opposite side is the height, the adjacent side is the 30 m, and tangent is the only ratio that joins those two without the hypotenuse.
• Exact form: after substituting tan 30^∘=13, the height is 303, and multiplying top and bottom by 3 turns it into 103.
• Decimal last: only at the very end, if a decimal is asked for, do you write the rounded value ≈ 17.32 m.
• Carry the surd: keeping the exact form as long as possible makes the next part of any multi-step problem cleaner, since surds combine more neatly than rounded decimals.
• Sanity check: the tower should be shorter than the 30 m line of sight along the ground, and 17.32 m comfortably is.

103 m, kept in exact surd form before any decimal.

Watch which side is the hypotenuse. A frequent slip is to treat the string as the adjacent side and reach for cosine, when the string is the longest, slanting side and only sine links it to the height.

• Label first: once the triangle is labelled the rest is mechanical, with the height 60 m opposite the 60^∘ angle and the string as the hypotenuse.
• Apply sine: the value sin 60^∘=32 gives the string length =1203 directly.
• Rationalise: converting to 403 is essential, since a surd left in the denominator is marked incomplete.
• Check the size: the string must be longer than the 60 m height because it slants, and 403≈ 69.28 m is indeed longer, confirming the answer is reasonable.
• Justify in words: pointing out that the hypotenuse is always the longest side is a clean way to defend the sine choice.

403 m, longer than the 60 m height as a slant must be.

One combined formula for the walk. The distance walked is 28.5(cot 30^∘-cot 60^∘), which avoids computing the two distances on separate lines.

• Where it comes from: rearranging the tangent equation for the distance gives the distance as the usable height times the cotangent of its angle, so each horizontal distance can be written immediately without solving a separate equation.
• Simplify the bracket: since 3-13=3-13=23, the walk is 28.5·23=573=193 m once the surd in the denominator has been rationalised away.
• The key mark: the single biggest reasoning mark here comes from correctly subtracting the boy's height to work above eye level, so state that step in words rather than doing it silently.
• Direction check: as the boy walks closer the angle of elevation grows, which matches the near point with the larger angle sitting closer than the far point with the smaller angle.

193 m, as 28.5(cot 30^∘-cot 60^∘).

Find the shared distance first. Nailing x=20 from the easy 45^∘ angle turns the harder 60^∘ equation into a single unknown, which is the cleanest order of work.

• Shared base: both lines of sight start from the same point P and rest on the same vertical line, so the two triangles share the base x.
• Easy angle first: the lower angle 45^∘ fixes x=20 instantly because tan 45^∘=1.
• Then the height: feeding x=20 into tan 60^∘ gives the full height 203, and subtracting the known building height 20 leaves the tower as 20(3-1).
• Factor neatly: keeping the factor of 20 outside the bracket makes the surd answer compact and easy to mark.
• Reasonableness: the tower must be shorter than the building since the gap between 45^∘ and 60^∘ is small, and 14.64 m is indeed less than 20 m.

20(3-1) m, after fixing the shared base x=20 m first.

Use the 45^∘ angle to remove a variable. Setting x=h from the 45^∘ triangle collapses two unknowns into one, the standard tactic whenever one of the angles is 45^∘.

• Collapse first: after substituting the distance for the pedestal height, the elevation equation rearranges cleanly to the pedestal height times the bracket root three minus one equals one point six, leaving a single unknown.
• The delicate step: the only tricky part left is rationalising the conjugate, and multiplying top and bottom by root three plus one turns the messy denominator into a clean two, which gives the factored answer at once.
• Form expected: writing the answer in this factored surd form is what the board scheme wants, with the rounded decimal offered only as a final approximation right at the end.
• Check the height: the pedestal at 2.19 m should be larger than the 1.6 m statue on top, since the elevation gap from 45^∘ to 60^∘ is wider than the statue is tall, and it is.

0.8(3+1) m, found by rationalising 1.63-1.

Notice the neat scaling relationship. The unknown building height turns out to be the known tower height scaled by a fixed factor, and seeing this saves time and gives a clean memory hook.

• Find the base: the tower triangle gives the gap between the two feet as d=503, and this same horizontal distance is then reused for the building triangle.
• Carry it across: feeding that distance into the building triangle gives the building height as h=d3, which simplifies to the value 503 without any extra work.
• The scaling factor: a compact way to see the answer is that the building height equals the tower height multiplied by the product 30^∘60^∘, and because each of those factors is the same special value, the product is one third.
• Memorable check: stating that the building comes out at exactly one third of the tower is an easy way to confirm the arithmetic, and a wrong answer rarely lands on so clean a fraction.
• Where the marks are: the key reasoning mark comes from explaining that both sightings are measured across the very same horizontal base, which is precisely what lets you carry the distance from one triangle straight into the other.

503 m, exactly one-third of the 50 m tower.

Equate the two heights, not the two distances. The cleanest equation comes from writing h two ways and setting them equal, which removes h and leaves a simple linear equation in x.

• State the key fact: the single fact that the two poles are the same height is what makes the problem solvable, so say so explicitly in your working.
• Two expressions: from the near pole h=x3 and from the far pole h=80-x3, and equating them gives 3x=80-x, hence x=20.
• Back-substitute: the distances are then 20 m and 60 m, and putting x=20 back gives the common height h=203.
• Sanity check: the closer pole at 20 m seen at the steeper 60^∘ and the farther pole at 60 m seen at the gentler 30^∘ both yield the same height, which confirms the equal heights stated in the problem.
• Final mark: keeping the height as 203 in surd form rather than a decimal earns the closing mark.

Height 203 m; distances 20 m and 60 m from the point.

This is the same pattern as the two-pole problem. Whenever two elevation angles are taken from points a known distance apart along one line, write the height from each and subtract to eliminate it.

• Same target: both sightings look at the same tower top from the same straight line, so the height can be written from the near point and from the far point, and the two expressions must be equal.
• Clear the root: substituting the special values turns the equation into x3=x+203, and clearing the root gives 3x=x+20, so the canal width is ten metres and the height is ten root three.
• Geometry check: the height of about seventeen metres is larger than the ten-metre canal width, which fits a tall tower seen at a steep angle from the near bank.
• Final mark: keeping the surd form for the height rather than rounding it to a decimal secures the closing mark.

Tower 103 m high; canal 10 m wide.

Split the tower at roof level. The roof line cuts the tower into a lower piece equal to the building height and an upper piece, so find the distance, then the upper piece, then add.

• The crucial setup: draw the horizontal sightline from the building top across to the tower, which splits the tower into the 7 m matching the roof and the part above it.
• Distance from depression: the 45^∘ depression to the foot gives the horizontal distance d=7, because 45^∘=1 makes that distance equal the vertical 7 m drop.
• Upper piece from elevation: the 60^∘ elevation then gives the part above the roof as 73, and the full height is 7+73=7(3+1).
• Factor tidily: pulling the seven outside the bracket keeps the surd compact and easy for the examiner to read, and it also makes the structure of the answer, a roof piece plus an upper piece, clear at a glance.
• Quick check: the tower at about nineteen metres is taller than the seven-metre building, which is exactly what we expect since its top is seen by looking up rather than down.

7(3+1) m, the 7 m roof piece plus the 73 m upper piece.

Both ships, one shared height. The 75 m height is the common vertical side for both right triangles, so compute each horizontal distance from the foot and then subtract.

• Depression equals elevation: an angle of depression from the top equals the angle of elevation from the ship, so each ship sits in a right triangle with the 75 m lighthouse as the opposite side.
• Each distance: the nearer ship at 45^∘ is 75 m out, and the farther ship at 30^∘ is 753 m out, and their difference 75(3-1) is the gap between them.
• Do not shortcut: do not try to build one triangle between the two ships directly, since the clean route is two triangles sharing the same height.
• Useful check: the farther distance 753129.9 m is larger than the nearer 75 m, so the subtraction is positive and sensible.
• Keep it exact: presenting the answer as 75(3-1) rather than a raw decimal keeps the exactness the board scheme expects.

75(3-1) m, the difference of the two ship distances.

Factor the common height for speed. The travelled distance is 87(cot 30^∘-cot 60^∘), and computing the bracket once avoids two separate rationalisations.

• Same model: this is the walking-boy problem turned on its side, with the balloon moving horizontally while the line of sight swings from 60^∘ down to 30^∘.
• One bracket: each horizontal distance is 87cotθ, so the travel is 87(3-13)=87·23=583 after rationalising.
• The key step: subtract the 1.2 m eye height first so that the 87 m is measured from the correct level, as this is where the main reasoning mark sits.
• Direction check: as the balloon drifts away the angle shrinks from 60^∘ to 30^∘, which matches the farther position carrying the smaller angle.

583 m, as 87(cot 30^∘-cot 60^∘).

The tower height cancels, so do not look for it. Students often get stuck hunting for the height, but it never appears in the final answer because only the ratio of distances matters.

• Two distances: the far distance from the first sighting is the height times root three, and the near distance from the second sighting is the height divided by root three, both written straight from the two depression angles.
• Covered and remaining: the six-second leg covers the difference of those two distances, while the leg still to go is just the near distance, so both legs are written in terms of the same unknown height.
• Everything cancels: the ratio of the remaining leg to the covered leg works out to one half, with every height and every root dividing out, which is exactly why the tower height never matters.
• Time follows distance: state up front that uniform speed makes time proportional to distance, then the remaining time is one half of the six seconds, which is three seconds.
• Neat check: the car covers two-thirds of the journey in the first six seconds and the last third in three seconds, a clean split that confirms the answer is reasonable.

3 seconds, independent of the tower height h.