NCERT Exemplar Solutions Class 10 Maths Chapter 9 Some Applications of Trigonometry

Match the surd to a standard angle.

• One ratio: the whole problem collapses to opposite over adjacent, which is the pole height 6 over the shadow length 23, and nothing else in the figure matters.
• Tidy first: write the top as 6=32 and cancel the common factor of two, so the messy fraction reduces to the clean value 3 before you read any angle.
• Read the table: the only acute angle in the whole Class 10 list whose tangent equals 3 is sixty degrees, so the Sun must stand that far above the level ground.
• Stay exact: keep numbers as surds until the very last line, because switching to a decimal turns the answer into a messy 1.732 and you lose the clean match with the standard table.

Option (A), 60^∘.

Height is fixed, so angle and shadow pull in opposite ways.

• Pin the height: keep the tower height fixed and treat the shadow as the only quantity that is allowed to change in the whole question.
• Inverse link: the defining equation tanθ=h/x then says the tangent and the shadow are inversely tied, so pushing the shadow up must drag the tangent down.
• Angle follows: because the tangent always grows with the angle between zero and ninety degrees, a falling tangent forces a falling angle, so the Sun sinks lower as the shadow lengthens.
• Claim reversed: the statement asserts both rise together, which is exactly backwards, so it is false; the long evening shadow beside a low Sun is the quickest sanity check in the exam.

False; increasing shadow length forces the angle of elevation to decrease.

Reflection mirrors the water line, not the eye line.

• Mirror plane: a reflection in still water is symmetric about the lake surface, so the image sits exactly as far below the water as the cloud floats above it.
• Broken symmetry: the observer is raised three metres on a platform, so the mirror plane is not at eye height but three metres below the eye, and that one fact ruins the symmetry the statement quietly assumes is there.
• Measure from the eye: the cloud is therefore H-3 up and the image is H+3 down, with both reached over the same horizontal run to the vertical line of the cloud.
• Compare the tangents: same denominator but the depression carries the larger numerator, so the depression beats the elevation and the two angles are unequal whenever the eye sits off the water.
• Verdict: equality would need the platform to vanish, so on a real three metre platform the claim is simply false.

False; the platform height 3 m makes the depression of the reflection larger than the elevation of the cloud.

Double the opposite side, not the angle.

• What doubling does: doubling the height doubles only the value of the tangent, since the ratio is height over base and the base never changes when you stand at the same point.
• Undoing is nonlinear: the angle comes from reversing the tangent, and that step bends, because the tangent climbs ever more steeply as the angle opens out toward ninety degrees.
• So scaling fails: a factor of two on the ratio therefore buys far less than a factor of two on the angle, and the two simply cannot move in step with each other.
• The numbers: here the new tangent lands near forty nine degrees, comfortably short of the sixty degrees that genuinely doubling the original angle would have demanded.
• Verdict: the claim treats the angle as if it scaled with the side, which trigonometry never allows, so it is false.

False; the new elevation is about 49^∘, not the doubled 60^∘.

Same multiplier on both legs is a similar triangle.

• Shape, not size: the angle of elevation reads only the shape of the right triangle and never its size, and that shape is captured entirely by the ratio of height to base.
• Similar triangle: multiplying the height and the base by the identical factor produces a triangle similar to the first, so every angle inside it is preserved unchanged.
• Algebra agrees: the common factor of one point one sits in both the top and the bottom of the ratio, so it simply cancels and returns the original tangent untouched.
• Verdict: the statement is therefore true, and it would fail only if the height and the distance were increased by two different percentages instead of the same one.

True; equal scaling of height and distance keeps tanθ, and so the angle, fixed.

A self-similar triangle fixes the angle.

• Height drops out: because the shadow is a fixed multiple of the height, the right triangle keeps the same shape no matter what the actual height is, so its angles are settled once and for all.
• Read the ratio: the tangent works out to one over root three, and the only acute angle in the whole Class 10 table that matches that value is thirty degrees.
• Mirror of before: this is the exact opposite of the earlier pole problem, where the short shadow gave a high Sun at sixty; here the long shadow gives a low Sun at thirty.

The Sun's elevation is 30^∘.

Name the side before picking the ratio.

• The one decision: everything turns on whether the wall is opposite or adjacent to the given angle, and getting that single call right is what makes or breaks the whole problem.
• Wall is adjacent: since the sixty degree angle sits between the ladder and the wall, the wall touches that angle and is the adjacent side, so cosine is the ratio that links it to the ladder.
• Clean number: the cosine of sixty is one half, so the wall is exactly half the ladder, a tidy seven and a half metres with no surd left to simplify.
• Exam habit: mark the given angle on a quick sketch and write the words opposite and adjacent on the two sides before you ever reach for sine or cosine.

The wall is 7.5 m high.

Shift the baseline up to the eye.

• Raise the ground: the line of sight starts at the eye, one and a half metres up, so the cleanest move is to imagine the whole ground level lifted to that eye height before you do anything else.
• Shorter rise: from this raised baseline the tower no longer climbs the full twenty two metres; its bottom slice now lies below the new line and contributes nothing, leaving a rise of twenty and a half metres.
• Run unchanged: the horizontal distance the observer stands back is still twenty and a half metres, so the triangle has a rise that exactly equals its run.
• Clean angle: rise equal to run means the tangent is exactly one, and the only standard angle whose tangent is one is forty five degrees, so the answer falls out at once.
• The trap: the matching numbers are no accident, but they only line up for students who subtract the eye height first; skip that step and you wrongly use the full height over the distance.

The angle of elevation is 45^∘.

Two stations, one shared height.

• Shared link: the tower height is the single quantity common to both viewpoints, so it is the thread that ties the two separate sightings together into one solvable system.
• Two equations: write a tangent equation at each station and treat the height and the near base as the two unknowns; two equations in two unknowns can always be solved exactly.
• Convert the angle: the question gives an increase of fifteen degrees, not the new angle, so first add it on to reach forty five degrees before any algebra begins.
• Free relation: the forty five degree station is a gift, because its tangent is one and it hands you the near base equal to the height for nothing.
• Finish cleanly: feeding that into the other equation isolates the height, and rationalising the surd denominator leaves the tidy closed form, roughly twenty seven metres.
• Label carefully: the twenty metre walk is what breaks the deadlock, so label the two bases as x and x+20 rather than guessing either, since that labelling is where most students slip.

The tower is 10(3+1)27.32 m high.

Write each shadow in terms of the one height.

• One unknown: the same tower throws both shadows, so the height is the single unknown that runs right through the problem and everything else should be written in terms of it.
• Shadow as base: a shadow is the base of the right triangle, so it equals the height divided by the tangent of the Sun's elevation, which lets you size each one from the height alone.
• Two shadows: the high Sun at sixty degrees gives the short base, while the low Sun at thirty degrees gives the long base, which is three times as long.
• Set the difference: the problem says the long shadow beats the short one by fifty metres, so their difference over the common root three denominator collapses to a simple equation in the height.
• Order matters: subtract short from long so the fifty stays positive, since the lower Sun always casts the longer shadow, and reversing the order quietly flips the sign.

The tower is 25343.3 m high.

Each foot looks at the opposite top.

• Crossed sightings: one line starts at the base of tower two and climbs to the top of tower one, while the other starts at the base of tower one and climbs to the top of tower two.
• Shared gap: both lines span the same horizontal distance between the towers, and that single shared distance is the key that unlocks the whole pair of equations.
• Known tower first: the sixty degree sighting looks at the already known thirty metre tower, so its equation holds only one unknown and pins the distance down right away.
• Then the height: with the distance settled, the thirty degree sighting has only the second height left, and substituting delivers it cleanly as ten metres.
• Tidy check: the second tower comes out at exactly one third the first, which mirrors the one to three ratio of the two tangents, a satisfying confirmation that the setup was right.

d=103 m and the second tower is 10 m high.

Turn depressions into elevations, then subtract.

• Alternate angles: the horizontal at the top of the tower is parallel to the ground, so each angle of depression copies straight down to an equal angle of elevation at the foot.
• Draw the line: that alternate angle step is the heart of every depression problem, so draw the parallel dashed line each time to make the equal angles plainly visible.
• One tangent each: once the depressions become elevations, every object sits one tangent away from its distance, since the tower is the opposite side and the ground gap is the adjacent side.
• Nearer is steeper: cotangent shrinks as the angle grows, so the larger depression gives the smaller distance, which means the steeper sighting points at the closer object.
• Subtract in order: the objects lie in a straight line with the foot, so the gap is just the difference of their distances, and keeping the larger one first stops the answer going negative.

The objects are h(cotα-cotβ) apart.

Same base, two heights above eye.

• Shared base: the two points lie on one vertical line, ten metres apart, so they sit at the same horizontal distance from the tower, and that shared distance bridges the two sightings.
• Full climb below: from the ground point the line of sight climbs the entire tower height at sixty degrees, which gives the base in terms of the full height over root three.
• Shorter climb above: from the upper point the tower rises only the height minus ten, and at forty five degrees the tangent of one makes the base equal to that reduced height outright.
• Eliminate the base: setting the two expressions equal clears the distance and leaves one equation in the height, which rationalising then solves cleanly.
• Watch the trap: the common error is keeping the full height at the upper point, so always cut it to the height minus ten there; the answer near twenty four metres comfortably clears the ten metre climb.

The tower is 5(3+3)23.66 m high.

Split the far house at window level.

• Cut in two: the clean way to see the proof is to slice the opposite house into two pieces at the height of the window, so the whole height becomes a simple sum of two parts.
• Lower piece: the part from the ground to window level is exactly the window height, because both buildings stand on the same flat ground, so this slice needs no extra work.
• Width first: the window height also serves as the known vertical drop in the downward sighting, so the depression to the far foot pins the lane width straight away.
• Upper piece: with the lane width known, the elevation to the top gives the rise above the window, and substituting the width expresses it through the two given angles.
• Add and factor: summing the lower and upper pieces and pulling out the common window height gives the required formula; the downward sighting must go first, as it is the only one with a fully known length.

Proved: the other house is h(1+tanβ) m tall.

One balloon, two eye levels, shared base.

• Shared run: the two windows sit on the same wall, one directly above the other, so both look out over the identical horizontal distance to the balloon, and that common run is what lets the two sightings be solved together.
• Ground heights first: settle every height against the ground, never against a window, so the lower window is two metres up and the upper window is six metres up before any tangent is written.
• Two bases: from the lower window the balloon rises H-2 at sixty degrees and gives the short base, while from the upper window it rises only H-6 at thirty degrees and gives the long base; the higher, gentler sighting making the longer base is a quick sanity check.
• Clear the base: equating the two bases removes the distance, and multiplying through by the surd collapses everything into the tidy linear equation H-2=3(H-6), which solves to eight metres.
• The trap: the recurring slip is mixing window-relative and ground-relative heights, so always convert the window heights to ground heights before writing the tangents.

The balloon is 8 m above the ground.