Build the observation geometry first, then use a complete toolkit of triangle properties
JEE Main · JEE Advanced95 min study guide12 indexed sections
BY THE END, YOU CAN
Model elevation and depression angles with horizontal distances, vertical differences and instrument heights.
Solve two-observer configurations while distinguishing same-side and opposite-side arrangements.
Derive and apply sine, cosine and area rules for non-right triangles.
Handle ambiguous SSA data and compute radii, medians and angle-bisector quantities with exact conditions.
BEFORE YOU START
Right-triangle trigonometry and special-angle values.
Algebraic equations, square roots and angle sums.
Basic geometry of perpendicular lines and triangles.
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
Selected exam questions & worked solutions
These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.
No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.
01 / UNDERSTAND & APPLY
1. The diagram determines what the tangent ratio measures
An angle of elevation is measured upward from the observer’s horizontal line to a target above that line. An angle of depression is measured downward from the horizontal to a target below it. The reference is horizontal, not vertical, and the observer’s position is the vertex of the measured angle.
For a vertical height difference H and horizontal separation d>0, tan θ=H/d when 0<θ<90°. The sloping line-of-sight distance is the hypotenuse and is not interchangeable with d. If the sloping distance L is given, use H=L sin θ and d=L cos θ.
When an instrument or eye is h₀ above the ground, the tangent relation usually gives the target’s height above that eye level. Add h₀ to recover height above the same ground datum. If the two ground points have different elevations, include that signed level difference instead of assuming a flat common baseline.
An elevation angle at one end equals the depression angle at the other only when both are measured relative to parallel horizontal lines. This follows from alternate interior angles. It does not equate arbitrary angles drawn at the two endpoints.
ORIGINAL WORKED EXAMPLE
Separate line-of-sight height from total height
An observer’s eye is 1.5 m above level ground, 20 m horizontally from a vertical pole. The elevation to its top is 60°. Find the pole height.
The right triangle’s vertical side is the pole top above eye level, not the full pole height.
Let the pole height be H. Then tan60°=(H−1.5)/20.
Since tan60°=√3, H=20√3+1.5 m.
The line-of-sight distance would be 20/cos60°=40 m, confirming why the given horizontal distance cannot be treated as the hypotenuse.
Answer Pole height 20√3+1.5 m.
02 / UNDERSTAND & APPLY
2. Two same-side observations share a height but not a distance
Suppose two observers lie on the same straight horizontal ray from a tower base, separated by a known distance s. Let x be the nearer horizontal distance, so the farther distance is x+s. For equal eye levels, the same vertical difference H appears in H=x tan β=(x+s)tan α, with β>α.
Solving gives x=s tan α/(tan β−tan α) and H=s tan α tan β/(tan β−tan α). An equivalent height formula is H=s/(cot α−cot β). These formulas require the same-side collinear geometry and a common observation level.
If the nearer and farther angles are interchanged, the algebra may produce a negative distance. That is a signal to revisit the geometry, not an instruction to take an absolute value automatically. A taller target at a smaller horizontal distance gives the larger elevation angle in this model.
If observers have unequal eye or instrument heights, write two separate vertical differences H−h₁ and H−h₂. The simple shared-height formula then no longer applies unchanged, but the two tangent equations still solve the problem.
ORIGINAL WORKED EXAMPLE
A measured approach distance determines the tower height
On level ground, a target top is seen at elevation 30° from a point. After moving 20 m directly toward its base, the elevation is 60°. Ignore eye height. Find the height and nearer distance.
Let x be the nearer distance and H the height. Then H=x√3 and H=(x+20)/√3.
Equating gives 3x=x+20, so x=10 m.
The height is H=10√3 m, and the farther distance is 30 m.
Check the original ratios: H/10=√3 and H/30=1/√3. Both observations are satisfied.
Answer Nearer distance 10 m; height 10√3 m.
03 / UNDERSTAND & APPLY
3. The baseline may be a sum rather than a difference
If a tower base lies between two observers on a straight level baseline of length s, their distances x and s−x add to s. With elevations α and β, H=x tan α=(s−x)tan β, giving H=s/(cot α+cot β). The plus sign is a geometric consequence of opposite sides.
A target above and a target below one observer’s horizontal line create two vertical segments. For a building viewed from a tower, an elevation to the building top and a depression to its base can determine its full height by adding those segments, provided the horizontal separation is the same.
Use a single signed vertical coordinate when the arrangement is complicated. If the observer is at height h and the target at height H on a common datum, tan θ=(H−h)/d for a signed angle θ. This keeps elevations positive and depressions negative without a separate memorized formula for each case.
Not every pair of observation points is collinear with the target base. In a genuinely two-dimensional ground layout, first solve the horizontal triangle from its baseline and bearings, then use the vertical right triangles. Treating an oblique ground distance as a straight approach distance can give a plausible but wrong height.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
First locate the tower relative to both observers, then decide whether horizontal distances add or subtract.
ORIGINAL WORKED EXAMPLE
Opposite-side observers use the full baseline
Two ground observers are 60 m apart with a vertical tower base between them. Their elevations to the top are 30° and 60°. Find the height and the two base distances.
Let x be the distance from the 30° observer to the base. Then H=x/√3 and H=(60−x)√3.
Equating gives x=3(60−x), so 4x=180 and x=45 m.
The other distance is 15 m and H=45/√3=15√3 m.
The two distances add to 60. Using a difference formula would describe a different placement of the observers.
Answer Height 15√3 m; base distances 45 m and 15 m.
ORIGINAL WORKED EXAMPLE
Elevation and depression combine two vertical parts
From the top of a 30 m tower, the angle of depression to a building base is 60° and elevation to its top is 30°. Both bases lie on the same level ground. Find building height.
Let d be the horizontal separation. The depression triangle gives tan60°=30/d, so d=10√3 m.
If building height is H, the elevation triangle gives tan30°=(H−30)/d.
Therefore H−30=(10√3)/√3=10 m, and H=40 m.
The common ground level and common horizontal separation are essential. The building top is 10 m above the tower top, not 10 m above ground.
Answer Building height 40 m.
04 / UNDERSTAND & APPLY
4. Fix the side-angle correspondence before using any rule
For a nondegenerate triangle ABC, use a=|BC| opposite angle A, b=|CA| opposite B and c=|AB| opposite C. All side lengths are positive, all interior angles lie strictly between 0 and π, and A+B+C=π.
The strict triangle inequalities a<b+c, b<c+a and c<a+b are necessary and sufficient for three positive lengths to form a nondegenerate triangle. Equality gives a flattened configuration with zero area, not an ordinary triangle with positive inradius.
The larger side lies opposite the larger angle. This ordering is a useful check after solving, especially when an inverse sine returns an acute angle although the data require an obtuse one. Equal sides correspond to equal opposite angles.
Let s=(a+b+c)/2 be the semiperimeter and Δ the positive area. The same triangle can be described by sides, angles, altitudes or circle radii. The formulas below connect those descriptions, but every one retains the nondegeneracy and side-angle conventions.
05 / UNDERSTAND & APPLY
5. The sine rule can produce two angle candidates
Dropping an altitude gives h=b sin A=a sin B, hence a/sin A=b/sin B. Repeating with another altitude gives a/sin A=b/sin B=c/sin C. Extending the triangle to its circumcircle yields the common value 2R, where R is the circumradius.
The sine rule is particularly useful when two angles and one side are known, or when an opposite side-angle pair is available. With only side-side-angle data where the angle is not included between the two sides, it can be ambiguous because sin B=sin(π−B).
If the calculated sine exceeds 1, no triangle exists. If it is in (0,1), consider both B=arcsin(value) and B=π−arcsin(value), then reject any candidate with A+B≥π. Do not assume the calculator’s principal acute output is the only geometric angle.
For acute given A and side b adjacent in the SSA description, the altitude b sin A provides a helpful classification: a below that altitude gives no triangle, equality gives one right triangle, altitude<a<b gives two, and a≥b gives one. This case rule depends on A being acute; the direct angle check works more generally.
ORIGINAL WORKED EXAMPLE
Two triangles satisfy the same SSA data
Given A=30°, a=7 and b=10, show that two triangles are possible and find the two possible values of c.
The sine rule gives sin B=b sin A/a=10·(1/2)/7=5/7. Its acute solution and its supplement are both less than 150°, so both leave a positive C.
For an exact side calculation, use the cosine rule at A: 7²=10²+c²−2·10·c cos30°.
This gives c²−10√3 c+51=0. The roots are c=5√3±2√6, both positive.
Each root satisfies the original side-angle relation and the triangle inequalities. The two values correspond to the acute and obtuse possibilities for B.
Answer Two triangles; c=5√3+2√6 or c=5√3−2√6.
06 / UNDERSTAND & APPLY
6. The cosine rule handles an included angle or three sides
Place one side of length c on the x-axis and the adjacent side of length b at angle A. Their endpoints are (c,0) and (b cos A,b sin A). The squared distance between them is c²+b²−2bc cos A, giving a²=b²+c²−2bc cos A.
The rule works for acute, right and obtuse angles. When A=90°, the cosine term vanishes and Pythagoras is recovered. Solving for the angle gives cos A=(b²+c²−a²)/(2bc), whose real value must lie in [−1,1] for consistent data.
With three valid sides, classify the largest angle by comparing the square of the largest side with the sum of squares of the other two: smaller gives acute, equal gives right and larger gives obtuse. This one comparison classifies whether the whole triangle is acute, right or obtuse.
Use the cosine rule for SAS or SSS data, then the sine rule if it simplifies the remaining work. The rules are complementary tools; choosing one solely from a memorized keyword without checking which angle is included can produce an incorrect equation.
ORIGINAL WORKED EXAMPLE
An included angle fixes the opposite side
Two sides of a triangle are 5 and 8 with included angle 60°. Find the third side and area.
Let the opposite side be a. The cosine rule gives a²=5²+8²−2·5·8·(1/2)=25+64−40=49.
Therefore a=7, taking the positive length.
The area is (1/2)·5·8·sin60°=20·√3/2=10√3.
The sides 5,7,8 satisfy the strict triangle inequalities, and the computed side 7 is consistent with the specified included angle.
Answer Third side 7; area 10√3.
07 / UNDERSTAND & APPLY
7. Area connects sides, angles and radii
The area formulas are Δ=(1/2)bc sin A=(1/2)ca sin B=(1/2)ab sin C. They follow from base times perpendicular altitude, and remain valid for obtuse triangles because the positive sine gives the altitude magnitude.
Heron’s formula is Δ=√[s(s−a)(s−b)(s−c)]. It can be derived by squaring the sine-area formula and using the cosine rule: 16Δ²=4b²c²−(b²+c²−a²)², which factors into (a+b+c)(−a+b+c)(a−b+c)(a+b−c).
If r is the inradius, joining the incenter to the vertices splits the area into ar/2+br/2+cr/2=rs. Thus r=Δ/s. The circumradius follows from the extended sine rule and area: R=abc/(4Δ). These are different circles with different distance definitions.
The exradius opposite A is r_a=Δ/(s−a), with analogous formulas cyclically. The denominator is positive by the strict triangle inequality. Degenerate data make Δ zero and can invalidate ratio formulas, so check the triangle before substituting.
ORIGINAL WORKED EXAMPLE
A single Heron calculation yields several geometric quantities
A triangle has sides 13,14,15. Find its area, inradius and circumradius.
The semiperimeter is s=(13+14+15)/2=21.
Heron gives Δ=√(21·8·7·6)=√7056=84.
The inradius is r=84/21=4.
The circumradius is R=(13·14·15)/(4·84)=2730/336=65/8. The exact fraction avoids unnecessary rounding.
Answer Area 84; inradius 4; circumradius 65/8.
08 / UNDERSTAND & APPLY
8. Different cevians satisfy different length relations
A median from A meets BC at its midpoint. Its length m_a satisfies m_a²=(2b²+2c²−a²)/4. One derivation places the midpoint at the origin with B,C at ±a/2 on an axis, then adds the two squared distances from A to B and C.
The internal angle bisector from A meets BC at D with BD:DC=c:b. This follows by comparing the areas of triangles ABD and ACD: they share the same altitude to BC, while their sine-area expressions use the equal half angles and adjacent sides c,b.
The angle-bisector length satisfies ℓ_a²=bc[1−a²/(b+c)²], or ℓ_a=2bc cos(A/2)/(b+c). These formulas assume the internal bisector and a nondegenerate triangle. An external bisector has a different signed division relation and should not be inserted into the same expression without rederivation.
A median, altitude and angle bisector need not coincide. In an isosceles triangle they coincide from the apex to the unequal side, but that is a consequence of symmetry, not a general definition. Identify the requested cevian before choosing its formula.
ORIGINAL WORKED EXAMPLE
Median and angle-bisector data in the same triangle
Let a=14,b=13,c=15. Find the median from A and the division of BC by its internal angle bisector.
The median square is [2·13²+2·15²−14²]/4=(338+450−196)/4=148.
Thus m_a=√148=2√37.
The angle-bisector ratio is BD:DC=c:b=15:13. Since BC=14, BD=14·15/28=15/2 and DC=13/2.
The median would divide BC into 7 and 7 instead, showing that the two cevians are different in this scalene triangle.
Answer Median 2√37; angle-bisector segments BD=15/2 and DC=13/2.
Because A+B+C=π, the half angles are positive and sum to π/2. The half-angle identities have unambiguous positive square roots: sin(A/2)=√[(s−b)(s−c)/(bc)] and cos(A/2)=√[s(s−a)/(bc)]. Their ratio gives tan(A/2)=r/(s−a).
The sine rule replaces side ratios by opposite sine ratios, which can simplify expressions mixing sides and angles. The cosine rule replaces cos A by a symmetric side expression when all side lengths are known. Choose the representation that reduces the number of independent variables.
For example, tan(A/2)tan(B/2)+tan(B/2)tan(C/2)+tan(C/2)tan(A/2)=1, because the half-angle sum is π/2 and the denominator in the tangent-sum expression vanishes while all three half angles are finite and positive.
The identity tan A+tan B+tan C=tan A tan B tan C holds when none of A,B,C is a right angle, so all tangents exist. A triangle angle sum does not authorize dividing by a cosine that could be zero. State that restriction rather than treating the right-triangle case as a finite tangent identity.
10 / UNDERSTAND & APPLY
10. Original practice: heights and distances
State whether a given distance is horizontal or along the line of sight, and whether the ground and eye levels are shared.
ORIGINAL PRACTICE · 1
A pole is viewed at elevation 45° from 18 m horizontally away on level ground. Ignore eye height. Find its height.
Reveal answer & reasoning
18 m.
tan45°=H/18=1, so H=18.
ORIGINAL PRACTICE · 2
A 10 m line of sight makes elevation 30°. Find the vertical and horizontal components.
Reveal answer & reasoning
Vertical 5 m; horizontal 5√3 m.
Use 10sin30° and 10cos30° because 10 is the hypotenuse, not the horizontal side.
ORIGINAL PRACTICE · 3
An eye 2 m above ground sees a top at elevation 45° from horizontal distance 12 m. Find total height.
Reveal answer & reasoning
14 m.
The top is 12tan45°=12 m above the eye, so add the eye height 2 m.
ORIGINAL PRACTICE · 4
From the top of a 20 m tower, depression to a ground point is 45°. Find horizontal distance.
Reveal answer & reasoning
20 m.
The downward vertical difference is 20 and tan45°=20/d, giving d=20.
ORIGINAL PRACTICE · 5
Two same-side observations are 10 m apart, with nearer elevation 45° and farther elevation 30°. Find height, ignoring eye level.
Reveal answer & reasoning
5(√3+1) m.
If near distance is x, then H=x and H=(x+10)/√3. Thus x=10/(√3−1)=5(√3+1).
ORIGINAL PRACTICE · 6
Two observers on opposite sides are 40 m apart and both see a top at elevation 45°. The base lies between them. Find height.
Reveal answer & reasoning
20 m.
Each base distance equals the common height, so 2H=40.
ORIGINAL PRACTICE · 7
Can a measured elevation angle alone determine a tower height?
Reveal answer & reasoning
No.
Without a distance or another independent geometric constraint, infinitely many similar right triangles have the same angle.
ORIGINAL PRACTICE · 8
A target is 6 m above an observer’s eye and 8 m horizontally away. Find the line-of-sight distance.
Reveal answer & reasoning
10 m.
The right-triangle hypotenuse is √(6²+8²)=10. The elevation tangent is 6/8, not 6/10.
11 / UNDERSTAND & APPLY
11. Original practice: properties of triangles
Use a,b,c opposite A,B,C. Check that the given data produce a nondegenerate triangle before using area or radius ratios.
ORIGINAL PRACTICE · 1
Can sides 3,4,7 form a nondegenerate triangle?
Reveal answer & reasoning
No.
The sum 3+4 equals 7, so the strict triangle inequality fails and the configuration is collinear.
ORIGINAL PRACTICE · 2
Classify a triangle with sides 5,6,8 as acute, right or obtuse.
Reveal answer & reasoning
Obtuse.
The largest square is 64, exceeding 25+36=61, so the opposite angle is obtuse.
ORIGINAL PRACTICE · 3
Find the area of a triangle with sides 6 and 10 including angle 30°.
Reveal answer & reasoning
15.
Use Δ=(1/2)·6·10·(1/2)=15.
ORIGINAL PRACTICE · 4
If A=30° and a=8, find the circumradius.
Reveal answer & reasoning
8.
The extended sine rule gives 2R=a/sinA=8/(1/2)=16.
ORIGINAL PRACTICE · 5
Find the inradius of a 3–4–5 right triangle.
Reveal answer & reasoning
1.
Area is 6 and semiperimeter is 6, so r=Δ/s=1.
ORIGINAL PRACTICE · 6
For a=14,b=13,c=15 with area 84, find tan(A/2).
Reveal answer & reasoning
4/7.
The semiperimeter is 21 and inradius is 84/21=4. Thus tan(A/2)=r/(s−a)=4/7.
ORIGINAL PRACTICE · 7
An equilateral triangle has side 6. Find its median length.
Reveal answer & reasoning
3√3.
The median is also an altitude, splitting the base into lengths 3 and 3. Its length is √(36−9)=3√3.
ORIGINAL PRACTICE · 8
A sine-rule calculation gives sinB=6/5. What does this mean?
Reveal answer & reasoning
No real triangle satisfies the data.
A real angle has sine at most 1. Do not round or force an inverse-sine value outside its domain.
12 / UNDERSTAND & APPLY
12. Geometry first, formulas second
For an observation problem, mark the observer’s horizontal line, the vertical target, the correct ground distances and any eye-height offsets. Decide whether a baseline is a sum, a difference or one side of an oblique ground triangle before writing a tangent equation.
For a general triangle, label each side opposite its angle, check strict triangle inequalities and choose the rule matching the known data. SSA can have two solutions; SAS and SSS use the cosine rule directly. A principal inverse-sine output is only one candidate in an ambiguous case.
Area is a useful bridge: it connects side-angle data, Heron’s formula, inradius and circumradius. Medians and angle bisectors have different definitions and length formulas. Half-angle formulas are safe because triangle half angles lie in the first quadrant, but full-angle tangent identities still need nonzero cosines.
Finish with scale and feasibility checks. Heights and lengths must be positive in the chosen geometry, side ordering must agree with angle ordering, and exact special-angle results should remain exact until an approximation is requested. A negative or impossible value usually points to a modelling or branch error before it points to a need for more calculation.
KEEP THE BIG IDEAS
Quick revision
Elevation and depression use horizontal reference lines and vertical differences from the observer.
Same-side and opposite-side baselines lead to different equations.
Sine and cosine rules require correct side-angle correspondence and SSA branch checks.
Area links side data to inradius, circumradius and half angles.
Medians, altitudes and angle bisectors coincide only under special symmetry.
The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.