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NCERT Class 10 Mathematics Exercise 9.1 Solutions: Some Applications of Trigonometry
AI-written, checked on 23 Oct 2026
Solutions written by Shishya, checked — not the board's official answers
Step-by-step working for every question of Exercise 9.1 in NCERT's Class 10 Mathematics book, chapter “Some Applications of Trigonometry”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.
How these were checked: the solutions were drafted with AI and then read and corrected on 23 Oct 2026. Every numerical final answer is recomputed from the question's own numbers by a calculator check that runs before each release; proofs and explanations were read, not machine-checked. The NCERT textbook is the authority — if something here disagrees with it, follow the book.
Tap “why?” under a step to see the textbook rule it uses — from Shishya's fixed list of NCERT Class 9-10 rules, not written by AI for this page. A step without one has no rule from the list yet.
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Ex 9.1 Q1
A 20 m rope runs tight from the top of a vertical pole to the ground, making 30° with the ground. Find the height of the pole.
Uses a figure printed in the NCERT book — open the chapter above to see it.
- Let the height of the pole be h m.
- The rope acts as the hypotenuse of a right-angled triangle with length 20 m.
why?
In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.
- The angle between the rope and the ground is 30°.
- Using the sine ratio, sin 30° = opposite / hypotenuse.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Substitute the values: sin 30° = h / 20.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- We know that sin 30° = 1/2.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- So, 1/2 = h / 20.
- Solving for h, h = 20 × 1/2.
- h = 10 m.
Answer: The height of the pole is 10 m.
Ex 9.1 Q2
A tree breaks in a storm; the top touches the ground 8 m from the foot, the broken part making 30° with the ground. Find the original height of the tree.
- Let the tree break at point B, with the top touching the ground at point C and the foot at point A.
- The broken part BC makes an angle of 30° with the ground AC.
- In right-angled triangle ABC, tan(30°) = AB / AC.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Substitute the given values: tan(30°) = AB / 8.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- We know that tan(30°) = 1/√3.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- So, 1/√3 = AB / 8.
- Solving for AB, we get AB = 8/√3.
- Rationalizing the denominator, AB = 8√3 / 3.
why?
Multiplying the top and bottom of a fraction by the same non-zero number keeps its value. Using the conjugate (a − √b for a + √b) clears the root from the denominator, because (a + √b)(a − √b) = a² − b.
- Now, find the length of the broken part BC using sec(30°) = BC / AC.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- We know that sec(30°) = 2/√3.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- So, 2/√3 = BC / 8.
- Solving for BC, we get BC = 16/√3.
- Rationalizing the denominator, BC = 16√3 / 3.
why?
Multiplying the top and bottom of a fraction by the same non-zero number keeps its value. Using the conjugate (a − √b for a + √b) clears the root from the denominator, because (a + √b)(a − √b) = a² − b.
- The original height of the tree is the sum of AB and BC.
- Original height = AB + BC = 8√3/3 + 16√3/3.
- Original height = 24√3/3 = 8√3 m.
Answer: 8√3 m
Ex 9.1 Q3
A slide for small children has its top 1.5 m high at 30° to the ground; one for older children is 3 m high at 60°. Find the length of each slide.
- For the slide for small children, the height is 1.5 m and the angle with the ground is 30°.
- Let the length of this slide be L₁.
- Using the sine ratio, sin 30° = height / length = 1.5 / L₁.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Substituting the value of sin 30° = 1/2, we get 1/2 = 1.5 / L₁.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Solving for L₁, we get L₁ = 1.5 × 2 = 3 m.
- For the slide for older children, the height is 3 m and the angle with the ground is 60°.
- Let the length of this slide be L₂.
- Using the sine ratio, sin 60° = height / length = 3 / L₂.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Substituting the value of sin 60° = √3/2, we get √3/2 = 3 / L₂.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Solving for L₂, we get L₂ = 3 × 2 / √3 = 6 / √3.
- Rationalizing the denominator, L₂ = (6√3) / 3 = 2√3 m.
why?
Multiplying the top and bottom of a fraction by the same non-zero number keeps its value. Using the conjugate (a − √b for a + √b) clears the root from the denominator, because (a + √b)(a − √b) = a² − b.
Answer: Length of the slide for small children = 3 m; Length of the slide for older children = 2√3 m
Ex 9.1 Q4
From a point 30 m from the foot of a tower, the angle of elevation of its top is 30°. Find the height of the tower.
- Let the height of the tower be h meters and the distance from the foot of the tower to the observation point be 30 m.
- The angle of elevation of the top of the tower is 30°.
- In the right-angled triangle formed by the tower, the ground, and the line of sight, tan 30° = h / 30.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Substitute the value of tan 30° which is 1/√3 into the equation: 1/√3 = h / 30.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Solve for h by multiplying both sides by 30: h = 30 / √3.
why?
Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.
- Rationalize the denominator by multiplying the numerator and denominator by √3: h = (30 × √3) / (√3 × √3).
why?
Multiplying the top and bottom of a fraction by the same non-zero number keeps its value. Using the conjugate (a − √b for a + √b) clears the root from the denominator, because (a + √b)(a − √b) = a² − b.
- Simplify the expression: h = 30√3 / 3.
why?
Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
- Calculate the final value: h = 10√3 m.
Answer: 10√3 m
Ex 9.1 Q5
A kite flies 60 m above the ground and its tight string makes 60° with the ground. Find the length of the string.
- Let the length of the string be x m.
- The height of the kite is 60 m and the angle with the ground is 60°.
- In the right-angled triangle, sin 60° = Perpendicular / Hypotenuse.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Substitute the values: sin 60° = 60 / x.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- We know that sin 60° = √3 / 2.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- So, √3 / 2 = 60 / x.
- Cross-multiply to get x√3 = 120.
why?
If a/b = c/d (b and d not zero), multiplying both sides by bd gives ad = bc — the same 'do the same to both sides' rule.
- Divide both sides by √3 to get x = 120 / √3.
why?
Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.
- Rationalize the denominator by multiplying numerator and denominator by √3.
why?
Multiplying the top and bottom of a fraction by the same non-zero number keeps its value. Using the conjugate (a − √b for a + √b) clears the root from the denominator, because (a + √b)(a − √b) = a² − b.
- x = (120 × √3) / (√3 × √3) = 120√3 / 3.
- Simplify the fraction to get x = 40√3.
why?
Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
Answer: 40√3 m
Ex 9.1 Q6
A 1.5 m tall boy walks towards a 30 m tall building and the angle of elevation of its top from his eyes rises from 30° to 60°. How far did he walk?
- Let the initial distance from the boy to the building be x m.
- The height of the building above the boy's eyes is 30 − 1.5 = 28.5 m.
- In the first position, tan 30° = 28.5 / x, so x = 28.5√3 m.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Let the distance walked be d m, so the new distance is (x − d) m.
- In the second position, tan 60° = 28.5 / (x − d), so x − d = 28.5/√3 m.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Simplifying 28.5/√3 gives 9.5√3 m.
why?
Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
- The distance walked is d = x − (x − d) = 28.5√3 − 9.5√3.
- Calculating the difference gives 19√3 m.
Answer: 19√3 m
Ex 9.1 Q7
From a point on the ground the angles of elevation of the bottom and top of a transmission tower on a 20 m high building are 45° and 60°. Find the height of the tower.
- Let the height of the building be 20 m and the height of the tower be h m.
- Let the horizontal distance from the observation point to the building be x m.
- In the right triangle formed by the observation point, the base of the building, and the bottom of the tower, the angle of elevation is 45°.
- Using tan 45° = opposite / adjacent, we have 20 / x = 1, so x = 20 m.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- In the right triangle formed by the observation point, the base of the building, and the top of the tower, the angle of elevation is 60°.
- The total height from the ground to the top of the tower is (20 + h) m.
- Using tan 60° = opposite / adjacent, we have (20 + h) / 20 = √3.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Solving for h, we get 20 + h = 20√3, so h = 20√3 − 20.
- Factor out 20 to get h = 20(√3 − 1) m.
Answer: 20(√3 - 1) m
Ex 9.1 Q8
A 1.6 m statue stands on a pedestal. From a point on the ground the elevation of the statue top is 60° and of the pedestal top 45°. Find the height of the pedestal.
- Let the height of the pedestal be h m and the distance from the point on the ground to the base of the pedestal be x m.
- The total height from the ground to the top of the statue is h + 1.6 m.
- In the right-angled triangle formed by the point on the ground, the base of the pedestal, and the top of the statue, tan 60° = (h + 1.6) / x.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Since tan 60° = √3, we have √3 = (h + 1.6) / x, which gives x = (h + 1.6) / √3.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- In the right-angled triangle formed by the point on the ground, the base of the pedestal, and the top of the pedestal, tan 45° = h / x.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Since tan 45° = 1, we have 1 = h / x, which gives x = h.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Equating the two expressions for x, we get h = (h + 1.6) / √3.
- Multiplying both sides by √3 gives √3 h = h + 1.6.
why?
Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.
- Rearranging the terms gives √3 h − h = 1.6.
- Factoring out h gives h(√3 − 1) = 1.6.
- Solving for h gives h = 1.6 / (√3 − 1).
- Rationalizing the denominator by multiplying numerator and denominator by (√3 + 1) gives h = 1.6(√3 + 1) / ((√3 − 1)(√3 + 1)).
why?
Multiplying the top and bottom of a fraction by the same non-zero number keeps its value. Using the conjugate (a − √b for a + √b) clears the root from the denominator, because (a + √b)(a − √b) = a² − b.
- The denominator becomes (√3)² − 1² = 3 − 1 = 2.
- So h = 1.6(√3 + 1) / 2 = 0.8(√3 + 1) m.
Answer: 0.8(√3 + 1) m
Ex 9.1 Q9
The elevation of a building top from the foot of a 50 m tower is 30°, and of the tower top from the foot of the building is 60°. Find the height of the building.
- Let the height of the building be h m and the distance between the tower and the building be d m.
- From the foot of the tower, the angle of elevation to the top of the building is 30°, so tan 30° = h/d.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- This gives 1/√3 = h/d, which implies d = h√3.
- From the foot of the building, the angle of elevation to the top of the tower is 60°, so tan 60° = 50/d.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- This gives √3 = 50/d, which implies d = 50/√3.
- Equating the two expressions for d, we get h√3 = 50/√3.
- Multiplying both sides by √3 gives 3h = 50.
why?
Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.
- Solving for h, we get h = 50/3.
Answer: The height of the building is 50/3 m.
Ex 9.1 Q10
Two equal poles stand on either side of an 80 m wide road. From a point between them on the road the elevations of their tops are 60° and 30°. Find the height of the poles and the distances of the point from them.
- Let the height of each pole be h metres.
- Let the distance from the observation point to the first pole be x metres.
- Since the road is 80 m wide, the distance to the second pole is (80 − x) metres.
- In the triangle formed by the first pole, tan 60° = h ÷ x, so h = x√3.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- In the triangle formed by the second pole, tan 30° = h ÷ (80 − x), so h = (80 − x) ÷ √3.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Equating the two expressions for h gives x√3 = (80 − x) ÷ √3.
- Multiplying both sides by √3 yields 3x = 80 − x.
why?
Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.
- Solving for x gives 4x = 80, so x = 20 m.
- Substituting x = 20 into h = x√3 gives h = 20√3 m.
why?
Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
- The distance to the second pole is 80 − 20 = 60 m.
Answer: Height of poles = 20√3 m; distances from the point = 20 m and 60 m
Ex 9.1 Q11
A TV tower stands on one bank of a canal. From the point directly opposite, its top is at 60° elevation; from a point 20 m further back in line, at 30°. Find the height of the tower and the width of the canal.
Uses a figure printed in the NCERT book — open the chapter above to see it.
- Let the height of the tower be h m and the width of the canal be x m.
- From the point directly opposite the tower, the angle of elevation to the top is 60°.
- In the right-angled triangle formed, tan 60° = h/x, so √3 = h/x, which gives h = x√3.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- From the point 20 m further back, the distance from the tower is (x + 20) m and the angle of elevation is 30°.
- In this triangle, tan 30° = h/(x + 20), so 1/√3 = h/(x + 20).
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Substitute h = x√3 into the equation to get 1/√3 = x√3/(x + 20).
why?
Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
- Cross-multiply to get x + 20 = 3x, which simplifies to 2x = 20, so x = 10 m.
why?
If a/b = c/d (b and d not zero), multiplying both sides by bd gives ad = bc — the same 'do the same to both sides' rule.
- Substitute x = 10 into h = x√3 to get h = 10√3 m.
why?
Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
Answer: height 10√3 m; width 10 m
Ex 9.1 Q12
From the top of a 7 m building, the elevation of the top of a cable tower is 60° and the depression of its foot is 45°. Find the height of the tower.
- Let the height of the building be AB = 7 m and the height of the tower be CD.
- Draw a horizontal line from the top of the building A to the tower, meeting CD at point E.
- Since the angle of depression of the foot of the tower is 45°, angle CAE is 45°.
- In right-angled triangle ACE, tan 45° = CE / AE.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Since tan 45° = 1, we have CE / AE = 1, which means CE = AE.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Because ABCE is a rectangle, CE is equal to the height of the building AB, so CE = 7 m.
- Therefore, the horizontal distance AE is also 7 m.
- The angle of elevation of the top of the tower is 60°, so angle DAE is 60°.
- In right-angled triangle ADE, tan 60° = DE / AE.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Substitute the values: √3 = DE / 7.
why?
Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
- Solving for DE gives DE = 7√3 m.
- The total height of the tower CD is the sum of CE and DE.
- CD = 7 + 7√3 = 7(√3 + 1) m.
Answer: 7(√3 + 1) m
Ex 9.1 Q13
From the top of a 75 m lighthouse the angles of depression of two ships, one behind the other on the same side, are 30° and 45°. Find the distance between the ships.
- Let the height of the lighthouse be AB = 75 m, with A at the top and B at the base on the shore.
- Let the two ships be at points C and D on the water, with C being the closer ship and D the farther ship.
- The angle of depression to the closer ship C is 45°, so the angle of elevation from C to A is also 45°.
- In right-angled triangle ABC, tan 45° = AB / BC, which gives 1 = 75 / BC, so BC = 75 m.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- The angle of depression to the farther ship D is 30°, so the angle of elevation from D to A is also 30°.
- In right-angled triangle ABD, tan 30° = AB / BD, which gives 1/√3 = 75 / BD, so BD = 75√3 m.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- The distance between the two ships is CD = BD - BC.
- Substituting the values, CD = 75√3 - 75 = 75(√3 - 1) m.
why?
Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
Answer: 75(√3 - 1) m
Ex 9.1 Q14
A 1.2 m tall girl sees a balloon moving horizontally at a height of 88.2 m; its elevation from her eyes changes from 60° to 30°. Find the distance the balloon travelled.
Uses a figure printed in the NCERT book — open the chapter above to see it.
- Height of balloon above ground = 88.2 m
- Height of balloon above girl's eyes = 88.2 − 1.2 = 87 m
- Let initial horizontal distance be x and final be y
- tan 60° = 87/x, so √3 = 87/x, giving x = 87/√3 = 29√3 m
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- tan 30° = 87/y, so 1/√3 = 87/y, giving y = 87√3 m
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Distance travelled = y − x = 87√3 − 29√3 = 58√3 m
Answer: 58√3 m
Ex 9.1 Q15
From the top of a tower a car approaching on a straight road is seen at a depression of 30°, and 6 seconds later at 60°. At the same speed, how long will it take to reach the foot of the tower?
- Let the height of the tower be h and the distance from the foot of the tower to the car at the first observation be x.
- In the right-angled triangle formed by the first observation, tan 30° = h/x, so h = x/√3.
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- After 6 seconds, the car has moved closer, and the new distance from the foot of the tower is x - d, where d is the distance covered in 6 seconds.
- In the right-angled triangle formed by the second observation, tan 60° = h/(x - d), so h = √3(x - d).
why?
In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
- Equating the two expressions for h gives x/√3 = √3(x - d).
- Simplifying the equation yields x = 3x - 3d, which leads to 2x = 3d, so x = 1.5d.
why?
Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
- The distance from the car's second position to the foot of the tower is x - d = 1.5d - d = 0.5d.
- Since the car covers distance d in 6 seconds, it covers 0.5d in 3 seconds.
- Therefore, the time taken to reach the foot of the tower from the second position is 3 seconds.
Answer: 3 seconds
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