SOLUTION OF RIGHT TRIANGLES - Questions & Answers
1. Multiple Choice Type : Choose the correct answer from the options given below.
(a) The given figure, shows an isosceles triangle with angle ABC = 90°, the measure of angle BAC is :
(i) 30° (ii) 45° (iii) 60° (iv) 40°
Step 1: In the given right-angled triangle ABC, the angle at B is 90°.
Step 2: Since it is an isosceles triangle, the perpendicular and base are equal in length, meaning AB = BC.
Step 3: Angles opposite to equal sides of a triangle are also equal, so angle BAC = angle BCA.
Step 4: The sum of angles in a triangle is 180°, so angle BAC + angle BCA + 90° = 180°.
Step 5: 2 × angle BAC = 90°, which gives angle BAC = 45°.
Answer: (ii) 45°
(b) In the given triangle, ∠ACB = 90° and angle CAB = 60°, the value of cot ∠ABC is :
(i) 2 (ii) 1 (iii) √3 (iv) 1/√3
Step 1: In triangle ABC, the sum of all internal angles is 180°.
Step 2: We are given ∠ACB = 90° and ∠CAB = 60°.
Step 3: Therefore, ∠ABC = 180° - (90° + 60°) = 30°.
Step 4: We need to find the value of cot ∠ABC, which is cot 30°.
Step 5: From standard trigonometric values, cot 30° = √3.
Answer: (iii) √3
(c) In the given triangle, the length of AB is :
(i) 4√2 m (ii) 4 m (iii) 8 m (iv) 6 m
Step 1: From the figure, triangle ABC is a right-angled triangle at B, so ∠ABC = 90°.
Step 2: We are given angle A = 45° and the side opposite to angle A (BC) = 8 m.
Step 3: We need to find the base AB. We can use the tangent function.
Step 4: tan(45°) = Perpendicular / Base = BC / AB.
Step 5: 1 = 8 / AB.
Step 6: Therefore, AB = 8 m.
Answer: (iii) 8 m
(d) From the information given in the triangle shown below, the length of CD is : (take √3 = 1.732)
(i) 7.32 m (ii) 27.32 m (iii) 10 m (iv) 17.32 m
Step 1: The figure shows two right-angled triangles, ABD and ABC, sharing the same vertical side AB = 10 m.
Step 2: In the smaller triangle ABD, ∠ADB = 45°.
Step 3: tan(45°) = AB / DB, which means 1 = 10 / DB, so DB = 10 m.
Step 4: In the larger triangle ABC, ∠ACB = 30°.
Step 5: tan(30°) = AB / CB, which means 1/√3 = 10 / CB, so CB = 10√3 m.
Step 6: We are given √3 = 1.732, so CB = 10 × 1.732 = 17.32 m.
Step 7: The length of CD = CB - DB.
Step 8: CD = 17.32 m - 10 m = 7.32 m.
Answer: (i) 7.32 m
(e) In the given triangle, the length of AB is :
(i) 160 cm (ii) 120 cm (iii) 60 cm (iv) 40 cm
Step 1: The figure shows a right-angled triangle ABD with the right angle at B.
Step 2: The hypotenuse AD is given as 80 cm.
Step 3: The angle opposite to the side AB is ∠ADB = 30°.
Step 4: We can use the sine ratio: sin(30°) = Opposite / Hypotenuse = AB / AD.
Step 5: 1/2 = AB / 80.
Step 6: Multiplying both sides by 80 gives AB = 80 / 2 = 40 cm.
Answer: (iv) 40 cm
2. Find 'x', if :
(i) [Triangle with angle 60°, hypotenuse 20, adjacent side x]
Step 1: In the given right-angled triangle, we know the hypotenuse = 20 and the angle = 60°.
Step 2: The side marked 'x' is adjacent to the 60° angle.
Step 3: Using the cosine ratio: cos(60°) = Base / Hypotenuse.
Step 4: 1/2 = x / 20.
Step 5: x = 20 / 2 = 10.
Answer: x = 10
(ii) [Triangle with angle 30°, hypotenuse 20, opposite side x]
Step 1: In this right-angled triangle, the hypotenuse = 20 and the angle = 30°.
Step 2: The side marked 'x' is opposite to the 30° angle.
Step 3: Using the sine ratio: sin(30°) = Opposite / Hypotenuse.
Step 4: 1/2 = x / 20.
Step 5: x = 20 / 2 = 10.
Answer: x = 10
(iii) [Triangle with angle 45°, adjacent side 20, opposite side x]
Step 1: In this right-angled triangle, the base (adjacent side) = 20 and the angle = 45°.
Step 2: The side marked 'x' is opposite to the 45° angle.
Step 3: Using the tangent ratio: tan(45°) = Opposite / Base.
Step 4: 1 = x / 20.
Step 5: x = 20 × 1 = 20.
Answer: x = 20
3. Find angle 'A' if :
(i) [Triangle with hypotenuse 20, opposite side 10]
Step 1: We are given the Opposite side = 10 and Hypotenuse = 20 with respect to angle A.
Step 2: sin(A) = Opposite / Hypotenuse.
Step 3: sin(A) = 10 / 20 = 1/2.
Step 4: The angle whose sine is 1/2 is 30°.
Answer: A = 30°
(ii) [Triangle with hypotenuse 10, opposite side 10/√2]
Step 1: We are given the Opposite side = 10/√2 and Hypotenuse = 10 with respect to angle A.
Step 2: sin(A) = Opposite / Hypotenuse.
Step 3: sin(A) = (10/√2) / 10 = 1/√2.
Step 4: The angle whose sine is 1/√2 is 45°.
Answer: A = 45°
(iii) [Triangle with adjacent side 10√3, opposite side 10]
Step 1: We are given the Opposite side = 10 and Adjacent side (Base) = 10√3 with respect to angle A.
Step 2: tan(A) = Opposite / Adjacent.
Step 3: tan(A) = 10 / 10√3 = 1/√3.
Step 4: The angle whose tangent is 1/√3 is 30°.
Answer: A = 30°
4. Find angle 'x' if : [Figure with two connected right triangles]
Step 1: The figure shows two right-angled triangles sharing a common vertical side (altitude).
Step 2: Let's find the height using the right-side triangle, which has a hypotenuse of 20 and a top angle of 60°.
Step 3: Using the cosine ratio (since height is adjacent to the top angle): cos(60°) = Height / 20.
Step 4: 1/2 = Height / 20, which gives Height = 10.
Step 5: Now, look at the left-side triangle. Its base is given as 30 and its height is 10.
Step 6: tan(x) = Perpendicular / Base = Height / Base.
Step 7: tan(x) = 10 / 30 = 1/3.
Step 8: Angle x is tan⁻¹(1/3), which using tables is approximately 18° 26'.
Answer: x = 18° 26'
5. Find AD, if :
(i) [Figure with vertical line segment AC, point B on AC, etc.]
Step 1: Based on the figure, A-B-C forms a vertical line, and C-D is a horizontal line, making ∠ACD = 90°.
Step 2: A horizontal line is drawn from B to E on AD, meaning ∠ABE = 90°.
Step 3: In the right-angled triangle ABE, ∠AEB is given as 45°, so ∠BAE is also 45°, meaning AB = BE.
Step 4: Because BC and BE are parallel to the vertical and horizontal axes respectively, BE equals the distance CD = 50 m.
Step 5: Thus, AB = 50 m. We are given BC = 10 m.
Step 6: The total vertical length AC = AB + BC = 50 + 10 = 60 m.
Step 7: In the large right-angled triangle ACD, we can apply Pythagoras theorem: AD² = AC² + CD².
Step 8: AD² = 60² + 50² = 3600 + 2500 = 6100.
Step 9: AD = √6100 = 10√61 m.
Answer: AD = 10√61 m
(ii) [Figure with two triangles, AB = 100 m]
Step 1: Triangle ABC is a right-angled triangle at C. We are given hypotenuse AB = 100 m.
Step 2: Let the vertical side AC be 'h' and the horizontal side BC be 'y'.
Step 3: The figure has tick marks indicating BC = CD. So, CD = 'y' as well.
Step 4: In triangle ACD, tan(60°) = AC / CD = h / y.
Step 5: √3 = h / y, which means h = y√3.
Step 6: In triangle ABC, apply Pythagoras theorem: AB² = AC² + BC².
Step 7: 100² = h² + y² = (y√3)² + y² = 3y² + y² = 4y².
Step 8: 10000 = 4y², which means y² = 2500, so y = 50 m.
Step 9: Now in triangle ACD, we need to find the hypotenuse AD. cos(60°) = CD / AD.
Step 10: 1/2 = 50 / AD.
Step 11: AD = 50 × 2 = 100 m.
Answer: AD = 100 m
6. Find the length of AD. Given : ∠ABC = 60°, ∠DBC = 45° and BC = 40 cm.
Step 1: The figure shows two right-angled triangles, ABC and DBC, sharing the same horizontal base BC = 40 cm.
Step 2: In the smaller triangle DBC, tan(45°) = DC / BC.
Step 3: 1 = DC / 40, which means DC = 40 cm.
Step 4: In the larger triangle ABC, tan(60°) = AC / BC.
Step 5: √3 = AC / 40, which means AC = 40√3 cm.
Step 6: From the figure, the length of AD is the difference between the full vertical line AC and the segment DC.
Step 7: AD = AC - DC = 40√3 - 40.
Step 8: Factoring out 40, AD = 40(√3 - 1) cm.
Answer: AD = 40(√3 - 1) cm
7. Find lengths of diagonals AC and BD. Given AB = 60 cm and ∠BAD = 60°.
Step 1: The figure is a rhombus ABCD, meaning all sides are equal (AB = BC = CD = DA = 60 cm).
Step 2: The diagonals of a rhombus bisect each other at right angles (90°) and bisect the vertex angles.
Step 3: Let the diagonals intersect at point O. In right-angled triangle AOB, ∠AOB = 90°.
Step 4: Angle OAB is half of ∠BAD, so ∠OAB = 60° / 2 = 30°.
Step 5: Using sine ratio: sin(30°) = OB / AB => 1/2 = OB / 60 => OB = 30 cm.
Step 6: The total diagonal BD is twice OB: BD = 2 × 30 = 60 cm.
Step 7: Using cosine ratio: cos(30°) = OA / AB => √3/2 = OA / 60 => OA = 30√3 cm.
Step 8: The total diagonal AC is twice OA: AC = 2 × 30√3 = 60√3 cm.
Answer: AC = 60√3 cm, BD = 60 cm
8. Find AB.
Step 1: The figure shows a polygon bounded by horizontal line segment A-C-D-B and vertical segments FC = 20 and ED = 30.
Step 2: In right-angled triangle ACF, tan(45°) = FC / AC => 1 = 20 / AC => AC = 20.
Step 3: In right-angled triangle BDE, tan(60°) = ED / DB => √3 = 30 / DB => DB = 30 / √3 = 10√3.
Step 4: Draw a horizontal line from F to a point G on ED. FG is parallel to CD, so FG = CD.
Step 5: The segment EG = ED - GD. Since GD = FC = 20, EG = 30 - 20 = 10.
Step 6: In right-angled triangle FGE, the angle with the horizontal is given as 60°.
Step 7: tan(60°) = EG / FG => √3 = 10 / FG => FG = 10 / √3 = 10√3 / 3.
Step 8: Since CD = FG, CD = 10√3 / 3.
Step 9: The total length AB = AC + CD + DB = 20 + 10√3/3 + 10√3 = 20 + 40√3/3.
Answer: AB = 20 + 40√3/3 cm
9. In trapezium ABCD, as shown, AB // DC, AD = DC = BC = 20 cm and ∠A = 60°. Find : (i) length of AB (ii) distance between AB and DC.
Step 1: Drop perpendiculars from D and C to points E and F on AB. DE is the distance between parallel lines.
Step 2: In the right-angled triangle ADE, hypotenuse AD = 20 cm and ∠A = 60°.
Step 3: (ii) Distance DE = AD × sin(60°) = 20 × (√3/2) = 10√3 cm.
Step 4: Base of this triangle, AE = AD × cos(60°) = 20 × (1/2) = 10 cm.
Step 5: Since AD = BC, the trapezium is isosceles. Triangle BCF is congruent to triangle ADE, so FB = 10 cm.
Step 6: The central shape CDEF is a rectangle, so EF = DC = 20 cm.
Step 7: (i) Total length AB = AE + EF + FB = 10 + 20 + 10 = 40 cm.
Answer: (i) AB = 40 cm, (ii) Distance = 10√3 cm
10. Use the information given to find the length of AB.
Step 1: The figure shows two right-angled triangles resting on a common horizontal line segment A-P-B.
Step 2: In the left triangle QAP, the vertical side QA = 10 cm and ∠QPA = 30°.
Step 3: tan(30°) = QA / AP => 1/√3 = 10 / AP => AP = 10√3 cm.
Step 4: In the right triangle RBP, the vertical side RB = 8 cm and ∠RPB = 45°.
Step 5: tan(45°) = RB / PB => 1 = 8 / PB => PB = 8 cm.
Step 6: The total length AB is the sum of AP and PB.
Step 7: AB = 10√3 + 8 cm.
Answer: AB = 10√3 + 8 cm
TEST YOURSELF
1. Multiple Choice Type : Choose the correct answer from the options given below.
(a) The value of angle A is :
(i) 45° (ii) 30° (iii) 60° (iv) none of these
Step 1: The figure displays a right-angled triangle ABC with a right angle at B.
Step 2: Tick marks on sides AB and BC indicate that they are equal in length.
Step 3: In an isosceles right-angled triangle, the two acute angles are equal.
Step 4: (180° - 90°) / 2 = 45°.
Answer: (i) 45°
(b) The length of AB is :
(i) 20 cm (ii) 10(√3 - 1)cm (iii) 10√3 cm (iv) 10(√3 + 1)cm
Step 1: The figure shows a triangle divided into two right-angled triangles by a vertical altitude of 10 cm.
Step 2: In the left right-angled triangle, the angle is 45°. Base = Height / tan(45°) = 10 / 1 = 10 cm.
Step 3: In the right right-angled triangle, the angle is 30°. Base = Height / tan(30°) = 10 / (1/√3) = 10√3 cm.
Step 4: The total length of AB is the sum of these two bases.
Step 5: AB = 10 + 10√3 = 10(1 + √3) cm.
Answer: (iv) 10(√3 + 1)cm
(c) The length of AB is :
(i) (10 - √3)cm (ii) 10(√3 - 1)cm (iii) 10√3 cm (iv) 10(√3 + 1)cm
Step 1: The figure shows two right-angled triangles sharing a vertical height of 10 cm at the right.
Step 2: Let the points be C (top), D (bottom right, 90°), B (middle), and A (bottom left).
Step 3: In the smaller triangle CBD, tan(45°) = CD / BD => 1 = 10 / BD => BD = 10 cm.
Step 4: In the larger triangle CAD, tan(30°) = CD / AD => 1/√3 = 10 / AD => AD = 10√3 cm.
Step 5: The length of segment AB is the difference between AD and BD.
Step 6: AB = AD - BD = 10√3 - 10 = 10(√3 - 1) cm.
Answer: (ii) 10(√3 - 1)cm
(d) Length of AB is equal to :
(i) 20√3 cm (ii) 20/√3 cm (iii) 20 cm (iv) none of these
Step 1: The figure shows a right-angled triangle ADB at B.
Step 2: A point C is on the base DB such that DC = CB, as indicated by the tick marks.
Step 3: The label "20 cm" spans the entire length of DB, so DB = 20 cm.
Step 4: The angle ∠ADB = 30°.
Step 5: tan(30°) = AB / DB => 1/√3 = AB / 20.
Step 6: Solving for AB gives AB = 20 / √3 cm.
Answer: (ii) 20/√3 cm
(e) Statement (1) : At a particular time, the length of the shadow of a 50 m tower is 50√3 m.
Statement (2) : The sun's altitude is 30°
(i) Both the statements are true. (ii) Both the statements are false. (iii) Statement 1 is true, and statement 2 is false. (iv) Statement 1 is false, and statement 2 is true.
Step 1: The tangent of the sun's altitude angle equals the tower's height divided by its shadow's length.
Step 2: tan(θ) = 50 / (50√3) = 1 / √3.
Step 3: The angle whose tangent is 1/√3 is 30°.
Step 4: Therefore, if the shadow is 50√3 m, the altitude is indeed 30°, making both statements consistent and true.
Answer: (i) Both the statements are true.
(f) Statement (1) : The distance between B and C = 5 m.
Statement (2) : BC = 3√3 m
(i) Both the statements are true. (ii) Both the statements are false. (iii) Statement 1 is true, and statement 2 is false. (iv) Statement 1 is false, and statement 2 is true.
Step 1: The figure shows two vertical poles AB = 3 m and DC = 2 m on a horizontal line B-P-C.
Step 2: Angles of elevation from P are 60° to A and 30° to D.
Step 3: In triangle ABP, tan(60°) = AB / BP => √3 = 3 / BP => BP = 3 / √3 = √3 m.
Step 4: In triangle DCP, tan(30°) = DC / PC => 1/√3 = 2 / PC => PC = 2√3 m.
Step 5: The total distance BC = BP + PC = √3 + 2√3 = 3√3 m.
Step 6: Statement 1 says BC = 5 m, which is false. Statement 2 says BC = 3√3 m, which is true.
Answer: (iv) Statement 1 is false, and statement 2 is true.
(g) Assertion (A) : In rhombus ABCD, angle ABC = 120° and length of its each side is 20 cm. The length of diagonal BD = 20 cm.
Reason (R) : In ΔAOB, cos 60° = OB / 20 cm and BD = 2 × OB
(i) A is true, R is false. (ii) A is false, R is true. (iii) Both A and R are true and R is the correct reason for A. (iv) Both A and R are true and R is the incorrect reason for A.
Step 1: In a rhombus, the diagonals bisect the vertex angles and intersect at 90° (at point O).
Step 2: If ∠ABC = 120°, then the diagonal BD bisects it, making ∠ABO = 60°.
Step 3: In the right-angled triangle AOB, side AB (hypotenuse) is 20 cm.
Step 4: cos(∠ABO) = Adjacent / Hypotenuse => cos(60°) = OB / 20.
Step 5: This means OB = 20 × (1/2) = 10 cm.
Step 6: The full diagonal BD = 2 × OB = 2 × 10 = 20 cm. The reason exactly demonstrates this calculation.
Answer: (iii) Both A and R are true and R is the correct reason for A.
(h) Assertion (A) : The length of the line AB is 100 m.
Reason (R) : tan 45° = 50 / PB => PB = 50 m and AB = 50 m + 50 m = 100 m
(i) A is true, R is false. (ii) A is false, R is true. (iii) Both A and R are true and R is the correct reason for A. (iv) Both A and R are true and R is the incorrect reason for A.
Step 1: In the given figure, triangle CBP is a right-angled triangle at B.
Step 2: We are given vertical side CB = 50 m and ∠CPB = 45°.
Step 3: tan(45°) = CB / PB => 1 = 50 / PB => PB = 50 m. This matches the first part of Reason R.
Step 4: The figure contains tick marks indicating that segment AP is equal in length to segment CB.
Step 5: Thus, AP = 50 m.
Step 6: The total length AB = AP + PB = 50 m + 50 m = 100 m.
Step 7: The reasoning provided leads directly and correctly to the assertion.
Answer: (iii) Both A and R are true and R is the correct reason for A.
2. Find the value of cot x.
Step 1: The given figure is a right-angled triangle with an angle x.
Step 2: The side opposite to angle x is √3, and the hypotenuse is 2.
Step 3: Let the adjacent side (base) be 'b'. By Pythagoras theorem, 2² = (√3)² + b².
Step 4: 4 = 3 + b², which gives b² = 1, so b = 1.
Step 5: cot(x) = Adjacent / Opposite = 1 / √3.
Answer: cot x = 1/√3
3. Find the length of AB.
Step 1: The figure shows a vertical line AB, with a perpendicular horizontal line from point D meeting AB at E.
Step 2: A vertical line from D goes down to C, meeting the horizontal line BC from B. So, DEBC forms a rectangle.
Step 3: Therefore, DE = CB = 30 cm, and ∠AED = 90°, ∠DEB = 90°.
Step 4: In the top right-angled triangle ADE, ∠ADE = 45°. tan(45°) = AE / DE => 1 = AE / 30 => AE = 30 cm.
Step 5: In the right-angled triangle BDE, ∠BDE = 60°. tan(60°) = EB / DE => √3 = EB / 30 => EB = 30√3 cm.
Step 6: The total length AB = AE + EB = 30 + 30√3 = 30(1 + √3) cm.
Answer: AB = 30(1 + √3) cm
4. In the given figure, AB and EC are parallel to each other. Sides AD and BC are 2 cm each and are perpendicular to AB. Given that ∠AED = 60° and ∠ACD = 45°; calculate : (i) AB (ii) AC (iii) AE
Step 1: Since AB and EC are parallel, and AD and BC are perpendicular to AB, ABCD forms a rectangle.
Step 2: Therefore, DC = AB, and AD = BC = 2 cm. Both triangles ADE and ADC are right-angled at D.
Step 3: (i) In triangle ADC, ∠ACD = 45°. tan(45°) = AD / DC => 1 = 2 / DC => DC = 2 cm. Since AB = DC, AB = 2 cm.
Step 4: (ii) In triangle ADC, apply Pythagoras theorem: AC = √(AD² + DC²) = √(2² + 2²) = √8 = 2√2 cm.
Step 5: (iii) In triangle ADE, ∠AED = 60°. sin(60°) = AD / AE => √3/2 = 2 / AE => AE = 4/√3 cm.
Answer: (i) AB = 2 cm, (ii) AC = 2√2 cm, (iii) AE = 4/√3 cm
5. In the given figure, ∠B = 60°, AB = 16 cm and BC = 23 cm. Calculate : (i) BE (ii) AC.
Step 1: The figure shows a triangle ABC with a perpendicular AE dropped to the base BC, creating a right-angled triangle ABE.
Step 2: (i) In triangle ABE, cos(60°) = BE / AB => 1/2 = BE / 16 => BE = 8 cm.
Step 3: To find AC, we first need the altitude AE. sin(60°) = AE / 16 => √3/2 = AE / 16 => AE = 8√3 cm.
Step 4: The remainder of the base is EC = BC - BE = 23 - 8 = 15 cm.
Step 5: (ii) In the right-angled triangle AEC, apply Pythagoras theorem: AC² = AE² + EC².
Step 6: AC² = (8√3)² + 15² = 192 + 225 = 417.
Step 7: AC = √417 cm.
Answer: (i) BE = 8 cm, (ii) AC = √417 cm
6. Find : (i) BC (ii) AD (iii) AC
Step 1: The figure depicts a large right-angled triangle ABC with right angle at B. We are given AB = 12 cm and ∠C = 30°.
Step 2: (i) In triangle ABC, tan(30°) = AB / BC => 1/√3 = 12 / BC => BC = 12√3 cm.
Step 3: (iii) In triangle ABC, sin(30°) = AB / AC => 1/2 = 12 / AC => AC = 24 cm.
Step 4: A perpendicular BD is drawn to AC, creating right-angled triangle ADB. The angle ∠A = 180° - 90° - 30° = 60°.
Step 5: (ii) In triangle ADB, cos(60°) = AD / AB => 1/2 = AD / 12 => AD = 6 cm.
Answer: (i) BC = 12√3 cm, (ii) AD = 6 cm, (iii) AC = 24 cm
7. In right-angled triangle ABC; ∠B = 90°. Find the magnitude of angle A, if :
(i) AB is √3 times of BC.
Step 1: We are given AB = √3 × BC.
Step 2: tan(A) = Opposite / Adjacent = BC / AB.
Step 3: Substitute AB: tan(A) = BC / (√3 × BC) = 1/√3.
Step 4: Angle A = 30°.
Answer: A = 30°
(ii) BC is √3 times of AB.
Step 1: We are given BC = √3 × AB.
Step 2: tan(A) = Opposite / Adjacent = BC / AB.
Step 3: Substitute BC: tan(A) = (√3 × AB) / AB = √3.
Step 4: Angle A = 60°.
Answer: A = 60°
8. A ladder is placed against a vertical tower. If the ladder makes an angle of 30° with the ground and reaches upto a height of 15 m of the tower; find the length of the ladder. (HOTS)
Step 1: This situation forms a right-angled triangle where the tower is the perpendicular (15 m) and the ladder is the hypotenuse (L).
Step 2: The angle between the ladder and the ground is 30°.
Step 3: sin(30°) = Perpendicular / Hypotenuse = 15 / L.
Step 4: 1/2 = 15 / L.
Step 5: L = 15 × 2 = 30 m.
Answer: Length of ladder = 30 m
9. A kite is attached to a 100 m long string. Find the greatest height reached by the kite when its string makes an angle of 60° with the level ground. (LS)
Step 1: This situation forms a right-angled triangle where the string is the hypotenuse (100 m).
Step 2: The height of the kite is the perpendicular side, and the angle of elevation is 60°.
Step 3: sin(60°) = Height / Hypotenuse = Height / 100.
Step 4: √3/2 = Height / 100.
Step 5: Height = 100 × (√3 / 2) = 50√3 m.
Answer: Greatest height = 50√3 m
10. Find AB and BC, if :
(i) [Figure with two connected right triangles, right angle at B, angles 30° and 45°]
Step 1: Let AB = h and BC = x. In the smaller right triangle ABC, tan(45°) = h / x => 1 = h / x => h = x.
Step 2: In the larger right triangle ABD, the base is (x + 20) and the angle is 30°.
Step 3: tan(30°) = h / (x + 20) => 1/√3 = x / (x + 20).
Step 4: x√3 = x + 20 => x(√3 - 1) = 20.
Step 5: x = 20 / (√3 - 1). Rationalizing the denominator: x = 20(√3 + 1) / (3 - 1) = 10(√3 + 1).
Step 6: Thus, BC = 10(√3 + 1) cm, and since AB = h = x, AB = 10(√3 + 1) cm.
Answer: AB = 10(√3 + 1) cm, BC = 10(√3 + 1) cm
(ii) [Figure with altitude AC, triangles on opposite sides]
Step 1: Let the vertical altitude be AC. The figure forms two right-angled triangles ACD and ACB, right-angled at C.
Step 2: In left triangle ACD, tan(30°) = AC / 20 => 1/√3 = AC / 20 => AC = 20/√3 cm.
Step 3: In right triangle ACB, we need to find BC and AB.
Step 4: tan(60°) = AC / BC => √3 = (20/√3) / BC => BC = 20 / (√3 × √3) = 20/3 cm.
Step 5: sin(60°) = AC / AB => √3/2 = (20/√3) / AB => AB = 40 / 3 cm.
Answer: AB = 40/3 cm, BC = 20/3 cm
(iii) [Figure with right angle at B, angles 60° and 45°]
Step 1: Let AB = h and BC = x. In the smaller triangle ABC, tan(60°) = h / x => √3 = h / x => h = x√3.
Step 2: In the larger triangle ABD, the base is (x + 20) and angle is 45°.
Step 3: tan(45°) = h / (x + 20) => 1 = h / (x + 20) => h = x + 20.
Step 4: Equating both expressions for h: x√3 = x + 20 => x(√3 - 1) = 20 => x = 10(√3 + 1) cm.
Step 5: AB = h = x + 20 = 10√3 + 10 + 20 = 10√3 + 30 = 10(3 + √3) cm.
Answer: AB = 10(3 + √3) cm, BC = 10(√3 + 1) cm
11. Find PQ, if AB = 150 m, ∠P = 30° and ∠Q = 45°
(i) [Points P and Q are on opposite sides of base B]
Step 1: The vertical height AB = 150 m. Triangles ABP and ABQ are right-angled at B.
Step 2: In triangle ABQ, tan(45°) = AB / BQ => 1 = 150 / BQ => BQ = 150 m.
Step 3: In triangle ABP, tan(30°) = AB / PB => 1/√3 = 150 / PB => PB = 150√3 m.
Step 4: Since P and Q are on opposite sides, the total distance PQ = PB + BQ.
Step 5: PQ = 150√3 + 150 = 150(√3 + 1) m.
Answer: PQ = 150(√3 + 1) m
(ii) [Points P and Q are on the same side of base B]
Step 1: The vertical height AB = 150 m. Triangles ABP and ABQ are right-angled at B.
Step 2: In triangle ABQ, tan(45°) = AB / BQ => 1 = 150 / BQ => BQ = 150 m.
Step 3: In triangle ABP, tan(30°) = AB / PB => 1/√3 = 150 / PB => PB = 150√3 m.
Step 4: Since P and Q are on the same side, the distance PQ is the difference between PB and BQ.
Step 5: PQ = PB - BQ = 150√3 - 150 = 150(√3 - 1) m.
Answer: PQ = 150(√3 - 1) m
12. If tan x° = 5/12, tan y° = 3/4 and AB = 48 m; find the length of CD.
Step 1: Let the vertical length CD be 'h' and the horizontal segment BC be 'z'. The triangles ACD and BCD are right-angled at C.
Step 2: From triangle BCD, tan y° = CD / BC => 3/4 = h / z => z = 4h / 3.
Step 3: From triangle ACD, tan x° = CD / AC = CD / (AB + BC).
Step 4: 5/12 = h / (48 + z) => 5/12 = h / (48 + 4h/3).
Step 5: Cross-multiplying: 12h = 5 × (48 + 4h/3) => 12h = 240 + 20h/3.
Step 6: Multiply the entire equation by 3 to clear the fraction: 36h = 720 + 20h.
Step 7: 16h = 720, which gives h = 45 m.
Answer: Length of CD = 45 m
13. The perimeter of a rhombus is 96 cm and obtuse angle of it is 120°. Find the lengths of its diagonals.
Step 1: Since all 4 sides of a rhombus are equal, the length of one side is 96 / 4 = 24 cm.
Step 2: Adjacent angles in a rhombus sum to 180°, so the acute angle is 180° - 120° = 60°.
Step 3: The diagonals bisect the vertex angles and cross at 90°. Let them cross at O. This forms a right-angled triangle AOB.
Step 4: The hypotenuse AB = 24 cm, and the angle OAB is half of 60°, which is 30°.
Step 5: sin(30°) = OB / 24 => 1/2 = OB / 24 => OB = 12 cm.
Step 6: The short diagonal is 2 × OB = 24 cm.
Step 7: cos(30°) = OA / 24 => √3/2 = OA / 24 => OA = 12√3 cm.
Step 8: The long diagonal is 2 × OA = 24√3 cm.
Answer: Lengths of diagonals are 24 cm and 24√3 cm
Case-Study Based Question
A school authority constructed a slide for its children below the age of 12 years. (LS)
The constructed slide has a height of 4 m above the ground and is inclined at an angle of 30° to the ground. Use the given information to answer each of the following :
(i) the length of slide AB is :
(a) 8 m (b) 6 m (c) 5 m (d) 10 m
Step 1: The slide setup forms a right-angled triangle. Height (Opposite) = 4 m. Angle = 30°.
Step 2: The length of the slide AB represents the hypotenuse.
Step 3: sin(30°) = Height / AB => 1/2 = 4 / AB.
Step 4: AB = 4 × 2 = 8 m.
Answer: (a) 8 m
(ii) the value of sin² 30° + cos² 60° is :
(a) 1/4 (b) 1/2 (c) 3/4 (d) 3/2
Step 1: Write down the standard trigonometric values.
Step 2: sin(30°) = 1/2 and cos(60°) = 1/2.
Step 3: Square the values: (1/2)² + (1/2)².
Step 4: Evaluate the sum: 1/4 + 1/4 = 2/4 = 1/2.
Answer: (b) 1/2
(iii) if cos A = 1/2, then the value of 12 cot² A - 2 is :
(a) 5 (b) 4 (c) 3 (d) 2
Step 1: We are given cos A = 1/2. From standard trigonometric tables, angle A = 60°.
Step 2: The value of cot(60°) = 1/√3.
Step 3: Square this value: cot²(60°) = (1/√3)² = 1/3.
Step 4: Substitute into the expression: 12(1/3) - 2.
Step 5: Calculate the result: 4 - 2 = 2.
Answer: (d) 2
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Quick Review Flashcards - Click to flip and test your knowledge!
Question
What does it mean to 'solve' a right-angled triangle?
Answer
It means finding the values of all remaining angles and remaining sides.
Question
Name the two specific scenarios under which a right-angled triangle can be solved according to the source material.
Answer
When one side and one acute angle are given, or when two sides are given.
Question
In general, the ratio of the unknown side to the known side in a right-angled triangle equals the _____.
Answer
Corresponding trigonometrical ratio of the given angle.
Question
Ratio: $\frac{\text{Base}}{\text{Perpendicular}}$
Answer
$\cot \theta$
Question
Ratio: $\frac{\text{Perpendicular}}{\text{Hypotenuse}}$
Answer
$\sin \theta$
Question
Ratio: $\frac{\text{Base}}{\text{Hypotenuse}}$
Answer
$\cos \theta$
Question
Ratio: $\frac{\text{Perpendicular}}{\text{Base}}$
Answer
$\tan \theta$
Question
What is the numerical value of $\cot 30^\circ$?
Answer
$\sqrt{3}$
Question
What is the approximate decimal value of $\sqrt{3}$ used in the calculations?
Answer
$1.732$
Question
In Example 2, the angle $x^\circ$ is found using the ratio $\frac{\text{Perp.}}{\text{Hypt.}} = \frac{5}{10}$. What is the value of $x$?
Answer
$30^\circ$
Question
In Example 2, the angle $y^\circ$ is found using the ratio $\frac{\text{Perp.}}{\text{Base}} = \frac{10}{10} = 1$. What is the value of $y$?
Answer
$45^\circ$
Question
In a right-angled triangle $ABD$, if $\tan 45^\circ = \frac{AD}{BD}$ and $AD = 10\text{ cm}$, what is the length of $BD$?
Answer
$10\text{ cm}$
Question
If $\sin \theta = \frac{1}{2}$, what is the value of $\theta$ in degrees?
Answer
$30^\circ$
Question
If $\tan \theta = \frac{1}{\sqrt{3}}$, what is the value of $\theta$ in degrees?
Answer
$30^\circ$
Question
How do the diagonals of a rhombus intersect each other?
Answer
They bisect each other at right angles ($90^\circ$).
Question
Besides bisecting each other, what do the diagonals of a rhombus do to the vertex angles?
Answer
They bisect the angle of the vertex.
Question
If a rhombus $ABCD$ has a vertex angle $\angle A = 60^\circ$, what is the measure of the angle $\angle OAB$ (where $O$ is the intersection of diagonals)?
Answer
$30^\circ$
Question
Formula: Length of diagonal $AC$ in a rhombus with intersection $O$
Answer
$2 \times OA$
Question
Formula: Length of diagonal $BD$ in a rhombus with intersection $O$
Answer
$2 \times OB$
Question
In the rocket problem, if the rocket rises $40\text{ km}$ at $60^\circ$ to the vertical, the angle made with the horizontal is _____.
Answer
$30^\circ$
Question
How is the total height of the rocket at point $B$ calculated if it first rises vertically ($PA$) and then at an angle ($AB$)?
Answer
By adding the vertical height of the first stage ($PA$) to the vertical component of the second stage ($BD$).
Question
In a trapezium $ABCD$ where $AB$ is parallel to $DC$ and $BC$ is a transversal, what is the sum of $\angle B$ and $\angle C$?
Answer
$180^\circ$
Question
How is the length of side $AB$ determined in a trapezium when a perpendicular $CE$ is drawn to $AB$?
Answer
It is the sum of the segment $BE$ and the segment $AE$ (where $AE = CD$).
Question
What is the general formula for the area of a trapezium used in Example 8?
Answer
$\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{perpendicular distance between them}$
Question
In an isosceles right-angled triangle, what is the measure of each acute angle?
Answer
$45^\circ$
Question
If $\tan \theta = 1$, the triangle must be a/an _____ right-angled triangle.
Answer
Isosceles
Question
In Exercise 2(i), given a hypotenuse of $20$ and an angle of $60^\circ$, what ratio is used to find the base $x$?
Answer
$\cos 60^\circ = \frac{x}{20}$
Question
In Exercise 3(ii), given a hypotenuse of $10$ and a perpendicular of $\frac{10}{\sqrt{2}}$, what is the value of $\sin A$?
Answer
$\frac{1}{\sqrt{2}}$
Question
If $\sin A = \frac{1}{\sqrt{2}}$, what is the measure of angle $A$?
Answer
$45^\circ$
Question
In a right-angled triangle, if one acute angle is $60^\circ$, the other acute angle must be _____.
Answer
$30^\circ$
Question
What is the value of $\sin 60^\circ$?
Answer
$\frac{\sqrt{3}}{2}$
Question
What is the value of $\cos 60^\circ$?
Answer
$\frac{1}{2}$
Question
In the context of the slide problem, which part of the triangle represents the length of the slide?
Answer
The hypotenuse ($AB$).
Question
If the height of a slide is $4\text{ m}$ and it is inclined at $30^\circ$ to the ground, what is the length of the slide?
Answer
$8\text{ m}$
Question
Calculate the value of $\sin^2 30^\circ + \cos^2 60^\circ$.
Answer
$\frac{1}{2}$
Question
If $\cos A = \frac{1}{2}$, what is the value of $\cot A$?
Answer
$\frac{1}{\sqrt{3}}$
Question
In a right-angled triangle, if the side adjacent to angle $A$ is $\sqrt{3}$ times the side opposite to it, what is the measure of angle $A$?
Answer
$30^\circ$
Question
When finding the sun's altitude based on a tower's shadow, the shadow length represents the _____ of the right-angled triangle.
Answer
Base
Question
If a $50\text{ m}$ tower casts a shadow of $50\sqrt{3}\text{ m}$, what is the sun's altitude?
Answer
$30^\circ$
Question
In a ladder problem, if a ladder reaches $15\text{ m}$ up a tower and makes a $30^\circ$ angle with the ground, what ratio finds the ladder's length?
Answer
$\sin 30^\circ = \frac{15}{\text{length of ladder}}$
Question
What is the value of $\tan 60^\circ$?
Answer
$\sqrt{3}$
Question
In Example 7, why is $\angle BCD = 90^\circ$?
Answer
Because it is given that $DC \perp BC$.
Question
In a right-angled triangle, the sum of the two acute angles is always _____.
Answer
$90^\circ$
Question
In the trapezium problem, the distance between parallel sides $AD$ and $BC$ is represented by which segment?
Answer
$CE$ (the perpendicular height).
Question
If the perimeter of a rhombus is $96\text{ cm}$, what is the length of each side?
Answer
$24\text{ cm}$
Question
In a rhombus with a side of $10\text{ cm}$ and a vertex angle of $60^\circ$, the shorter diagonal has a length of _____.
Answer
$10\text{ cm}$
Question
If $\tan x^\circ = \frac{5}{12}$, what is the value of $\cot x^o$?
Answer
$\frac{12}{5}$
Question
In the kite problem, the string length represents which part of the right-angled triangle?
Answer
The hypotenuse.
Question
How is the height of a kite calculated if the string length is $100\text{ m}$ and the angle with the ground is $60^\circ$?
Answer
$100 \times \sin 60^\circ$
Question
In Figure 10(i), if $CD = 20\text{ cm}$, how is the base $BC$ expressed in terms of $BD$?
Answer
$BC = BD - 20$