Solution of Right Triangles
ICSE Class 9 · Mathematics · Chapter 22
What this chapter covers
- Why some angles are called standard angles
- Trigonometrical ratios of 45°
- Trigonometrical ratios of 30° and 60°
- Trigonometrical ratios of 0° and 90°
- The table of standard values
- Evaluating expressions involving standard-angle ratios
- Finding an angle from a known ratio
- Trigonometrical ratios of complementary angles
- Direct application of the complementary-angle relations
- Solving a right-angled triangle
- Simple 2-D problems involving one right-angled triangle
- Common errors and how to avoid them
Key points
- The standard angles are 0°, 30°, 45°, 60° and 90°; their ratios are found exactly from geometry.
- A square cut by its diagonal gives sin 45° = cos 45° = (1)/(√(2)) and tan 45° = 1.
- Half an equilateral triangle has sides in the ratio 1 : √(3) : 2, giving sin 30° = cos 60° = (1)/(2) and sin 60° = cos 30° = (√(3))/(2).
- sin 0° = 0, cos 0° = 1, sin 90° = 1, cos 90° = 0; tan 90°, sec 90°, cot 0° and cosec 0° are not defined.
- The sine row is (√(0))/(2), (√(1))/(2), (√(2))/(2), (√(3))/(2), (√(4))/(2); the cosine row is the same list reversed.
- To evaluate an expression, substitute exact values, simplify, and rationalise; sin2 A means (sin A)2.
- An acute angle can be read back from a standard value: tan θ = √(3) gives θ = 60°.
- For complementary angles: sin (90° - A) = cos A, cos (90° - A) = sin A, tan (90° - A) = cot A, cot (90° - A) = tan A, sec (90° - A) = cosec A, cosec (90° - A) = sec A.
- If a ratio of one acute angle equals the co-ratio of another acute angle, the two angles add up to 90°.
- Solving a right triangle needs two parts, at least one a side; choose the ratio that links what is known to what is wanted.
- In a word problem the hypotenuse is the slanting side facing the right angle: the ladder, the plank, the diagonal, the string.
Questions answered
Define complementary angles and give one pair that are standard angles.
Complementary angles are two angles whose sum is 90°; 30° and 60° form such a pair.
Using a square of side a, explain why tan 45° = 1.
The diagonal of a square makes 45° with each side, and in the right-angled triangle so formed the opposite and adjacent sides are both a, so tan 45° = (a)/(a) = 1.
Why is tan 90° not defined?
As the angle approaches 90° the adjacent side shrinks to 0 while the opposite side approaches the hypotenuse, so tan 90° would be (1)/(0), and division by zero has no meaning.
Evaluate 2 sin2 45° + 3 cos2 30° - tan2 60°.
2 × (1)/(2) + 3 × (3)/(4) - 3 = 1 + (9)/(4) - 3 = (1)/(4).
Evaluate (sin 30° + cos 60°)/(sec 0° + cosec 90°).
((1)/(2) + (1)/(2))/(1 + 1) = (1)/(2).
Find the acute angle θ if √(3) tan θ = 1.
tan θ = (1)/(√(3)), so θ = 30°.
Find x if 2 cos 3x = 1, given that 3x is acute.
cos 3x = (1)/(2), so 3x = 60° and x = 20°.
Evaluate (cos 47°)/(sin 43°) + (cot 25°)/(tan 65°).
cos 47° = sin 43° and cot 25° = tan 65°, so the value is 1 + 1 = 2.
Past board exam questions
ICSE 2019 sample paper · 3 marks
Without using tables, evaluate: 4tan 60° sec 30° + (sin 31° sec 59° + cot 59° cot 31°)/(8sin2 30° - tan2 45°)
Show answer
4tan 60° sec 30° = 4 × √(3) × (2)/(√(3)) = 8
sin 31° sec 59° = (sin 31°)/(cos(90°-31°)) = (sin 31°)/(sin 31°) = 1; cot 59° cot 31° = tan 31° cot 31° = 1
Denominator = 8 × (1)/(4) - 1 = 1
Value = 8 + (1+1)/(1) = 10
ICSE 2019 sample paper · 3 marks
Solve for x; where 0° ≤ x ≤ 90° sin2 x + cos2 30° = (5)/(4)
Show answer
cos2 30° = ((√(3))/(2))2 = (3)/(4)
sin2 x = (5)/(4) - (3)/(4) = (1)/(2) ⇒ sin x = (1)/(√(2))
x = 45°
ICSE 2019 sample paper · 4 marks
In △ABC, CD is perpendicular to AB, with D on AB. AC = 10 m, CB = 5√(2) m, ∠ACD = 60°, CD = x and ∠CBA = y°. Evaluate x and y.
Show answer
In right △ADC: cos 60° = (CD)/(AC) ⇒ x = 10 × (1)/(2) = 5 m.
In right △BDC: sin y° = (CD)/(CB) = (5)/(5√(2)) = (1)/(√(2)) ⇒ y = 45.
x = 5 m, y = 45°.