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Land surveying How is the sine law used in land surveying ? Land surveying makes an extensive use of the sine law. The idea is to subdivide the land into many triangles and to measure one side and two angles of each triangle. With the sine law the other two sides can be computed. Mount Everest was found by this method to be the highest mountain on planet earth

The Law of Sines

The law of sines (also known as: sine rule, sine law, or sine formula) is an equation relating the lengths of the sides of any triangle to the ‘sines’ of its angles. Let Δ ABC be any triangle. If A, B, and C are the measures of the angles of Δ ABC and if 'a', 'b', and 'c' are the lengths of the sides opposite the corresponding angles, then

The law of sines can be used to find the remaining sides of a triangle when the measures of two angles and any side are known (AAS or ASA). It can also be used to find the remaining angles of a triangle when the measures of two sides and an angle opposite one of these sides are known (SSA). In some such cases, the formula gives two possible values for the enclosed angle, leading to an ambiguous case. The law of sines is very useful in solving direct and indirect measurement applications.

Sound waves Sound waves A sound wave arises because of a repeating pattern of high pressure and low pressure regions moving through a medium. It is sometimes referred to as a pressure wave. Sound waves are measured using the law of cosine to find the difference in observed and emitted frequencies in relativistic Doppler Effect.
The Law of Cosines

The law of cosines (also known as: cosine rule, cosine law, or cosine formula) is an equation relating the lengths of the sides of any triangle to the cosine of one of its angles. Let Δ ABC be any triangle. If A, B, and C are the measures of the angles of Δ ABC and if a, b, and c are the lengths of the sides opposite the corresponding angles, then:

The law of cosines can be used to find the remaining sides of a triangle when the measures of two sides and their included angle are known (SAS). It can also be used to find the measures of angles of a triangle when the measures of three sides are known (SSS).

Tangent Rule

The Tangent rule is also known as Napier analogy.

In any Δ ABC,

Half-angle Formulae

You very well know that the perimeter of a triangle whose sides are a, b, c is equal to a + b + c. The half-perimeter of a triangle is denoted by "s".
i.e., s = (a + b + c)/2
or 2s = a + b + c

In any Δ ABC, the half-angles can be expressed in terms of 's', 'a', 'b' and 'c' as:

Now can you write similar expressions for:
i) sin (B/2) and sin (C/2)
ii) cos (B/2) and cos (C/2)
iii) tan (B/2) and tan (C/2)

Projection formulae

Projection of AB on BC = BD = c cos B

Projection of AC on BC = CD = b cos C

Now, BC = a = BD + DC = c cos B + b cos C

i.e, a = b cos C + c cos B

Similarly, other formulas follow as:

b = c cos A + a cos C

c = a cos B + b cos A

Applications to heights and distances

Problems based on sine rule

Example: The angle of elevation of the top point P of the vertical tower PQ of height 'h' from a point is 30° and from a point B, the angle of elevation is 45°, where B is a point at a distance x from the point A measured along the line AB which makes an angle 15° with AQ. Find x value in terms of h.
Sol:
It is given that ∠PAQ = 30° and ∠BAQ = 15°.
Therefore, ∠BAP = 15°.
In ΔAQP, we have ∠PAQ = 30°, ∠PQA = 90° and ∠APQ = 60°.
In ΔBRP, we have ∠PBR = 45°, ∠PRB = 90° and ∠BPR = 45°.
Now, ∠APQ = 60° and ∠BPR = 45°.
Therefore ∠ABP = 150°.
Using sine rule in triangle ABP, we get
⇒ x = (√6 – √2)h (∵ h = PQ)
Applications based on cosine rule
Example: Two ships leave a port at the same time. One goes 20 km per hour in the direction N 60° E and the other travels 30 km per hour in the direction S 60° E. Find the distance between the ships at the end of 3 hours.
Sol:
Let P and Q be the positions of the two ships at the end of the 3 hours. Then
OP = 3 × 20 = 60 km and OQ = 3 × 30 = 90 km
Using cosine formulae.
In ΔOPQ, we get
PQ2 = OP2 + OQ2 – 2 . OP . OQ . cos 60°
      = 602 + 902 – 2(60)(90)(1/2)
      = 3600 + 8100 – 5400
      = 6300
PQ = √6300 km = 79.37 km
Hence, the distance between the ships at end of 3 hours is 79.37 km.
Solutions of triangles

The three sides a, b, c and three angles A, B, C are called the elements of the triangle ABC.
When any three of these six elements (except all the three angles), the triangle is known completely i.e, the other three elements can be expressed in terms of the given elements and can be evaluated.
This process is called the solution of a triangle.

Note: When the three angles are given, only the ratios of the lengths of the sides can be determined. The shape of the triangle will thus be known but not its exact size.

I) Solution of right-angled triangle


In right angled triangle one angle is equal to 90°. So, if any two of the remaining five elements are given, atleast one of which is side, then the other elements can be determined by using trigonometric formulae.
If A = 90°, side b and angle B are given, then

and also, tan B = b/c
⇒ c = b cot B
The angle 'C' can be determined by using the angle sum property of a triangle i.e, A + B + C = 180°.

II) Solution of oblique triangles

An oblique triangle has no right angle. It has three acute angles or one obtuse angle and two acute angles.

i) Given two angles and side opposite to one of them:


Suppose c, A, and C are given.
To find 'a', use sine rule,

To find B, use A + B + C = 180°
To find b, use sine rule,

iii) Given two sides and the included angle:


Suppose b, c and A are given.
To find a, use cosine rule,
a2 = b2 + c2 – 2bc cosA
To find B, C use sine rule

To check, use angle sum property of a triangle i.e, A + B + C = 180°

iv) Given the three sides:

When a, b, c are given, solve using the law of cosines for each of the angles.
To find the angles, use

To verify, use A + B + C = 180°

v) Given two sides and the angle opposite to one of them:

click Explore:

Area of a Triangle

The symbol Δ is also used to represent the area of a triangle.

The basic formula for the area of a triangle is: Δ = (1/2) × base × altitude

For a right-angled triangle, the right angle (90°) is contained by the base and altitude. So if a, b, c are its sides and 'c' is the hypotenuse, then:
Area of a right-angled triangle,
(i) Δ = (1/2) ab
Also (ii) Δ = (1/2) bc sin A = (1/2) ca sin B = (1/2) ab sin C
(iii) Δ =
Note that formula (iii) for the area of a triangle is valid for any triangle.
(iv) Δ = 2R2 sin A sin B sin C

Circumcenter and circumradius:
In a triangle, the point of concurrence of the perpendicular bisectors of the sides is called as circumcenter (represented by 'O'). The circumcenter is equidistant from the three vertices.

i.e., OA = OB = OC = R (say)

So if we draw a circle with 'O' as center and R as radius, it passes through the three vertices. The circle is called circumcircle and 'R' is called as circumradius.

The area of a triangle in terms of 'a', 'b', 'c and 'R' is: Δ = .

In-radius Ex-radii
In-radius and Ex-radii

The point of concurrence of internal bisectors of a triangle is called its in-center (rep. by I). We can draw a circle with 'I' as center touching the three sides of a triangle. The circle is said to be inscribed in the triangle. Such a circle is called in-circle and its radius is called in-radius (denoted by r).

A circle that touches one side of a triangle internally and the extensions of the other two sides is called ex-circle. Its radius is termed ex-radius and its center as ex-center. Thus there can be three ex-circles and correspondingly 3 ex-centers (I1, I2, I3) and 3 ex-radii (r1, r2, r3). These are summarized below:

Ex-circle Internally
touching side
Externally
touching sides
Opposite
angle
Ex-center Ex-radii
1 BC AB, AC A I1 r1
2 CA BC, BA B I2 r2
3 AB CA, CB C I3 r3

Then we have:
I) (i) r = ; (ii) r1 = ; (iii) r2 = ; (iv) r3 =

II) (i) r = 4R sin (A/2) sin (B/2) sin (C/2)
     (ii) r1 = 4R sin (A/2) cos (B/2) cos (C/2)
     (iii) r2 = 4R cos (A/2) sin (B/2) cos (C/2)
     (iv) r3 = 4R cos (A/2) cos (B/2) sin (C/2)

Formulae Involving Inradius and Ex-radii

In Δ ABC, I1, I2, I3 are ex-centers and r1, r2, r3 are ex-radii of the ex-circles opposite to the angles A, B, C respectively, then:

  • rr1r2r3 = Δ2, where Δ = area of the triangle ABC
  • r1 + r2 + r3 – r = 4R, where R = circumradius
  • r – r1 + r2 + r3 = 4R cos A
  • r + r1 – r2 + r3 = 4R cos B
  • r + r1 + r2 – r3 = 4R cos C

Few other fascinating formulae:

  • r(r1r2 + r2r3 + r3r1) = r1r2r3 = sΔ, where 's' is semi-perimeter of the triangle ABC
  • r(r1 + r2 + r3) = ab + bc + ca – s2
  • rr1 + rr2 + rr3 + r1r2 + r2r3 + r3r1 = ab + bc + ca
  • r1r2 + r2r3 + r3r1 – (rr1 + rr2 + rr3) = (1/2)(a2 + b2 + c2)

What is 9 point circle?  

Miscellaneous

m-n cot theorem

Let D be a point on the side BC of a triangle ABC such that BD : DC = m : n.
Let ∠ADC = θ, ∠BAD = α, ∠CAD = β, then
(i) (m + n)cot θ = m cot α – n cot β
(ii) (m + n)cot θ = n cot B – m cot C

Example: A tower leans towards west making an angle θ1 with the vertical. The angular elevation of B, the top most point of the tower, is θ2 as observed from a point C due east of A at a distance d from A. If the angular elevation of B from a point due east of C at a distance 2d from C is θ3, then 2 tan θ1 =
Sol:
By using m-n cot theorem of trigonometry, we have
(d + 2d) cot θ2 = d cot θ3 – 2d cot ( + θ1)
3 cot θ2 = cot θ3 + 2 tan θ1
∴ 2 tan θ1 = 3 cot θ2 – cot θ3
Pedal triangle



Let ABC be any triangle and AD, BE, CF be perpendiculars from A, B, C to the opposite sides of the triangle.

The triangle DEF is called pedal triangle of triangle ABC. Let T be the orthocenter of triangle ABC.

i. TB = 2R cos B, TA = 2R cos A, TC = 2R cos C.

ii. TD = 2R cos B cos C, TE = 2R cos C cos A, TF = 2R cos A cos B

iii. Sides of pedal triangle; DE = c cos C, DF = b cos B, EF = a cos A.

iv. ∠EDF = 180° – 2A, ∠DEF = 180° – 2B, ∠DFE = 180° – 2C.

v. Area of pedal triangle = 2Δ cos A cos B cos C.

vi. Radius of circumcircle of pedal triangle = R/2.

vii. Radius of in-circle of pedal triangle DEF = 2R cos A cos B cos C.

T-ratios applied to a regular polygon

A polygon is said to be regular polygon if all its sides are equal in measure (length) and the measure of all the angle are same.

Sum of all angles of a regular polygon = (2n – 4) radian, where n = number of sides.

Measure of each angle = radian
where,
r = radius of in-circle of polygon = (a/2) cot π/n
R = radius of circum circle of polygon = a/2 cosec π/n

Area of n-sided regular polygon is given as

In a regular polygon, the centroid, circumcenter and in-center coincide.

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