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We will discuss here about the law of sines or the sine rule which is required for solving the problems on triangle.
In any triangle the sides of a triangle are proportional to the sines of the angles opposite to them.
That is in any triangle ABC,
asinA = bsinB = csinC
Proof:
Let ABC be a triangle.
Now will derive the three different cases:
Case I: Acute angled triangle (three angles are acute): The triangle ABC is acute-angled.
Now, draw AD from A which is perpendicular to BC. Clearly, D lies on BC
Now from the triangle ABD, we have,
sin B = AD/AB
⇒ sin B = AD/c, [Since, AB = c]
⇒ AD= c sin B ……………………………………. (1)
Again from the triangle ACD we have,
sin C = AD/AC
⇒ sin C = AD/b, [Since, AC = b]
⇒ AD = b sin C ...………………………………….. (2)
Now, from (1) and (2) we get,
c sin B = b sin C
⇒ b/sin B = c/sin c………………………………….(3)
Similarly, if we draw a perpendicular to AC from B, we will get
a/sin A = c/sin c………………………………….(4)
Therefore, from (3) and (4) we get,
asinA = bsinB = csinC
Case II: Obtuse angled triangle (one angle is obtuse): The triangle ABC is obtuse angled.
Now, draw AD from A which is perpendicular to produced BC. Clearly, D lies on produced BC.
Now from the triangle ABD, we have,
sin ∠ABD = AD/AB
⇒ sin (180 - B) = AD/c, [Since ∠ABD = 180 - B and AB = c]
⇒ sin B = AD/c, [Since sin (180 - θ) = sin θ]
⇒ AD = c sin B ……………………………………. (5)
Again, from the triangle ACD, we have,
sin C = AD/AC
⇒ sin C = AD/b, [Since, AC = b]
⇒ AD = b sin C ……………………………………. (6)
Now, from (5) and (6) we get,
c sin B = b sin C
b/sin B = c/sin C ……………………………………. (7)
Similarly, if we draw a perpendicular to AC from B, we will get
a/sin A = b/sin B ……………………………………. (8)
Therefore, from (7) and (8) we get,
asinA = bsinB = csinC
Case III: Right angled triangle (one angle is right angle): The triangle ABC is right angled. The angle C is a right angle.
Now from triangle ABC, we have,
sin C = sin π/2
⇒ sin C = 1, [Since, sin π/2 = 1], ……………………………………. (9)
sin A = BC/AB
⇒ sin A = a/c, [Since, BC = a and AB = c]
⇒ c = a/sin A ……………………………………. (10)
and sin B = AC/AB
⇒ sin B = b/c, [Since, AC = b and AB = c]
⇒ c = b/sin B ……………………………………. (11)
Now from (10) and (11) we get,
a/sin A = b/sin B = c
⇒ a/sin A = b/sin B = c/1
Now from (9) we get,
⇒ asinA = bsinB = csinC
Therefore, from all three cases, we get,
asinA = bsinB = csinC. Proved.
Note:
1. The sine rule or the law of sines can be expressed as
sinAa = sinBb = sinCc
2. The sine rule or the law of sines is a very useful rule to
express sides of a triangle in terms of the sines of angles and vice-versa in
the following manner.
We have asinA = bsinB = csinC = k1 (say)
⇒ a = k1 sin A, b = k1 sin B and c = k1 sin C
Similarly, sin A/a = sin B/b = sin C/c = k2 (say)
⇒ sin A = k2 a, sin B = k2 b and sin C = k2 c
Solved problem using the law of sines:
The triangle ABC is isosceles; if ∠A = 108°, find the value of a : b.
Solution:
Since the triangle ABC is isosceles and A = 108°, A + B + C = 180°, hence it is evident that B = C.
Now, B + C = 180° - A = 180° - 108°
⇒ 2B = 72° [Since, C = B]
⇒ B = 36°
Again, we have, asinA = bsinB
Therefore, ab = sinAsinB = sin108°sin36° = cos18°sin36°
Now, cos 18° = √1−sin218°
= √1−(√5−14)2
= ¼√10+2√5
and sin 36° = √1−cos236°
= √1−(√5+14)2
= ¼√10−2√5
Therefore, a/b = 14√10+2√514√10−2√5
= √10+2√5√10−2√5
= √(10+2√5)2102−(2√5)2
= 10+2√5√80
⇒ ab = 2√5(√5+1)4√5
⇒ ab = √5+12
Therefore, a : b = (√5 + 1) : 2
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