1. In a triangle ABC, if a = 2, b = 3 and sinA = 2/3, find ∠B.
2. If the angles of a triangle are in the ratio 1 : 2 : 3, prove that the corresponding sides are in the ratio 1 : √3 : 2.
3. The angles of a triangle ABC are in arithmetic sequence If b/c = √3/√2, find ∠A.
4. If in a triangle ABC, ∠A = 45°, ∠B = 60° and ∠C = 75°, find the ratio of its sides.
5. In ΔABC, ∠B = 45°, ∠C = 105° and a = 2, find b.
1. Answer :
Using Law of Sines,
a/sinA = b/sinB
Substitute.
2/(2/3) = 3/sinB
3 = 3/sinB
3sinB = 3
sinB = 1
∠B = 90°
2. Answer :
From the ratio 1 : 2 : 3, the angles of a triangle are assumed to be x, 2x and 3x.
x + 2x + 3x = 180°
6x = 180°
x = 30°
So, the angles are 30°, 60° and 90°.
Let a, b and c be the corresponding side lengths of the angles 30°, 60° and 90° respectively.
Using Law of Sines,
a/sin30° = b/sin60° = c/sin90°
Let a/sin30° = b/sin60° = c/sin90° = k.
a/sin30° = k
a = ksin30°
a = k/2
b/sin60° = k
b = ksin60°
b = k√3/2
c/sin90° = k
c = ksin90°
c = k
a : b : c = k/2 : k√3/2 : k
= k : k√3 : 2k
= 1 : √3 : 2
3. Answer :
It is given that the angles ∠A, ∠B and ∠C are in arithmetic sequence.
∠B - ∠A = ∠C - ∠B
2∠B = ∠A + ∠C ----(1)
Add ∠B to both sides.
3∠B = ∠A + ∠B + ∠C
3∠B = 180°
∠B = 60°
Using Law of Sines,
In (1), substitute ∠B = 60° and ∠C = 45° and solve for ∠A.
2∠B = ∠A + ∠C
2(60°) = ∠A + 45°
120° = ∠A + 45°
75° = ∠A
4. Answer :
Law of Sines :
a/sinA = b/sinB = c/sinC
a/sin45° = b/sin60° = c/sin75°
Let a/sin45° = b/sin60° = c/sin75° = k
a/sin45° = k
a = ksin45°
a = k√2/2
b/sin60° = k
b = ksin60°
b = k√3/2
c/sin90° = k
c = ksin75°
Find the value of sin75°.
sin75° = sin(45° + 30°)
= sin(45° + 30°)
= sin45°cos30° + cos45°sin30°
5. Answer :
In ΔABC,
∠A + ∠B + ∠C = 180°
∠A + 45° + 105° = 180°
∠A + 150° = 180°
∠A = 30°
By Law of Sines :
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