A triangle is to be drawn with one side 8cm and an angle on it 50°.What should be the maximum length of the side opposite this angle?
Answers
Given : A triangle is to be drawn with one side 8cm and an angle on it 50°.
To Find : What should be the maximum length of the side opposite this angle?
( correct Question should be minimum length)
Solution:
One side = 8 cm
let say adjacent side with angle = a
Then side opposite to angle would be
= √a² + 8² - 2a.8cos50°
= √a² + 64 - 16acos50°
√a² + 64 - 16acos50° is max when a² + 64 - 16acos50° is maximum
z = a² + 64 - 16acos50°
dZ/da = 2a - 16cos50°
dZ/da = 0
2a - 16cos50° = 0
=> a = 8cos50°
d²Z/da² = 2 > 0
Hence a = 8cos50° will give minimum not maximum
= √(8cos50°)² + 64 - 16(8cos50°)cos50°
= √(8cos50°)² + 64 - 16(8cos50°)cos50°
= √64 - 64cos²50°
= √64Sin²50°
= 8 Sin50°
We can get minimum length of the side opposite this angle
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Answer:
Given : A triangle is to be drawn with one side 8cm and an angle on it 50°.
To Find : What should be the maximum length of the side opposite this angle?
( correct Question should be minimum length)
Solution:
One side = 8 cm
let say adjacent side with angle = a
Then side opposite to angle would be
= √a² + 8² - 2a.8cos50°
= √a² + 64 - 16acos50°
√a² + 64 - 16acos50° is max when a² + 64 - 16acos50° is maximum
z = a² + 64 - 16acos50°
dZ/da = 2a - 16cos50°
dZ/da = 0
2a - 16cos50° = 0
=> a = 8cos50°
d²Z/da² = 2 > 0
Hence a = 8cos50° will give minimum not maximum
= √(8cos50°)² + 64 - 16(8cos50°)cos50°
= √(8cos50°)² + 64 - 16(8cos50°)cos50°
= √64 - 64cos²50°
= √64Sin²50°
= 8 Sin50°
We can get minimum length of the side opposite this angle