A triangle is to be drawn with one side 8cm and an angle on it 80°.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 80°.
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.8cos80°
= √a² + 64 - 16acos80°
√a² + 64 - 16acos80° is max when a² + 64 - 16acos80° is maximum
z = a² + 64 - 16acos80°
dZ/da = 2a - 16cos80°
dZ/da = 0
2a - 16cos80° = 0
=> a = 8cos80°
d²Z/da² = 2 > 0
Hence a = 8cos80° will give minimum not maximum
= √(8cos80°)² + 64 - 16(8cos80°)cos80°
= √(8cos80°)² + 64 - 16(8cos80°)cos80°
= √64 - 64cos²80°
= √64Sin²80°
= 8 Sin80°
We can get minimum length of the side opposite this angle
Learn More:
this problem is about maxima and minima. find the area of the ...
https://brainly.in/question/13355753
examine the maxima and minima of the function f(x)=2x³-21x²+36x ...
https://brainly.in/question/1781825
Answer:
Given : A triangle is to be drawn with one side 8cm and an angle on it 80°.
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.8cos80°
= √a² + 64 - 16acos80°
√a² + 64 - 16acos80° is max when a² + 64 - 16acos80° is maximum
z = a² + 64 - 16acos80°
dZ/da = 2a - 16cos80°
dZ/da = 0
2a - 16cos80° = 0
=> a = 8cos80°
d²Z/da² = 2 > 0
Hence a = 8cos80° will give minimum not maximum
= √(8cos80°)² + 64 - 16(8cos80°)cos80°
= √(8cos80°)² + 64 - 16(8cos80°)cos80°
= √64 - 64cos²80°
= √64Sin²80°
= 8 Sin80°
We can get minimum length of the side opposite this angle