Introduction for trigonometry math help:
Greek Mathematician Ptolemy, Father of trigonometry proved the equation sin2A+cos2A=1 using geometry involving a relationship between the chords of a circle. A degree is subdivided into 60 equal parts. Each part is called a minute and it is denoted by 1'. Again a minute is subdivided into 60 equal parts. Each part is called a second and it is denoted by 1?. The trigonometry math help example problems and practice problems are given below.
Example problems for trigonometry math help:
Example problem 1:
Prove that tan3A - tan2A - tanA = tanA tan2A tan3A
Solution:
tan3A = tan(A + 2A) = (tanA + tan2A) /1 - tanA tan2A
tan 3A (1 - tanA tan2A) = tanA + tan2A
tan3A - tanA tan2A tan3A = tanA + tan2A
tan3A - tan2A - tanA = tanA tan2A tan3A
Example problem 2:
An aeroplane at an height of 2500 metres observes the angles of depression of opposite points on the two banks of a river to be 41° 20' and 52°10'. Find the width of the river.
Solution:
Let A represented by position of the aeroplane at an instant and D be the point vertically below A on the ground. Then DA = 2500 m. Let B and C be the two opposite points on the two banks of a river such that B, C and D all line in a straight line. Then ?DBA = 41°20'. ?DCA = 52°10'. Let CD = x. Let BC = y be the width of the river.
In a right angled triangle CDA
Cot 52°10' = CD/AD= x/2500m
x = 2500 cot 52°10' m ----------(1)
In a right angled triangle BDA
cot 41 20' = BD/AD =( x + y)/2500m
x + y = 2500 cot 41°20' m --------(2)
(2) – (1) x+y–x = 2500 cot 41°20'–2500 cot 52°10'm
y = 2500 (cot 41°20'–cot 52°10')m
= 2500 (tan 48° 40' – tan 37°50')m
= 2500 (1.1370 – 0.7766) = 2500 x 0.3604 m
y = 901 m
Practice problems for trigonometry math help:
Practice problem 1:
Two men standing on the 20m apart observed the top of a tree in between them at angles of elevation of 30° and 45°. Find the height of the tree.
Answer: 7.32 m
Practice problem 2:
From the top of a hill 240m altitude the angles of the depression of the top and the bottom of a tower are 30° and 45°. Find the height of the tower.
Answer: 101.44 m
These example problems and practice problems are very helpful to study the trigonometry.
Greek Mathematician Ptolemy, Father of trigonometry proved the equation sin2A+cos2A=1 using geometry involving a relationship between the chords of a circle. A degree is subdivided into 60 equal parts. Each part is called a minute and it is denoted by 1'. Again a minute is subdivided into 60 equal parts. Each part is called a second and it is denoted by 1?. The trigonometry math help example problems and practice problems are given below.
Example problems for trigonometry math help:
Example problem 1:
Prove that tan3A - tan2A - tanA = tanA tan2A tan3A
Solution:
tan3A = tan(A + 2A) = (tanA + tan2A) /1 - tanA tan2A
tan 3A (1 - tanA tan2A) = tanA + tan2A
tan3A - tanA tan2A tan3A = tanA + tan2A
tan3A - tan2A - tanA = tanA tan2A tan3A
Example problem 2:
An aeroplane at an height of 2500 metres observes the angles of depression of opposite points on the two banks of a river to be 41° 20' and 52°10'. Find the width of the river.
Solution:
Let A represented by position of the aeroplane at an instant and D be the point vertically below A on the ground. Then DA = 2500 m. Let B and C be the two opposite points on the two banks of a river such that B, C and D all line in a straight line. Then ?DBA = 41°20'. ?DCA = 52°10'. Let CD = x. Let BC = y be the width of the river.
In a right angled triangle CDA
Cot 52°10' = CD/AD= x/2500m
x = 2500 cot 52°10' m ----------(1)
In a right angled triangle BDA
cot 41 20' = BD/AD =( x + y)/2500m
x + y = 2500 cot 41°20' m --------(2)
(2) – (1) x+y–x = 2500 cot 41°20'–2500 cot 52°10'm
y = 2500 (cot 41°20'–cot 52°10')m
= 2500 (tan 48° 40' – tan 37°50')m
= 2500 (1.1370 – 0.7766) = 2500 x 0.3604 m
y = 901 m
Practice problems for trigonometry math help:
Practice problem 1:
Two men standing on the 20m apart observed the top of a tree in between them at angles of elevation of 30° and 45°. Find the height of the tree.
Answer: 7.32 m
Practice problem 2:
From the top of a hill 240m altitude the angles of the depression of the top and the bottom of a tower are 30° and 45°. Find the height of the tower.
Answer: 101.44 m
These example problems and practice problems are very helpful to study the trigonometry.
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