Exercise 1

Determine the height, h, of the drawing below:

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Exercise 2

Calculate the distance from point A to point B.

Exercise 3

Calculate the distance between points A and B.

 

Exercise 4

Calculate the radius of the circle circumscribed in a triangle, where A = 45°, B = 72°, and a = 20m.

Exercise 5

The radius of a circle measures 25 m. Calculate the angle between the tangents to the circle, drawn at the ends of a chord with a length of 36 m.

Exercise 6

The diagonals of parallelogram ABCD measure 10 cm (AC) and 12 cm (BD), and the angle that they form in the center is 48° 15'. Calculate the length of the sides AD and AB.

 

Solution of exercise 1

Determine the height, h, of the drawing below:

ABC = 180 - (72\circ 18' + 60\circ 32') = 47\circ 10'

\frac { c }{ \sin { (60 \circ 32') }} = \frac { 500 }{ \sin { 47 \circ 10'} } \qquad c = 593.62 m

h = 593.62 \times \sin { 62 \circ 5' } = 524.54 m

 

Solution of exercise 2

Calculate the distance from point A to point B.

B = 180 \circ - (61 \circ 28' + 54 \circ 53 ') = 63 \circ 39'

\frac { c }{ \sin { 54 \circ 53' } } = \frac { 200 }{ \sin { 63 \circ 39' }}

c = 182.565 m

 

Solution of exercise 3

Calculate the distance between points A and B.

\frac { AC }{ \sin { 43 \circ 52' }} = \frac { 450 }{ \sin { 67 \circ 57 ' }} \qquad AC = 336.45 m

\frac { CB }{ \sin { 80 \circ 40' }} = \frac { 450 }{ \sin { 66 \circ 44' } } \qquad CB = 483.35m

{ x }^{ 2 } = { 336.45 }^{ 2 } + { 483.35 }^{ 2 } - 2 \times 336.45 \times 483.35 \times \cos { (68 \circ 11' - 32 \circ 36' ) }

x = 286.902 m

 

Solution of exercise 4

Calculate the radius of the circle circumscribed in a triangle, where A = 45°, B = 72°, and a = 20m.

\frac { a }{ \sin A } = 2R

R =\frac { 20 }{ 2 \sin { 45 }} = 14.14m

Solution of exercise 5

The radius of a circle measures 25 m. Calculate the angle between the tangents to the circle, drawn at the ends of a chord with a length of 36 m.

{ 36 }^{ 2 } = { 25 }^{ 2 } + { 25 }^{ 2 } - 2 \times 25 \times 25 \cos { O } \qquad O = 92 \circ 6' 32''

In the quadrilateral, AOBT, angles A and B are right angles.

O + T = 180 \circ

T = 180 \circ - 92 \circ 6' 32'' = 87 \circ 53' 28''

 

Solution of exercise 6

The diagonals of parallelogram ABCD measure 10 cm (AC) and 12 cm (BD), and the angle that they form in the center is 48° 15'. Calculate the length of the sides AD and AB.

AD = \sqrt { { 5 }^{ 2 } + { 6 }^{ 2 } - 2 . 5 . 6 \cos { (48 \circ 15') } } = 4.5877 cm

180 \circ - 48 \circ 15' = 131 \circ 45'

AB = \sqrt { { 5 }^{ 2 } + { 6 }^{ 2 } - 2 . 5 . 6 \cos { (131 \circ 45' ) } } = 10.047 cm

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Emma

I am passionate about travelling and currently live and work in Paris. I like to spend my time reading, gardening, running, learning languages and exploring new places.