Exercise 1

The known data for a right triangle ABC is a = 5 m and B = {41.7}^{\circ }. Solve the triangle.

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Exercise 2

The known data for a right triangle ABC is b = 3 m and B = { 54.6 }^{\circ }. Solve the triangle.

Exercise 3

The known data for a right triangle ABC is a = 6 m and b = 4 m. Solve the triangle.

Exercise 4

The known data for a right triangle ABC is b = 3 m and c = 5 m. Solve the triangle.

Exercise 5

A tree 50 m tall casts a shadow 60 m long. Find the angle of elevation of the sun at that time.

Exercise 6

An airship is flying at an altitude of 800 m when it spots a village in the distance with a depression angle of {12}^{\circ }. How far is the village from where the plane is flying over?

Exercise 7

Find the radius of a circle knowing that a chord of 24.6 m has a corresponding arc of {70}^{\circ }.

Exercise 8

Calculate the area of a triangular field, knowing that two of its sides measure 80 m and 130 m and between them is an angle of {70}^{\circ }.

Exercise 9

Calculate the height of a tree, knowing that from a point on the ground the top of the tree can be seen at an angle of {30}^{\circ } and from 10 m closer the top can be seen at an angle of {60}^{\circ }.

Exercise 10

The length of the side of a regular octagon is 12 m. Find the radii of the inscribed and circumscribed circles.

Exercise 11

Calculate the length of the side and the apothem of a regular octagon inscribed in a circle with a radius of 49 centimeters.

Exercise 12

Three towns A, B, and C are connected by roads that form a triangle. The distance from A to C is 6 km and from B to C, 9 km. The angle between these roads is {120}^{\circ }. How far are towns A and B from each other?

 

Solution of exercise 1

The known data for a right triangle ABC is a = 5 m and B = {41.7}^{\circ }. Solve the triangle.

 

C = { 90 }^{\circ } - {41.7}^{\circ } = {48.3}^{\circ }

\sin { B } = \frac { b }{ a } \qquad b = a . \sin { B } \qquad b = 5 . \sin { {41.7}^{\circ } } = 3.326 m

\cos { B } = \frac { c }{ a } \qquad c = a . \cos { B } \qquad c = 5 . \cos { {41.7}^{\circ } } = 3.733 m

 

Solution of exercise 2

The known data for a right triangle ABC is b = 3 m and B = { 54.6 }^{\circ }. Solve the triangle.

 

C = { 90 }^{ \circ } - { 54.6 }^{ \circ } = { 35.4 }^{ \circ }

\tan { B } = \frac { c }{ b } \qquad c = \frac { b }{ \tan { B }} \qquad c = \frac { 3 }{ \tan { { 54.6 }^{ \circ } }} = 2.132 m

\sin { B } = \frac { b }{ a } \qquad a = \frac { b }{ \sin { B } } \qquad a = \frac { 3 }{ \sin { { 54.6 }^{ \circ } }} = 3.68 m

 

Solution of exercise 3

The known data for a right triangle ABC is a = 6 m and b = 4 m. Solve the triangle.

\cos { C } = \frac { AC }{ BC }

C = \arccos { (\frac { 4 }{ 6 }) } = { 48.19 }^{ \circ }

B = { 90 }^{ \circ } - { 48.19 }^{ \circ } = { 41.81 }^{ \circ }

\sin { C } = \frac { a }{ c } \qquad c = a . \sin { C } \qquad c = 6 . \sin { { 48.19 }^{ \circ } } = 4.47 m

 

Solution of exercise 4

The known data for a right triangle ABC is b = 3 m and c = 5 m. Solve the triangle.

 

\tan { C } = \frac { AB }{ AC }

C = \arctan { \frac { 5 }{ 3 } } = { 59.04 }^{ \circ }

B = { 90 }^{ \circ } - { 59.04 }^{ \circ } = { 30.96 }^{ \circ }

\sin { C } = \frac { a }{ c } \qquad a = \frac { 5 }{ \sin { { 59.04 }^{ \circ } }} = 5.831 m

 

Solution of exercise 5

A tree 50 m tall casts a shadow 60 m long. Find the angle of elevation of the sun at that time.

 

\tan { \alpha } = \frac { 50 }{ 60 }

\alpha = \arctan { \frac { 5 }{ 6 } }

\alpha = { 39.80 }^{ \circ }

 

Solution of exercise 6

An airship is flying at an altitude of 800 m when it spots a village in the distance with a depression angle of {12}^{\circ }. How far is the village from where the plane is flying over?

 

\tan { { 12 }^{ \circ } } = \frac { 800 }{ d }

d = \frac { 800 }{ \tan { { 12 }^{ \circ } }}

d = 3, 763.7 m

 

Solution of exercise 7

Find the radius of a circle knowing that a chord of 24.6 m has a corresponding arc of {70}^{\circ }.

 

\sin { { 35 }^{ \circ } } = \frac { 12.3 }{ OA }

OA = \frac { 12.3 }{ \sin { { 35 }^{ \circ } } } = 21.44 cm

 

Solution of exercise 8

Calculate the area of a triangular field, knowing that two of its sides measure 80 m and 130 m and between them is an angle of {70}^{\circ }.

 

\sin { { 70 }^{ \circ } } = \frac { h }{ 80 } \qquad h = 80 . \sin { { 70 }^{ \circ } }

A = \frac { 130 . h }{ 2 } = \frac { 130 . 80 . \sin { { 70 }^{ \circ } } }{ 2 } = 4886.40 { m }^{ 2 }

 

Solution of exercise 9

Calculate the height of a tree, knowing that from a point on the ground the top of the tree can be seen at an angle of {30}^{\circ } and from 10 m closer the top can be seen at an angle of {60}^{\circ }.

 

\tan { { 30 }^{ \circ } } = \frac { h }{ 10 + x } \times \frac { \sqrt { 3 } }{ 3 } = \frac { h }{ 10 + x } \rightarrow equation 1

\tan { { 60 }^{ \circ } } = \frac { h }{ x } \qquad \sqrt { 3 } = \frac { h }{ x } \qquad x = \frac { h }{ \sqrt { 3 } } \rightarrow equation 2

Putting the value of x in equation 1:

\frac { \sqrt { 3 } }{ 3 } = \frac { h }{ 10 + x }

\frac { \sqrt { 3 } }{ 3 } = \frac { h }{ 10 + \frac { h }{ \sqrt { 3 } } }

\frac { \sqrt { 3 } }{ 3 } = \frac { h }{ \frac { h + 10 \sqrt { 3 } }{ \sqrt { 3 } } }

\frac { \sqrt { 3 } }{ 3 } \times \frac { h + 10 \sqrt { 3 } }{ \sqrt { 3 } }= h

\frac { h + 10 \sqrt { 3 } }{ 3 } = h

\frac { h }{ 3 } + \frac { 10 \sqrt { 3 } }{ 3 } = h

\frac { 10 \sqrt { 3 } }{ 3 } = h - \frac { h }{ 3 }

\frac { 10 \sqrt { 3 } }{ 3 } = \frac { 2h }{ 3 }

\frac { 10 \sqrt { 3 } }{ 3 } \times \frac { 3 }{ 2 } = h

5 \sqrt { 3 } = h

 

Solution of exercise 10

The length of the side of a regular octagon is 12 m. Find the radii of the inscribed and circumscribed circles.

 

O = \frac { { 54.6 }^{ \circ } }{ 8 } = { 45 }^{ \circ } \qquad \frac { O }{ 2 } = { 22.5 }^{ \circ }

Radius of the inscribed circle.

OC = \frac { AC }{ \tan { { 22.5 }^{ \circ } }} \qquad OC =\frac { 6 }{ 0.4142 } = 14.49

Radius of the circumscribed circle.

OA = \frac { AC }{ \sin { 22.5 } } \qquad OC =\frac { 6 }{ 0.3827 } =15.68

 

Solution of exercise 11

Calculate the length of the side and the apothem of a regular octagon inscribed in a circle with a radius of 49 centimeters.

 

O = \frac { { 360 }^{ \circ } }{ 8 } = { 45 }^{ \circ } \qquad \frac { O }{ 2 } = { 22.5 }^{ \circ }

\frac { I }{ 2 } = 49 . \sin { 22.5 } = 18.75 \qquad I = 37.5 cm

ap = 49 . \cos { 22.5 } = 45.27 cm

 

Solution of exercise 12

Three towns A, B, and C are connected by roads that form a triangle. The distance from A to C is 6 km and from B to C, 9 km. The angle between these roads is {120}^{\circ }. How far are towns A and B from each other?

 

 

CH = 9 \cos { { 60 }^{ \circ } } \qquad BH = 9 \sin { { 60 }^{ \circ } }

AB = \sqrt { { (6 + 9 \cos { { 60 }^{ \circ } }) }^{ 2 } + { (9 \sin { { 60 }^{ \circ } }) }^{ 2 } } = 13.077 km

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Hamza

Hi! I am Hamza and I am from Pakistan. My hobbies are reading, writing and playing chess. Currently, I am a student enrolled in the Chemical Engineering Bachelor program.