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The Navigation Problem Trigonometry Mr. Brennan

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A plane flies for 2.25 hours (from an airport) at a speed of 240 km/hr Then on a course of 300 degrees the plane flies for 3.5 hours at a speed of 300 km/hr. At this time, how far is the plane from the airport? The Navigation Problem on a course of 203 degrees. A plane flies for t 1 hours (from an airport) at a speed of s 1 km/hr Then on a course of c 2 degrees the plane flies for t 2 hours at a speed of s 2 km/hr. At this time, how far is the plane from the airport? General Case on a course of c 1 degrees.

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A plane flies for t 1 hours (from an airport) at a speed of s 1 km/hr A c1° Then on a course of c 2 degrees the plane flies for t 2 hours at a speed of s 2 km/hr. At this time, how far is the plane from the airport? The Navigation Problem c2° on a course of c 1 degrees.

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A plane flies for t 1 hours (from an airport) at a speed of s 1 km/hr A c1° The Navigation Problem on a course of c 1 degrees. N Note: Remember that the navigation problems do not measure angles in standard position, but are oriented from North. Whenever there angle measures you can draw in a North-South line to help you orient the angle. Remember to draw a picture.

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A plane flies for t 1 hours (from an airport) at a speed of s 1 km/hr on a course of c 1 degrees. A d1 = s1 * t1 c1° At this time the plane has traveled a distance of d1 = s1 * t1 km This distance becomes the first side of a triangle N

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A plane flies for t 1 hours (from an airport) at a speed of s 1 km/hr A c1° Then on a course of c 2 degrees the plane flies for t 2 hours at a speed of s 2 km/hr. The Navigation Problem c2° on a course of c 1 degrees. d1 The distance covered by the second direction d2 = s2 * t2 km d2 becomes the second side of the triangle NN

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A plane flies for t 1 hours (from an airport) at a speed of s 1 km/hr A c1° Then on a course of c 2 degrees the plane flies for t 2 hours at a speed of s 2 km/hr. At this time, how far is the plane from the airport? The Navigation Problem c2° on a course of c 1 degrees. d1 d2 d3 We want to find the distance d3, which is the third side of the triangle. B We can start by finding the measure of angle B, then with sides d1 and d2 we an use the law of cosines to find side d3.

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A c1° The Navigation Problem c2° d1 d2 d3 We want to find the distance d3, which is the third side of the triangle. B We can start by finding the measure of angle B, then with sides d1 and d2 we an use the law of cosines to find side d3. We need to find the measure of angle B. We can do this by finding the sum of the two smaller angles that make up angle B. First – look at the shaded region from course C1. The measure of the shaded region is C1 - 180°. (C1 - 180°) N N Because the lines going North are parallel, the right half of angle B also has a measure of (C1 - 180°.) [Alternate Interior Angles are congruent. ] As shown, the measure of the left side of angle B is (360 – c2)° (360 –c 2 °) The measure of angle B = (360 – c2) + (c1 – 180) = 180 + c1 – c2

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A c1° The Navigation Problem c2°c2° d1d1 d2d2 d3d3 We want to find the distance d3, which is the third side of the triangle. B We can start by finding the measure of angle B, then with sides d1 and d2 we an use the law of cosines to find side d3. (C1 - 180°) N N The measure of angle B = (360 – c2) + (c1 – 180) = 180 + c1 – c2 The law of cosines says (d 3 ) 2 = (d 1 ) 2 + (d 2 ) 2 – 2(d 1 )(d 2 ) cos (B) So d3 = (d 1 ) 2 + (d 2 ) 2 – 2(d 1 )(d 2 ) cos (B)

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A plane flies for 2.25 hours (from an airport) at a speed of 240 km/hr Then on a course of 300 degrees the plane flies for 3.5 hours at a speed of 300 km/hr. At this time, how far is the plane from the airport? The Navigation Problem on a course of 203 degrees.

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A c1° The Navigation Problem c2° d1d1 d 1 = speed * time 240 km/hr * 2.25 hr = 540 km d 2 = 1050 km A plane flies for 2.25 hours (from an airport) at a speed of 240 km/hr Then on a course of 300 degrees the plane flies for 3.5 hours at a speed of 300 km/hr. At this time, how far is the plane from the airport? on a course of 203 degrees. The distance covered by the second direction d 2 = speed * time d 2 = s 2 * t 2 d 2 = 300 km/hr * 3.5 hr = 1050 km = 540 km

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A c1° The Navigation Problem c2° d1d1 From course 1 the measure of the right half of angle B = (203 – 180) = 23° d 2 = 1050 km A plane flies for 2.25 hours (from an airport) at a speed of 240 km/hr Then on a course of 300 degrees the plane flies for 3.5 hours at a speed of 300 km/hr. At this time, how far is the plane from the airport? on a course of 203 degrees. From course 2 the measure of the left half of angle B = (360 – 300) = 60° = 540 km B 23° 60° Calculate the measure of angle B. The measure of angle B = 23° + 60° = 83°

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A c1° The Navigation Problem c2° d1d1 d 2 = 1050 km A plane flies for 2.25 hours (from an airport) at a speed of 240 km/hr Then on a course of 300 degrees the plane flies for 3.5 hours at a speed of 300 km/hr. At this time, how far is the plane from the airport? on a course of 203 degrees. = 540 km B 23° 60° = 83° The law of cosines says (d 3 ) 2 = (d 1 ) 2 + (d 2 ) 2 – 2(d 1 )(d 2 ) cos (B) So d3 = (d 1 ) 2 + (d 2 ) 2 – 2(d 1 )(d 2 ) cos (B) d3d3 D 3 = 540 2 + 1050 2 – 2(540)(1050) cos (83) D 3 =1120.67 km

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