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Q: We know how to solve right triangles using trig, but how can we use trig to solve any triangle?
A: The Law of Sines (7.1/7.2) and The Law of Cosines (7.3).
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Q: We know how to solve right triangles using trig, but how can we use trig to solve any triangle?
A: The Law of Sines (7.1/7.2) and The Law of Cosines (7.3).
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Ex 2.The bearing of a lighthouse from a ship was found to be . After the ship sailed 5.8 km due south, the new bearing was . Find the distance between the ship and the lighthouse at each location.
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To solve a triangle, we’ve needed to know some of its sides and angles. We can classify the possibilities like this (A = Angle, S = Side):
Case 1: ASA or SAA
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Note: The Law of Sines helps us solve cases 1 and 2. The Law of Cosines helps us solve cases 3 and 4.
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What if we’re given the following info and asked to solve the triangle? This is Case 2 (SSA).But if we try todraw the triangle we get…
Side is too short! We can see this with the Law of Sines.
Pain in the SSA
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What if we’re given the following info and asked to solve the triangle? This is Case 2 (SSA).But if we try todraw the triangle we get…
Side is too short! We can see this with the Law of Sines.
Pain in the SSA
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What if we’re given the following info and asked to solve the triangle? This is Case 2 (SSA).But if we try todraw the triangle we get…
Side is too short! We can see this with the Law of Sines.
Pain in the SSA
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What if we’re given the following info and asked to solve the triangle? This is Case 2 (SSA).But if we try todraw the triangle we get…
Side is too short! We can see this with the Law of Sines.
Pain in the SSA
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You can imagine if side is in between (and just right), we might get two possible triangles:
This is called the ambiguous case, and it only happens with Case 2 (SSA).
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You can imagine if side is in between (and just right), we might get two possible triangles:
This is called the ambiguous case, and it only happens with Case 2 (SSA).
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You can imagine if side is in between (and just right), we might get two possible triangles:
This is called the ambiguous case, and it only happens with Case 2 (SSA).
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You can imagine if side is in between (and just right), we might get two possible triangles:
This is called the ambiguous case, and it only happens with Case 2 (SSA).
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Note that and are the two sides that make the angle . Also note that the area formula could be written: or
Area of a Triangle (SAS)
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Note that and are the two sides that make the angle . Also note that the area formula could be written: or
Area of a Triangle (SAS)
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Note that and are the two sides that make the angle . Also note that the area formula could be written: or
Area of a Triangle (SAS)
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Note that and are the two sides that make the angle . Also note that the area formula could be written: or
Area of a Triangle (SAS)
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Proof that :We can drop a perpendicular down to the base (see right). So, the area is: But so that .Replacing with , we get: QED
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Proof that :We can drop a perpendicular down to the base (see right). So, the area is: But so that .Replacing with , we get: QED
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Proof that :We can drop a perpendicular down to the base (see right). So, the area is: But so that .Replacing with , we get: QED
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Proof that :We can drop a perpendicular down to the base (see right). So, the area is: But so that .Replacing with , we get: QED
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Proof that :We can drop a perpendicular down to the base (see right). So, the area is: But so that .Replacing with , we get: QED
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Proof that :
Using the triangle to the right, we get: and
So, and Thus, Dividing, we get, By making perpendicular segments from and , and using similar reasoning, we get and . Thus, .QED
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Proof that :
Using the triangle to the right, we get: and
So, and Thus, Dividing, we get, By making perpendicular segments from and , and using similar reasoning, we get and . Thus, .QED
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Proof that :
Using the triangle to the right, we get: and
So, and Thus, Dividing, we get, By making perpendicular segments from and , and using similar reasoning, we get and . Thus, .QED
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Proof that :
Using the triangle to the right, we get: and
So, and Thus, Dividing, we get, By making perpendicular segments from and , and using similar reasoning, we get and . Thus, .QED
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Proof that :
Using the triangle to the right, we get: and
So, and Thus, Dividing, we get, By making perpendicular segments from and , and using similar reasoning, we get and . Thus, .QED
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Proof that :
Using the triangle to the right, we get: and
So, and Thus, Dividing, we get, By making perpendicular segments from and , and using similar reasoning, we get and . Thus, .QED
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Proof that :
Using the triangle to the right, we get: and
So, and Thus, Dividing, we get, By making perpendicular segments from and , and using similar reasoning, we get and . Thus, .QED