have right angles Canndt Pyth standard Trig · 2020-02-21 · Section 6.1 The Law of Sines Oblique...

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Section 6.1 The Law of Sines Oblique triangles - Standard notation You can find missing parts of your triangle is you know two angles and any side (AAS or ASA) or two sides and an angle opposite one of them (SSA). Law of Sines: AAS ASA *SSA* (see last example) Area of a Triangle: Note: This is SAS… two adjacent sides and an angle in between. Example: Solve the following triangle. Round answers to the nearest tenth. B = 33°, C = 46°, b = 4 C c B b A a sin sin sin = = A bc Area sin 2 1 = have no right angles Canndt use Pyth Theorem or standard Trig A is an angle B A is a side length c a A b O Tum ar a co A A B c A or E ca sinB A b C B L 5.3330 a 7.2 180 33 46 1010 c s Sink A O b 4 460 C Sind Sina sin433 sin4I a sina 33 sinI ummshwowrl.cl C sin 330 4 sin 46 a SF 30 a sin 33 4 sin lol c a 5.3 axt.TW

Transcript of have right angles Canndt Pyth standard Trig · 2020-02-21 · Section 6.1 The Law of Sines Oblique...

Section 6.1 The Law of Sines Oblique triangles - Standard notation

You can find missing parts of your triangle is you know two angles and any side (AAS or ASA) or two sides and an angle opposite one of them (SSA). Law of Sines:

AAS

ASA

*SSA* (see last example)

Area of a Triangle:

Note: This is SAS… two adjacent sides and an angle in between.

Example: Solve the following triangle. Round answers to the nearest tenth. B = 33°, C = 46°, b = 4

Cc

Bb

Aa

sinsinsin==

AbcArea sin21=

have no right angles Canndt use PythTheoremor standard Trig

A is an angle BA is a side length c a

A b O

Tum ar a coA

A

Bc A

or E c asinB A b C

BL5.3330 a 7.2

180 33 46 1010 c

s Sink A Ob 4

460 C

Sind Sinasin433 sin4I a

sina33 sinIummshwowrl.clC sin330 4 sin 46 a

SF 30a sin33 4 sin lol

c a 5.3axt.TW

Example: Find the area of the triangle below:

In this case... I don’t know angle A. But I don’t need it. As long as you have SAS you can use the formula from the front.

!"#$ = 1 2( $) sin - ! = 1 2( (7)(10) sin 25°

! = 14.8units:

Example: Solve the following triangle. Round answers to the nearest tenth. b = 12.5, c = 20.1, B = 37.3°

Now, this is called the ambiguous case (SSA) since when I did my inverse sine I got an answer in Quad I, I must figure out my possible angle in Quad II (since we were looking for where sine is positive) which would be 180 – 77.0 = 103.0°. Now continue the problem by finding an additional A value (180 – 37.3 – 103) which is 39.7°. Then use that new A to find a new a which would be 13.2. Now do a sketch to ensure that is possible. Remember that the smallest angle must be across from the smallest side, largest angle across from the largest side, etc

Option 1

Option 2

AbcArea sin21=

aw

twosidelengths

0A 21.10 7 sin 250 00angle

Abetween

BJ c 3730,8

sin 37.30 since A 65 no Cb12.512.5 20 I

12.5since 20.1 sin32.3180 37.3 77

Tas sina.glsin65.7

12.5 a

since 0.9744 deEomdta 18.8

sin ANSI 770 isthis

reallyLc

or isit areference

forLC

18077 1030

B B sin37.537.5 sina39ly20 18.8 375

20.1 ASin37.5 12ssin39.7in Tin Tins

A 65T12.5

77 C 39.7 105 32A 125 C

This was an SSA setuporiginally When this happens you willb f

have Two possible triangles for your answersI