JhamareeElam
  • JhamareeElam
How do you solve a right triangle when only given either one angle or one side ? Like my professor did not do a good job teaching this . Like i may have the 90 degree angle and another angle of 28 degrees . Then I may have side b which may be 7 and then the 90 degree angle . How do I solve this ?
Trigonometry
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schrodinger
  • schrodinger
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Nnesha
  • Nnesha
if you know one angle and one side then to find 2nd side you can use trig. to find 3rd angle use the fact all triangles add up to 180 degree
Nnesha
  • Nnesha
|dw:1449467247822:dw| i know two angles 90+66 to find 3rd angle i can add both angles and then subtract it from 180 \[\rm 90+66+x=180 \] solve for x then u will get x=24
Nnesha
  • Nnesha
i hope you're familiar with sin,cos , tan definitions \[\rm sin \rm \theta = \frac{ opposite }{ hypotenuse }~~~~ \cos \theta = \frac{ adjacent }{ hypotenuse } ~~\tan \theta = \frac{ opposite }{ adjacent }\] to find one of the side you can use trig ratios there are 3 but it depends on the question (which angle is given and which would be the correct ratio to find missing side )

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Nnesha
  • Nnesha
make sense ? or no
Nnesha
  • Nnesha
|dw:1449467553635:dw| like for this question the given is hypotenuse and i need to find x so i should use cos function bec it's equal to hypotenuse over adjacent (x is adjacent side of 66 angle ) so \[\rm \cos (66)=\frac{x}{5}\] and then just simply solve for x and if we have to find an angle then we should take invs of the ratio
Nnesha
  • Nnesha
i don't know if that make sense 2u
JhamareeElam
  • JhamareeElam
This is what i mean and an example I am talknig about. How would you solve these two triangles?|dw:1449469097458:dw|
JhamareeElam
  • JhamareeElam
Then there are ones like this |dw:1449469186536:dw|
JhamareeElam
  • JhamareeElam
This is all the information we get and we have to solve the whole triangle... I dont know how or even where to start. Google doesn't even help!
JhamareeElam
  • JhamareeElam
@Nnesha this is what I'm referring to ^^
Johan14th
  • Johan14th
Depending on whether your teacher is Bc or not then you can use (A/sin(a) )=(B/sin(b))=(C/sin(c)) This is called the law of sin Were all of the capital letters are lenghts and the lower case are the angles
Johan14th
  • Johan14th
For example like you said if you have 2 angles 90 and 28 with a lenght of 7 as long as you know where they correspond to the triangle when you draw it you van use this equation. (7/sin(90))=(B/sin(28)) you can then solve for B

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