How do you find the length of a leg on a triangle?

add the other lenghts togethere and subtrack those 2 from 180 and bam

Step-by-step explanation:

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Step-by-step explanation:

The method for finding the unknown length of a side of a triangle depends on what other information is given. In general, you need one side, one angle, and at least one other side or angle.

Solving a triangle usually is introduced with the Pythagorean theorem. For a right triangle, it tells you ...

  the square of the hypotenuse is equal to the sum of the squares of the other two sides.

If the side lengths are a, b and the hypotenuse is c, then ...

  c² = a² + b²

Solving for the hypotenuse, c, you have ...

  c = √(a² +b²)

Solving for one side, a, you have ...

  a = √(c² -b²)

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Occasionally, you're asked to find a measure of a triangle using the formula for area.

  A = 1/2bh

Solving for the base or height gives you ...

  b = 2A/h

  h = 2A/b

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Once you learn trigonometry, additional methods are available for solving triangles. The Law of Sines tells you ...

  a/sin(A) = b/sin(B) = c/sin(C)

where angles A, B, C are opposite sides a, b, c, respectively.

And the Law of Cosines tells you ...

  c² = a² +b² -2ab·cos(C) . . . . . . . . where sides and angles are as above

This can be rearranged by interchanging a, b, c, keeping the appropriate angle. For C = 90°, this reduces to the Pythagorean theorem.

For right triangles, the various trig relations can also be used to solve triangles. These are conveniently summarized in the mnemonic SOH CAH TOA. It is intended to remind you ...

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In summary, depending on the given information, you choose an applicable formula, fill in the given information, and solve for the unknown. If you have more than one unknown, you may need to choose a different formula.

i can't see the picture : /

1/3

step-by-step explanation:

to find this, you can divide 12 by 4 which is 3 or you can multiply 4 until you reach 12 (4x3=12). both should give you 3, meaning theres three times less bananas than there are oranges.