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# Geometry: Special right triangles

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Does anyone know the formulas involved? If you could walk me through this problem i would really appreciate that! Thank you.

Mar 12, 2018

### Best Answer

#1
+18431
+2

sin x  = opposite / hypotenuse    in a right triangle

remember sin 45 dgrees = sqrt2 / 2

so for the triangle on the LEFT    sin 45 = sqrt(2) /2  = a/(7 sqrt(2) )  solve for a

(7 * sqrt2) * sqrt(2)/2 = a

14/2 = a = 7

Mar 12, 2018
edited by ElectricPavlov  Mar 12, 2018

### 1+0 Answers

#1
+18431
+2
Best Answer

sin x  = opposite / hypotenuse    in a right triangle

remember sin 45 dgrees = sqrt2 / 2

so for the triangle on the LEFT    sin 45 = sqrt(2) /2  = a/(7 sqrt(2) )  solve for a

(7 * sqrt2) * sqrt(2)/2 = a

14/2 = a = 7

ElectricPavlov Mar 12, 2018
edited by ElectricPavlov  Mar 12, 2018