Given: △dmn, dm=10 3 m∠m=75°, m∠n=45° find: perimeter of △dmn

The Perimeter of △DMN is 35.9

First step

sin 45 / 10 = sin 60/ x

x = sin 60 × 10 / sin 45

= √(3)/2 × 10 × 2 / √(2)

= √(3) / 2 × 10 ×√(2)

= 5 × √(6)

Second step

sin 45 / 10 = sin 75 / y

y = sin 75 × 10 / sin 45

= sin 75 × 10 × 2 / √(2)

= sin 75 × 10 ×√(2)

= 13.066054

Now let determine Perimeter of △DMN

Perimeter of △DMN=10 + 5*√(6) + 13.06605425

Perimeter of △DMN= 35.90770275                      

Perimeter of △DMN=35.9

Inconclusion  Perimeter of △DMN is 35.9                                

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Angle D is 180° -75° -45° = 60°. Drawing altitude MX to segment DN divides the triangle into ΔMDX, a 30°-60°-90° triangle, and ΔMNX, a 45°-45°-90° triangle. We know the side ratios of such triangles (shortest-to-longest) are ...

... 30-60-90: 1 : √3 : 2

... 45-45-90: 1 : 1 : √2

The long side of ΔMDX is 10√3, so the other two sides are

... MX = MD(√3/2) = 15

... DX = MD(1/2) = 5√3

The short side of ΔMNX is MX = 15, so the other two sides are

... NX = MX(1) = 15

... MN = MX(√2) = 15√2

Then the perimeter of ΔDMN is ...

... P = DM + MN + NX + XD

... P = 10√3 +15√2 + 15 + 5√3

... P = 15√3 +15√2 +15 . . . . perimeter of ΔDMN

in problems involving proportions, we can use cross products to test whether two ratios are equal and form a proportion. to find the cross products of a proportion, we multiply the outer terms, called the extremes, and the middle terms, called the means.

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