Question

Given: △DMN, DM=10
m∠M=75°, m∠N=45°
Find: Perimeter of △DMN

Answer 1

35.91 units ( approx )

Explanation:

Given,

In triangle DMN,

DM = 10

, m∠M = 75°, m∠N = 45°,

By the law of sine,

`(\sin M)/(DN)=(\sin N)/(MD)=(\sin D)/(MN)----(1)`

∵ m∠M + m∠N + m∠D = 180°,

75° + 45° + m∠D = 180°,

120° + m∠D = 180°

⇒ m∠D = 60°,

From equation (1),

`(\sin 75)/(DN)=(\sin 45)/(10)=(\sin 60)/(MN)`

`(\sin 75)/(DN)=(\sin 45)/(10)`

`\implies DN = (10* \sin 75)/(\sin 45)\approx 13.66\text{ unit}`,

`(\sin 45)/(10)=(\sin 60)/(MN)`

`\implies MN = (10* \sin 60)/(sin 45)\approx 12.25\text{ unit}`

Hence, the perimeter of the triangle DMN = DM + MN + DN

= 10 + 13.66 + 12.25

= 35.91 units.

Answer 2

≈ 35.91

Explanation:

The law of sines lets you find the other sides:

DN/sin(M) = MN/sin(D) = DM/sin(N)

Angle D is 180° -75° -45° = 60°, so the remaining sides are ...

DN = sin(M)/sin(N)×DM = sin(75°)/sin(45°)×10 ≈ 13.66

MN = sin(D)/sin(N)×DM = sin(60°)/sin(45°)×10 ≈ 12.25

The perimeter is the sum of the side lengths, so is ...

10 + 13.66 +12.25 = 35.91