The obtuse angle of a parallelogram is 125º and the sides measure 3 ft and 5 ft.?
How long is the longer diagonal?
What is the measure of the angle the long diagonal makes with the short side of the parallelogram?
and what is the area?
Thanks!
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So that you can follow along with a diagram, please draw parallelogram ABCD with ∡A = ∡C = 125° and ∡B = ∡D = 55°, and AB = CD = 3 and BC = AD = 5.
To find the longer diagonal BD, we can use Law of Cosines in ∆ABD:
BD² = 5² + 3² - 2*5*3*cos125°
BD = 7.156 ft.
The second question is asking for the measure of ∡ABD. We can use Law of Sines within ∆ABD:
sin ∡ABD /5 = sin 125° / 7.156
sin ∡ABD = 5sin 125° / 7.156 = .5724
∡ABD = arcsin .5724 = 34.91°
There are a couple ways to find the area, but I think the simplest is to find the area of ∆ABD and double it:
Area of ∆ABD = ½·3·5·sin125° = 6.144 ft²·2 = 12.288 ft²
formulation for component to parallelogram =length x top so to your question is as persist with length(section) = 176ft & top (altitude) =290ft so your answer is 176 x290 = 51040sq.feet you may take shorter section as length with the aid of fact the altitude is to the shorter section. you additionally can define length as base........