here you go someone already has answered this question before <3
What are both of the ranges
1 answer:
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See diagram
I wrote that the degrees are 70 each because it is isocolese and 140/2=70
a.
to solve for the diagonal, I'm going to use trig to find the height and the base then use pythagorean theorem to find the diagonal
so
see I've drawn an auxiliry line to make a right triangle on the leftmost side
so
we want to find the height and base
sin(70°)=h/7
so 7sin(70°)=h≈6.57785
cos(70°)=base/7
so 7cos(70°)=base≈2.39414
now
the entire long base is 22, so that big triangle with the diagonal as the hyptonuse is 22-2.39414=19.6059
use pythagorean theorem, like 6.57785²+19.6059²=diagonal²
we get that the diagonal is about 20.68ft
b.
we got that the base was 7cos(70°)
2 of those bases so 22-14cos(70°)=17.2117ft or rounded 17.21ft
a. 20.68ft
b. 17.21ft
I wrote that the degrees are 70 each because it is isocolese and 140/2=70
a.
to solve for the diagonal, I'm going to use trig to find the height and the base then use pythagorean theorem to find the diagonal
so
see I've drawn an auxiliry line to make a right triangle on the leftmost side
so
we want to find the height and base
sin(70°)=h/7
so 7sin(70°)=h≈6.57785
cos(70°)=base/7
so 7cos(70°)=base≈2.39414
now
the entire long base is 22, so that big triangle with the diagonal as the hyptonuse is 22-2.39414=19.6059
use pythagorean theorem, like 6.57785²+19.6059²=diagonal²
we get that the diagonal is about 20.68ft
b.
we got that the base was 7cos(70°)
2 of those bases so 22-14cos(70°)=17.2117ft or rounded 17.21ft
a. 20.68ft
b. 17.21ft