If the legs of a right triangle are lengths a and b and the hytponuse is legnth c then
a²+b²=c²
so if the hyptonuse is 20 then
a²+b²=20²
a²+b²=400
hmm, what combos can we do?
find the values of a and b such taht they are whole numbers and a²+b²=400
looking at all the squares from 1 to 20
1²=1
2²=4
3²=9
4²=16
5²=25
6²=36
7²=49
8²=64
9²=81
10²=100
11²=121
12²=144
13²=169
14²=196
15²=225
16²=256
17²=289
18²=324
19²=361
20²=400
which pairs add up to 400?
the only pair is 144+256 which is 12²+16²
the legs are length 12 units and 16 units
Answer:
Step-by-step explanation:
Let the hypotenuse of the smaller triangle be h units.
Then; from the Pythagoras Theorem.
From the smaller triangle;
and
From the second triangle, let the other other shorter leg of the second triangle be s units.
Then;
and
We now use the double angle property;
we plug in the values to obtain;
Sin x cos x ( tan x - 1) = 0
so one solution is tan x = 1 giving x = 45, 225 degrees ( for x 0 to 360)
sin x cos x = 0 so either sinx = 0 giving x = 0, 180, 360 degrees
or cos x = 0 giving x = 90, 270
answer is 0, 45, 90, 180, 225, 270, 360 degrees ( for 0 =< x >= 360)
Answer: 200,000 + 10,000 + 8,000 + 700 + 60 + 4
Step-by-step explanation: