∠ACB is an inscribed angle, so
m∠ACB= (1/2)mAB =(1/2)*50=25⁰
m∠ACB= 25⁰
Answer:
It's 6.
Step-by-step explanation:
this was so confusing
The first step is to determine the distance between the points, (1,1) and (7,9)
We would find this distance by applying the formula shown below
![\begin{gathered} \text{Distance = }\sqrt[]{(x2-x1)^2+(y2-y1)^2} \\ \text{From the graph, } \\ x1\text{ = 1, y1 = 1} \\ x2\text{ = 7, y2 = 9} \\ \text{Distance = }\sqrt[]{(7-1)^2+(9-1)^2} \\ \text{Distance = }\sqrt[]{6^2+8^2}\text{ = }\sqrt[]{100} \\ \text{Distance = 10} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Ctext%7BDistance%20%3D%20%7D%5Csqrt%5B%5D%7B%28x2-x1%29%5E2%2B%28y2-y1%29%5E2%7D%20%5C%5C%20%5Ctext%7BFrom%20the%20graph%2C%20%7D%20%5C%5C%20x1%5Ctext%7B%20%3D%201%2C%20y1%20%3D%201%7D%20%5C%5C%20x2%5Ctext%7B%20%3D%207%2C%20y2%20%3D%209%7D%20%5C%5C%20%5Ctext%7BDistance%20%3D%20%7D%5Csqrt%5B%5D%7B%287-1%29%5E2%2B%289-1%29%5E2%7D%20%5C%5C%20%5Ctext%7BDistance%20%3D%20%7D%5Csqrt%5B%5D%7B6%5E2%2B8%5E2%7D%5Ctext%7B%20%3D%20%7D%5Csqrt%5B%5D%7B100%7D%20%5C%5C%20%5Ctext%7BDistance%20%3D%2010%7D%20%5Cend%7Bgathered%7D)
Distance = 10 units
If one unit is 70 meters, then the distance between both entrances is
70 * 10 = 700 meters
Answer:
155°
Step-by-step explanation:
The obtuse angle of the large (outside) triangle is the supplement of 60°, so is ...
180° -60° = 120°
The angle x is the sum of the remote interior angles of that large triangle:
x = 35° +120° = 155°
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<em>Check</em>
The other acute angle in the smaller (left) right triangle is 90° -35° = 55°. Then the top acute angle in the larger (bottom, right) right triangle is ...
180° -55° -60° = 65°
The other acute angle in that triangle is 90° -65° = 25°. It is supplementary to angle x. Hence angle x is 180° -25° = 155°, as above. (Note that x is also the sum of 90° and 65°, the remote interior angles of the nearest right triangle to x.)
