Answer:
101 meters
Step-by-step explanation:
The direct path across the stream and perpendicular to both sides of the stream is one leg of a right triangle. The 20 meter distance on the arriving side of the stream is another leg of the right triangle. The path he actually took is the hypotenuse of the right triangle.
leg 1 = 99 m
leg 2 = 20 m
hypotenuse = ?
We have a right triangle, and we know the lengths of the two legs. We need to find the length of the hypotenuse, so we can use the Pythagorean theorem.
a^2 + b^2 = c^2
99^2 + 20^2 = c^2
9801 + 400 = c^2
c^2 = 10201
c = sqrt(10201) = 101
Answer: 101 meters
Assume ladder length is 14 ft and that the top end of the ladder is 5 feet above the ground.
Find the distance the bottom of the ladder is from the base of the wall.
Picture a right triangle with hypotenuse 14 feet and that the side opposite the angle is h. Then sin theta = h / 14, or theta = arcsin 5/14. theta is
0.365 radian. Then the dist. of the bot. of the lad. from the base of the wall is
14cos theta = 14cos 0.365 rad = 13.08 feet. This does not seem reasonable; the ladder would fall if it were already that close to the ground.
Ensure that y ou have copied this problem accurately from the original.
147 63 82 101 155 160 175 92 116 138 74 93 110 162 154 105 97
The frequency to her third group is 80 - 99.
Acute would be less than 90 degrees, obtuse is greater than 90 degrees and right is 90 degrees
You can calculate it using the law of cosines: c^2=a^2+b^2-2*a*b*cos(C)
your triangle is
CD=15=a
CE=?=b
DE=CE+3=b+3=c
and C=90°
-> insert those values, with c substituted with b+3 to remove c
c^2=a^2+b^2-2*a*b*cos(C)
(b+3)^2=15^2+b^2-2*15*b*cos(90)
cos(90)=0->
(b+3)^2=15^2+b^2
b^2+2*3*b+3^2=225+b^2
6b+9=225
6b=216
b=36=CE
DE=CE+3=36+3=39