In this case we know the three sides of the triangle, then this is a SSS triangle (Side Side Side). To solve this case, first we must use the Law of Cosines, applied to the opposite side to the angle we want to find.
We want to find angle W, and its opposite side is XV, then we apply the Law of Cosines to the side XV:
XV^2=XW^2+WV^2-2(XW)(WV)cos W
Replacing the known values:
116^2=96^2+89^2-2(96)(89)cos W
Solving for W
13,456=9,216+7,921-17,088 cos W
13,456=17,137-17,088 cos W
13,456-17,137=17,137-17,088 cos W-17,137
-3,681=-17,088 cos W
(-3,681)/(-17,088)=(-17,088 cos W)/(-17,088)
0.215414326=cos W
cos W = 0.215414326
Solving for W:
W= cos^(-1) 0.215414326
Using the calculator:
W=77.56016397°
Rounded to one decimal place:
W=77.6°
Answer: Third option 77.6°
Answer:
yes sure continue with questions
Perimeter: x + (4x + 9) + (x + 11)
The value x is for the bottom of the triangle.
Simplified, it would be 6x + 20
If you're talking about Egyptian, you have to put 9 "ten" symbols on the paper and you get 90.
The "ten" symbol looks like a tiny arch.
Answer:
The first one would be m^30n^20p^30
And the second one would be a^2 / d^4
Step-by-step explanation:
For 1, you have to multiply the inner exponents by the outer one so you get (m^12n^8) times (m^18n^12p^30). Then you mutiply the two together, so you add the exponents, where you get m^30n^20p^30.
For 2, since you're dividing, you have to subtract the exponents so you get a^2d^-4, but since you shouldn't have the negative you turn it into a^2 / d^4
Hope this helps :)
