Answer:
The top angle of the triangle is, X = 68°.
Step-by-step explanation:
Let there is a triangle named PQR with bottom corner angles ∠ P and ∠ Q and the ∠ R is the angle at the top.
Now, given that ∠ P = 31° and ∠ Q = 81° and we have to find out ∠ R i.e. X.
In Δ PQR, we know, ∠ P + ∠ Q + ∠ R = 180°
⇒ 31° + 81° + X = 180°
⇒ X = 68°
Therefore, the top angle of the triangle is 68°. (Answer)
Assume that the radius of the circle is 1.
The diagonal of the square is 2. By the Pythagorean Theorem, the side of the square of a right triangle will be 2/sqrt(2).
Finally, the ratio of the square to the circle, in area, will be
(2/sqrt(2))^2 : pi*(1)^2 = 2 : pi
We can conclude that the white area is about 35%.
1 4 . 8 3333
6 8 9 50
14.8333
Um
81^5=3x
3,486,784,401=3x
x=1,162,261,467
Answer:
2) 162°, 72°, 108°
3) 144°, 54°, 126°
Step-by-step explanation:
1) Multiply the equation by 2sin(θ) to get an equation that looks like ...
sin(θ) = <some numerical expression>
Use your knowledge of the sines of special angles to find two angles that have this sine value. (The attached table along with the relations discussed below will get you there.)
____
2, 3) You need to review the meaning of "supplement".
It is true that ...
sin(θ) = sin(θ+360°),
but it is also true that ...
sin(θ) = sin(180°-θ) . . . . the supplement of the angle
This latter relation is the one applicable to this question.
__
Similarly, it is true that ...
cos(θ) = -cos(θ+180°),
but it is also true that ...
cos(θ) = -cos(180°-θ) . . . . the supplement of the angle
As above, it is this latter relation that applies to problems 2 and 3.