Answer:
(d) 71°
Step-by-step explanation:
The desired angle in the given isosceles triangle can be found a couple of ways. The Law of Cosines can be used, or the definition of the sine of an angle can be used.
<h3>Sine</h3>Since the triangle is isosceles, the bisector of angle W is an altitude of the triangle. The hypotenuse and opposite side with respect to the divided angle are given, so we can use the sine relation.
sin(W/2) = Opposite/Hypotenuse
sin(W/2) = (35/2)/(30) = 7/12
Using the inverse sine function, we find ...
W/2 = arcsin(7/12) ≈ 35.685°
W = 2×36.684° = 71.37°
W ≈ 71°
<h3>Law of cosines</h3>The law of cosines tells you ...
w² = u² +v² -2uv·cos(W)
Solving for W gives ...
W = arccos((u² +v² -w²)/(2uv))
W = arccos((30² +30² -35²)/(2·30·30)) = arccos(575/1800) ≈ 71.37°
W ≈ 71°
Answer:
Perimeter of quadrilateral will be
Step-by-step explanation:
Given quadrilateral is a reflection of quadrilateral
over the line
.
So, all the corresponding sides of both the quadrilateral should be equal.
Also, we can see from the diagram that
As both image are reflection of same scale we can add corresponding sides.
So, the perimeter of will be
And we have
Now, perimeter of will be
Perimeter of quadrilateral is
ANSWER:
For the first picture, it is the first answer. This is because of the Exterior Angle Theorem.
For the second picture, use the Exterior Angle Theorem again, but this time, solve for x.
The m∡B = 39°