There is a photo attachedtan 270 is incorrect!

Mathematics

1 answer:

wlad13 [49]3 years ago

3 0

$\dfrac{1-\cos135^\circ}{\sin135^\circ}=\dfrac{1-\left(\cos^2 67.5^\circ-\sin^2 67.5^\circ\right)}{2\sin67.5^\circ\cos67.5^\circ}=\dfrac{1-\left(1-\sin^2 67.5^\circ-\sin^2 67.5^\circ\right)}{2\sin67.5^\circ\cos67.5^\circ}=\medskip\\=\dfrac{2\sin^2 67.5^\circ}{2\sin67.5^\circ\cos67.5^\circ}=\dfrac{\sin 67.5^\circ}{\cos 67.5^\circ}=\tan 67.5^\circ$

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In the figure above, quadrilateral ABCD is a parallelogram. Let x represent the measure of angle GBF, y represent the measure of

ira [324]

Answers:
measure angle x = 40°
measure angle y = 35°
measure angle z = 55°

Explanation:
Part (a): getting angle x:
In triangle BED, we have:
measure angle BED = 90°
measure angle BDE = 50°
Therefore:
measure angle DBE = 180 - (90+50) = 40°
Now, we have angle DBE and angle GBF vertically opposite angles.
This means that they are both equal. Therefore angle GBF = 40°
Since angle GBF is x, therefore:
x = 40°

Part (b): getting angle y:
We know that the sum of measures of angles on a straight line is 180.
This means that:
angle GBF + angle GBC + angle CBE = 180
We have:
angle GBF = 40°
angle GBC = 105°
angle CBE = y
Therefore:
40 + 105 + y = 180
y = 35°

Part (c): getting angle z:
In triangle BCE, we have:
measure angle BCE = z
measure angle BEC = 90°
measure angle CBE = 35°
Therefore:
z + 90 + 35 = 180
z = 55°

Hope this helps :)