Please help quickly! What is the measure of Arc CB. How did you determine this?

Mathematics

1 answer:

S_A_V [24]3 years ago

4 0

Answer:

As shown in picture,

Angle CDF = Angle BDE = 54 deg (pair of opposite angles)

Angle CDB = 180 - CDF = 180 - 54 = 126 deg (BF is a diameter).

=> Arc CB = Angle CDB = 126 deg (property of angle on circle)

Hope this helps!

You might be interested in

Marco can paddle his canoe at a rate of 6 miles per hour on unmoving water. However, today Marco is canoeing down (with the curr

Radda [10]

Answer:

S = v1 t1 = 7 t1 traveling downstream

S = v2 t2 = 5 t2 traveling upstream

7 t1 = 5 t2

7 (6 - t2) = 5 t2 since t1 + t2 = 6

42 - 7 t2 = 5 t2

t2 = 42 / 12 = 3.5 hrs so t1 = 2.5 hrs

S = 7 t1 = 7 * 2.5 = 17.5 mi

Also, S = 5 t2 = 5 * 3.5 = 17.5 mi

3 0

3 years ago

I don't need you to work it out, just theorem(s) i need to reach the answer :)

VikaD [51]

Not sure why such an old question is showing up on my feed...

Anyway, let $x=\tan^{-1}\dfrac43$ and $y=\sin^{-1}\dfrac35$ . Then we want to find the exact value of $\cos(x-y)$ .

Use the angle difference identity:

$\cos(x-y)=\cos x\cos y+\sin x\sin y$

and right away we find $\sin y=\dfrac35$ . By the Pythagorean theorem, we also find $\cos y=\dfrac45$ . (Actually, this could potentially be negative, but let's assume all angles are in the first quadrant for convenience.)

Meanwhile, if $\tan x=\dfrac43$ , then (by Pythagorean theorem) $\sec x=\dfrac53$ , so $\cos x=\dfrac35$ . And from this, $\sin x=\dfrac45$ .

So,

$\cos\left(\tan^{-1}\dfrac43-\sin^{-1}\dfrac35\right)=\dfrac35\cdot\dfrac45+\dfrac45\cdot\dfrac35=\dfrac{24}{25}$

7 0

4 years ago

Triangle ABC has vertices at A(−3, 4), B(4, −2), C(8, 3). The triangle translates 2 units up and 1 unit right. Which rule repres

Nataly [62]

Answer:

The translation statement is given by:

$(x,y)\rightarrow (x+1,y+2)$