Given: △ABC, BC>AC, D∈ AC , CD=CB Prove: m∠ABD is acute

1 answer:

5 0

I want you to imagine as you read this or you can draw through the help of my explanation and see yourself:

1↪Draw triangle ABC where BC>AC

2↪D is any point on AC such that CD=CB

3↪Roughly drawing , you can assume CD=CB and and join BD

4↪SO triangle ABC which is a big triangle is divided into Triangles ABD and BDC

5↪See in triangle BDC ,CD=CB so, base angles of isosceles triangle are equal:
<CDB=<CBD = x (assume) which means x is acute angle since CDB and CBD are are in same triangle with same measure and there can't be any two obtuse angle in any traingle. So x must be acute.

6↪Now see in traingle ABD,

<ADB=180-<CDB=180-x=obtuse angle
...check yourself ...just subtract any acute angle from 180 you will get only obtuse angle (ie angle greater than 90)
That means in triangle ABD , one angle ADB is obtuse which means remaining <ABD and < BAD are acute. [PROVED]

❇Main Concept Used Here:
↪In any triangle there can be maximum of one obtuse angle...so remaining two must be acute angle otherwise interior angles sum can't be equal to 180.

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How do you find the volume of the solid generated by revolving the region bounded by the graphs

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Answer:

About the x axis

$V = 4\pi[ \frac{x^5}{5}] \Big|_0^2 =4\pi *\frac{32}{5}= \frac{128 \pi}{5}$

About the y axis

$V = \pi [4y -y^2 +\frac{y^3}{12}] \Big|_0^8 =\pi *\frac{32}{3}= \frac{32 \pi}{3}$

About the line y=8

$V = \pi [64x -\frac{32}{3}x^3 +\frac{4}{5}x^5] \Big|_0^2 =\pi *(128-\frac{256}{3} +\frac{128}{5})= \frac{1024 \pi}{5}$

About the line x=2

$V = \frac{\pi}{2} [\frac{y^2}{2}] \Big|_0^8 =\frac{\pi}{4} *(64)= 16\pi$

Step-by-step explanation:

For this case we have the following functions:

$y = 2x^2 , y=0, X=2$

About the x axis

Our zone of interest is on the figure attached, we see that the limit son x are from 0 to 2 and on y from 0 to 8.

We can find the area like this:

$A = \pi r^2 = \pi (2x^2)^2 = 4 \pi x^4$

And we can find the volume with this formula:

$V = \int_{a}^b A(x) dx$

$V= 4\pi \int_{0}^2 x^4 dx$

$V = 4\pi [\frac{x^5}{5}] \Big|_0^2 =4\pi *\frac{32}{5}= \frac{128 \pi}{5}$

About the y axis

For this case we need to find the function in terms of x like this:

$x^2 = \frac{y}{2}$

$x = \pm \sqrt{\frac{y}{2}}$ but on this case we are just interested on the + part $x=\sqrt{\frac{y}{2}}$ as we can see on the second figure attached.