∆ABC is isosceles, AB = BC, and CH is an altitude. How long is AC, if CH = 84 cm and m∠HBC = m∠BAC +m∠BCH?

Mathematics

1 answer:

son4ous [18]2 years ago

3 0

I think I could help you
Have fun and feel free to ask me something new.
Or we can prove some properities without calculating by details

You might be interested in

What is 720,000 written in scientific notation?

n200080 [17]

Answer:

=7.2×10⁵

is in scientific notation

5 0

2 years ago

Read 2 more answers

What is theformula for finding the hight of a 30-60-90 degree triangle

miv72 [106K]

We don't know which side your triangle is sitting on, so the 'height'
could be any one of three values.

Here are the two simple tools that tell you everything you want to
know about a 30°-60°-90° triangle. These are worth memorizing:

-- The side opposite the 30° angle is 1/2 of the hypotenuse.

-- The side opposite the 60° angle is 1/2 of the hypotenuse times √3 .

I memorized them exactly 60 years ago. They've been very useful,
and as far as I know, they haven't changed.

7 0

3 years ago

Which would result in the largest amount of money?!

nignag [31]

Answer:

c 364

Step-by-step explanation:

this is the correct answer because the rest are lower answers

8 0

3 years ago

Read 2 more answers

Given:

diamong [38]

Answer:

10-5 $\sqrt{2}$

Step-by-step explanation:

As per the attached figure, right angled $\triangle MDL$ has an inscribed circle whose center is $I$ .

We have joined the incenter $I$ to the vertices of the $\triangle MDL$ .

Sides MD and DL are equal because we are given that $\angle M = \angle L = 45 ^\circ.$

Formula for <em>area</em> of a $\triangle = \dfrac{1}{2} \times base \times height$

As per the figure attached, we are given that side <em>a = 10.</em>

Using pythagoras theorem, we can easily calculate that side ML = 10 $\sqrt{2}$

Points P,Q and R are at $90 ^\circ$ on the sides ML, MD and DL respectively so IQ, IR and IP are heights of $\triangle$ MIL, $\triangle$ MID and $\triangle$ DIL.

Also,

$\text {Area of } \triangle MDL = \text {Area of } \triangle MIL +\text {Area of } \triangle MID+ \text {Area of } \triangle DIL$

$\dfrac{1}{2} \times 10 \times 10 = \dfrac{1}{2} \times r \times 10 + \dfrac{1}{2} \times r \times 10 + \dfrac{1}{2} \times r \times 10\sqrt2\\\Rightarrow r = \dfrac {10}{2+\sqrt2} \\\Rightarrow r = \dfrac{5\sqrt2}{\sqrt2+1}\\\text{Multiplying and divinding by }(\sqrt2 +1)\\\Rightarrow r = 10-5\sqrt2$