What is each product? For 5/10•10/3

A triangular lot has sides of 215m, 185m, and 125m. Find the measures of the angles at its corners

Vlada [557]

Answer:

The measures of the angles at its corners are $59.1\°,35.4\°,85.5\°$

Step-by-step explanation:

see the attached figure to better understand the problem

step 1

Find the measure of angle A

Applying the law of cosines

$185^{2}= 215^{2}+125^{2}-2(215)(125)cos(A)$

$2(215)(125)cos(A)= 215^{2}+125^{2}-185^{2}$

$cos(A)= [215^{2}+125^{2}-185^{2}]/(2(215)(125))$ $cos(A)=0.513953$

$A=arccos(0.513953)=59.1\°$

step 2

Find the measure of angle B

Applying the law of cosines

$125^{2}= 215^{2}+185^{2}-2(215)(185)cos(B)$

$2(215)(185)cos(B)= 215^{2}+185^{2}-125^{2}$

$cos(B)= [215^{2}+185^{2}-125^{2}]/(2(215)(185))$ $cos(B)=0.81489$

$B=arccos(0.81489)=35.4\°$

step 3

Find the measure of angle C

Applying the law of cosines

$215^{2}= 125^{2}+185^{2}-2(125)(185)cos(C)$

$2(125)(185)cos(C)= 125^{2}+185^{2}-215^{2}$

$cos(C)= [125^{2}+185^{2}-215^{2}]/(2(125)(185))$ $cos(C)=0.0784$

$C=arccos(0.0784)=85.5\°$

3 0

3 years ago

A 10ft by 20ft rectangular swimming pool is surrounded by a walkway of uniform width. If the total area of the walkway is 216ft^

Dmitry [639]

check the picture below.

so, we know the dimensions of the pool, is a 20x10, so its area is simply 200 ft², and we know the walkway is 216 ft², so the whole thing, including pool and walkway is really 200 + 216 ft².

now, as you see in the picture, the dimensions for the combined area is 20+2x and 10+2x, since the walkway is "x" long, therefore,

$\bf \stackrel{length}{(20+2x)}\stackrel{width}{(10+2x)}=200+216\implies \stackrel{FOIL}{4x^2+60x+200}=200+216 \\\\\\ 4x^2+60x=216\implies \stackrel{dividing~by~4}{x^2+15x=54} \\\\\\ x^2+15x-54=0\implies (x+18)(x-3)=0\implies x= \begin{cases} -18\\ \boxed{3} \end{cases}$