All categories

3 years ago

Use special right triangle ratios to find the lengths of the other leg and the hypotenuse

Mathematics

2 answers:

Amanda [17]3 years ago

8 0

Answer:

leg = 18

hypotenuse = 18 sqrt(2)

Step-by-step explanation:

We know that sin theta = opp side / hypotenuse

sin 45 = 18 / hyp

hyp sin 45 = 18

hyp = 18 / sin 45

hyp = 18 sqrt(2)

Since this is an isosceles triangle ( the two angles are the same measure), the two legs have to be the same length

leg = 18

morpeh [17]3 years ago

6 0

the lengths of the other leg and the hypotenuse

is 18 units and 18 $\sqrt{2}$ units respectively.

Answer:

Solution given:

Let <C=<B=45°

AB=18 units

BC=?

AC=?

again

By using

By usingspecial right triangle ratios

sin C=opposite/hypotenuse=AB/AC=18/AC

Sin 45=18/AC

AC=18/sin45

AC=hypotenuse=18 $\sqrt{2}$ units

again

Tan A=opposite/adjacent=BC/AB=BC/18

Tan45=BC/18

BC=Tan45*18

BC=length of another leg=18 units.

You might be interested in

I dont the know the answer to c-d=b for c

Vinvika [58]

C-d=b add to both sides d and will get

c-d+d=b+d

c=b+d

hope helped

4 0

3 years ago

How does the graph of g(x) = -(x + 3)4 compare to the parent function of f(x) = xº?

AnnyKZ [126]

Answer:

g(x) is shifted 3 units to the left and reflected over the x-axis.

Step-by-step explanation:

8 0

3 years ago

Ill give brainlist please i beg ulol

denis23 [38]

GHE i think because it’s the same degrees

5 0

3 years ago

Read 2 more answers

A plane takes off circles airport then lands at the same airport is this statement true or false

Leya [2.2K]

the statement is true.

3 0

3 years ago

Read 2 more answers

Write down the explicit solution for each of the following: a) x’=t–sin(t); x(0)=1

Kay [80]

Answer:

a) x=(t^2)/2+cos(t), b) x=2+3e^(-2t), c) x=(1/2)sin(2t)

Step-by-step explanation:

Let's solve by separating variables:

$x'=\frac{dx}{dt}$

a) x’=t–sin(t), x(0)=1

$dx=(t-sint)dt$

Apply integral both sides:

$\int {} \, dx=\int {(t-sint)} \, dt\\\\x=\frac{t^2}{2}+cost +k$

where k is a constant due to integration. With x(0)=1, substitute:

$1=0+cos0+k\\\\1=1+k\\k=0$

Finally:

$x=\frac{t^2}{2} +cos(t)$

b) x’+2x=4; x(0)=5

$dx=(4-2x)dt\\\\\frac{dx}{4-2x}=dt \\\\\int {\frac{dx}{4-2x}}= \int {dt}\\$

Completing the integral:

$-\frac{1}{2} \int{\frac{(-2)dx}{4-2x}}= \int {dt}$

Solving the operator:

$-\frac{1}{2}ln(4-2x)=t+k$

Using algebra, it becomes explicit:

$x=2+ke^{-2t}$

With x(0)=5, substitute:

$5=2+ke^{-2(0)}=2+k(1)\\\\k=3$

Finally:

$x=2+3e^{-2t}$

c) x’’+4x=0; x(0)=0; x’(0)=1

Let $x=e^{mt}$ be the solution for the equation, then:

$x'=me^{mt}\\x''=m^{2}e^{mt}$

Substituting these equations in c)

$m^{2}e^{mt}+4(e^{mt})=0\\\\m^{2}+4=0\\\\m^{2}=-4\\\\m=2i$

This becomes the solution m=α±βi where α=0 and β=2

$x=e^{\alpha t}[Asin\beta t+Bcos\beta t]\\\\x=e^{0}[Asin((2)t)+Bcos((2)t)]\\\\x=Asin((2)t)+Bcos((2)t)$

Where A and B are constants. With x(0)=0; x’(0)=1:

$x=Asin(2t)+Bcos(2t)\\\\x'=2Acos(2t)-2Bsin(2t)\\\\0=Asin(2(0))+Bcos(2(0))\\\\0=0+B(1)\\\\B=0\\\\1=2Acos(2(0))\\\\1=2A\\\\A=\frac{1}{2}$

Finally:

$x=\frac{1}{2} sin(2t)$

7 0

3 years ago