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Consider the given right triangle. If angle B = 43 degrees and c = 26.7 meters, find a.

Nady [450]

Answer: Side a equals 19.5 metres

Step-by-step explanation: Consider the right angled triangle as shown in the picture attached. The triangle has been drawn with angle measuring 43 degrees, side c (line AB) measuring 26.7 m and side a (line CB) is yet unknown.

A right angled triangle can be solved if at least one side and an angle are available. In this question we shall apply the trigonometric ratios since we have one angle which shall be the reference angle (43°). Also we have an hypotenuse (the side facing the right angle) and an unknown side which is the adjacent (which lies between the right angle and the reference angle).

Cos B = Adjacent/Hypotenuse

Cos 43 = a/26.7

Cos 43 x 26.7 = a

0.7314 x 26.7 = a

19.52714 = a

a ≈ 19.5 (rounded to the nearest tenth)

Therefore the length of side a equals 19.5 metres.

3 0

3 years ago

Read 2 more answers

Does BBC live matter why or why not?

almond37 [142]

Answer:

THIS IS NOT A MATH QUESTION HOWEVER GIVE MY. ANSWER AS BRAINLIST ANSWER

3 0

2 years ago

Read 2 more answers

Un grupo de compañeros de clase está planificando un viaje de cada uno de ellos hicieron aportación de 14 pesos para gastos de v

Crank

Answer:

$14.8

Step-by-step explanation:

There is a mistype in the question. It should say: ... si cada uno de ellos hicieron una contribución de dieciseis pesos...

So if everybody contributes $14, they make a total of 14*x, where x denotes the number of students. In this scenario, $4 are left, calling y to the cost of the trip, then:

14*x = y - 4 (eq. 1)

In the other scenario, everybody contributes $16 and $6 are in excess, so

16*x = y + 6 (eq. 2)

subtracting eq. 1 to eq. 2:

16*x - 14*x = y + 6 - (y - 4)

2*x = y + 6 - y + 4

2*x = 10

x = 10/2 = 5

Replacing this value in eq. 1:

14*5 = y - 4

70 + 4 = y

y = $74

If every student pays: 74/5 = $14.8, then they would get altogether the exact cost of the trip

7 0

2 years ago

An equilateral triangle has three congruent sides. If sides are congruent, the opposite angles are congruent.

VladimirAG [237]

Answer:

B)law of detachment

Step-by-step explanation:

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4 0

2 years ago

SOLVE THE QUESTION BELOW ASAP

qwelly [4]

Answer:

Part A) The graph in the attached figure (see the explanation)

Part B) 16 feet

Part C) see the explanation

Step-by-step explanation:

Part A) Graph the function

Let

h(t) ----> the height in feet of the ball above the ground

t -----> the time in seconds

we have

$h(t)=-16t^{2}+98$

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex is a maximum

To graph the parabola, find the vertex, the intercepts, and the axis of symmetry

Find the vertex

The function is written in vertex form

so

The vertex is the point (0,98)

Find the y-intercept

The y-intercept is the value of the function when the value of t is equal to zero

For t=0

$h(t)=-16(0)^{2}+98$

$h(0)=98$

The y-intercept is the point (0,98)

Find the t-intercepts

The t-intercepts are the values of t when the value of the function is equal to zero

For h(t)=0

$-16t^{2}+98=0$

$t^{2}=\frac{98}{16}$

square root both sides

$t=\pm\frac{\sqrt{98}}{4}$

$t=\pm7\frac{\sqrt{2}}{4}$

therefore

The t-intercepts are

$(-7\frac{\sqrt{2}}{4},0), (7\frac{\sqrt{2}}{4},0)$

$(-2.475,0), (2.475,0)$

Find the axis of symmetry

The equation of the axis of symmetry in a vertical parabola is equal to the x-coordinate of the vertex

so

$x=0$ ----> the y-axis

To graph the parabola, plot the given points and connect them

we have

The vertex is the point $(0,98)$

The y-intercept is the point $(0,98)$

The t-intercepts are $(-2.475,0), (2.475,0)$

The axis of symmetry is the y-axis

The graph in the attached figure

Part B) How far is the artifact fallen from the time t=0 to time t=1

we know that

For t=0

$h(t)=-16(0)^{2}+98$

$h(0)=98\ ft$

For t=1

$h(t)=-16(1)^{2}+98$

$h(1)=82\ ft$

Find the difference

$98\ ft-82\ ft=16\ ft$

Part C) Does the artifact fall the same distance from time t=1 to time t=2 as it does from the time t=0 to time t=1?

we know that

For t=1

$h(t)=-16(1)^{2}+98$

$h(1)=82\ ft$

For t=2

$h(t)=-16(2)^{2}+98$

$h(2)=34\ ft$

Find the difference

$82\ ft-34\ ft=48\ ft$

so

The artifact fall 48 feet from time t=1 to time t=2 and fall 16 feet from time t=0 to time t=1

therefore

The distance traveled from t=1 to t=2 is greater than the distance traveled from t=0 to t=1

8 0

2 years ago