All categories

3 years ago

WILL GIVE BRAINLIEST

1 answer:

6 0

I would say the vertex and the midpoint of the opposite side maybe? I know the vertex is the highest point of a triangle so I could definitely say Vertex.

You might be interested in

Need help with trig problem in pic

Sidana [21]

Answer:

a) $cos(\alpha)=-\frac{3}{5}\\$

b) $\sin(\beta)= \frac{\sqrt{3} }{2}$

c) $\frac{4+3\sqrt{3} }{10}\\$

d) $\alpha\approx 53.1^o$

Step-by-step explanation:

a) The problem tells us that angle $\alpha$ is in the second quadrant. We know that in that quadrant the cosine is negative.

We can use the Pythagorean identity:

$tan^2(\alpha)+1=sec^2(\alpha)\\(-\frac{4}{3})^2 +1=sec^2(\alpha)\\sec^2(\alpha)=\frac{16}{9} +1\\sec^2(\alpha)=\frac{25}{9} \\sec(\alpha) =+/- \frac{5}{3}\\cos(\alpha)=+/- \frac{3}{5}$

Where we have used that the secant of an angle is the reciprocal of the cos of the angle.

Since we know that the cosine must be negative because the angle is in the second quadrant, then we take the negative answer:

$cos(\alpha)=-\frac{3}{5}$

b) This angle is in the first quadrant (where the sine function is positive. They give us the value of the cosine of the angle, so we can use the Pythagorean identity to find the value of the sine of that angle:

$cos (\beta)=\frac{1}{2} \\\\sin^2(\beta)=1-cos^2(\beta)\\sin^2(\beta)=1-\frac{1}{4} \\\\sin^2(\beta)=\frac{3}{4} \\sin(\beta)=+/- \frac{\sqrt{3} }{2} \\sin(\beta)= \frac{\sqrt{3} }{2}$

where we took the positive value, since we know that the angle is in the first quadrant.

c) We can now find $sin(\alpha -\beta)$ by using the identity:

$sin(\alpha -\beta)=sin(\alpha)\,cos(\beta)-cos(\alpha)\,sin(\beta)\\$

Notice that we need to find $sin(\alpha)$ , which we do via the Pythagorean identity and knowing the value of the cosine found in part a) above:

$sin(\alpha)=\sqrt{1-cos^2(\alpha)} \\sin(\alpha)=\sqrt{1-\frac{9}{25} )} \\sin(\alpha)=\sqrt{\frac{16}{25} )} \\sin(\alpha)=\frac{4}{5}$

Then:

$sin(\alpha -\beta)=\frac{4}{5}\,\frac{1}{2} -(-\frac{3}{5}) \,\frac{\sqrt{3} }{2} \\sin(\alpha -\beta)=\frac{2}{5}+\frac{3\sqrt{3} }{10}=\frac{4+3\sqrt{3} }{10}$

Since $sin(\alpha)=\frac{4}{5}$

then $\alpha=arcsin(\frac{4}{5} )\approx 53.1^o$

4 0

3 years ago

A set of 3 consecutive integers has a sum of 27. which integers are they

Scilla [17]

They are 8, 9 and 10.

5 0

4 years ago

A particle moves along the x− axis in such a way that its acceleration at time t for t>0 is given a (t)=3t/2. When t=1, the p

lbvjy [14]

Answer:

A because

A particle moves along the x-axis so that at time t ≥ 0 its position is given by x(t)=2t^3−9t^2−60t+4. What is the total distance traveled by the particle over the time interval 0≤t≤7 ?

Step-by-step explanation:

make me brainliest

7 0

2 years ago

Will mark as Brainliest.

strojnjashka [21]

B is the answer because there are 52 weeks in a year so $52,000/52 gives 1,000 and 1,000/40 gives 25. It's not A because it can't be less that $25 but more than it.

4 0

3 years ago

Read 2 more answers

Plz help 10 points :)

bulgar [2K]

The first 3 answers:
With a unit ^3 (cubed) it means they took 3 side measurements and multiplied them to find volume of an object.

8 0

3 years ago