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2 years ago

Which angle refers to the same angle as \angle DEF∠DEFangle, D, E, F?

Mathematics

1 answer:

Sergio039 [100]2 years ago

8 0

Answer:

Option

Step-by-step explanation:

If you look at the angle, it's an acute angle so it narrows your choices down a lot. Another indicator is the green lines showing similar angles.

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What basic trigonometric identity would you use to verify that sinx cosx tanx=1-cos^2x

andre [41]

Answer:

$sin^2x+cos^2x=1$ is used to prove the given equation.

Step-by-step explanation:

We have given an equation:

$sinx\cdot cosx\cdot tanx=1-cos^2x$

We will consider right hand side first

tan x can be written as:

$tanx=\frac{sinx}{cosx}$

So, on substituting tan x in the left hand side of the given equation we get:

$sinx\cdot cosx\cdot \frac{sinx}{cosx}$

Cancel common term from denominator and numerator which is cosx we get

$sin^2x$

And we have an identity:

$sin^2x+cos^2x=1$

$\Rightarrow sin^2x=1-cos^2x$

Hence, right hand side is equal to $sin^2x$ .

5 0

3 years ago

Read 2 more answers

An airplane leaves an airport at 9:00 p.M. With a heading of 270 and a speed of 610 mph. At 10:00 p.M. The pilot changes the he

timama [110]

Answer:

2330.51 miles

Step-by-step explanation:

Given that the speed of the airplane = 610 mph.

The airplane leaves an airport at 9:00 P.M. with a heading of 270 degrees and at 10:00 P.M., the pilot changes the heading to 310 degrees.

So, for 1 hour the plane is heading at 270 degrees and for 3 hours, from 10:00 p.m to 1:00 a.m, the plane is heading at 310 degrees as shown in the figure.

As, distance = time x speed, so

The distance covered at 270 degrees, $d_1 = 1\times610=610$ miles.

The distance covered at 310 degrees, $d_2 = 3\times610=1830$ miles.

Total distance covered, d, is the magnitude of the sum of vectors $\vec{d_1}$ and $\vec{d_2}$ as shown in the figure.

The angle between the vectors $\vec{d_1}$ and $\vec{d_2}$ , $\theta=310-270=40$ degree.

Magnitude of sum of the vectors \vec{d_1} and \vec{d_2},

$d= \sqrt{|\vec{d_1}|^2+|\vec{d_2}|^2+2|\vec{d_1}|\;|\vec{d_2}|\cos\theta} \\\\\Rightarrow d=\sqrt{610^2+1830^2+2|\vec{d_1}|\;|\vec{d_2}|\cos\(40^{\circ})}$