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2 years ago

The three angles of a triangle measure x+37, 90, and x+67. Find the measurement of the smallest angle in degrees.

Mathematics

2 answers:

Kisachek [45]2 years ago

7 0

Step-by-step explanation:

so the smallest angle is angle A which is 30 degree

Gelneren [198K]2 years ago

3 0

Answer:

The smallest angle in the triangle is 30°

Step-by-step explanation:

You can solve this by including the fact that the sum of the angles in the triangle has to be 180 degrees. We're given values describing the angles, so we can simply add them together, say the equal 180, and solve for x:

$(x + 37) + 90 + (x + 67) = 180\\2x + 37 + 90 + 67 = 180\\2x + 194 = 180\\2x = 180 - 194\\2x = -14\\x = -7$

The smallest angle then is x + 37, or 37 - 7, giving us 30°

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2 years ago

which order pair is a solution of the equation? y-4=7(x-6) A. only (5,4) B. only (6,5) C. both (5,4) and (6,5) D. neither

Svet_ta [14]

Answer:

option D. Neither

Step-by-step explanation:

y-4=7(x-6)

using (5,4):

4-4=7(5-6)

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5-4=7(6-6)

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2 years ago

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Virty [35]

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2 years ago

A parallelogram has sides 10m and 12m and an angle of 45°. Find the distance between the 12m-sides.

stich3 [128]

Answer:

The distance between 12m-sides is $5\sqrt{2}m$ .

Step-by-step explanation:

It is given that the parallelogram has sides 10m and 12m and an angle of 45°.

Draw an altitude from one 12m side to another 12 m side as shown in below figure.

The opposite angles of parallelogram are same. Two angles are obtuse angles and two are acute angle.

Since angle C is acute angle therefore it must be 45 degree.

$\sin\theta=\frac{perpendicular}{hypotenuse}$

$\sin C=\frac{BE}{BC}$

$\sin(45^{\circ})=\frac{d}{10}$

$\frac{1}{\sqrt{2}}=\frac{d}{10}$

$\frac{10}{\sqrt{2}}=d$

$\frac{10}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}=d$

$\frac{10\sqrt{2}}{2}=d$

$5\sqrt{2}=d$

Therefore the distance between 12m-sides is $5\sqrt{2}m$ .

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2 years ago

Simplify 1- cos^2 Θ / cos^2 Θ Answers sin2 θ cos2 θ tan2 θ 1

Tems11 [23]

Answer:

Trigonometric identities

$\large \begin{aligned}\cos^2 \theta + \sin^2 \theta & = 1\\\implies \quad \quad \sin^2 \theta & = 1 - \cos^2 \theta\\\\\tan \theta=\dfrac{\sin \theta}{\cos \theta} \end{aligned}$

Therefore, using the identities, the given expression can be simplified as follows:

$\large\begin{aligned}\implies \dfrac{1- \cos^2 \theta}{\cos^2 \theta} & = \dfrac{\sin^2 \theta}{\cos^2 \theta}\\\\& = \tan^2 \theta\end{aligned}$

3 0

1 year ago

Read 2 more answers