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Neporo4naja [7]

2 years ago

5

An equilateral triangle ΔLMQ (see picture to the right) is drawn in the interior of square LMNP so that side LM is common. Find

m∠LQP.

1 answer:

Tom [10]2 years ago

4 0

Answer:

lqp=75 degrees

Step-by-step explanation:

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If mA=mB and mA + mC=D, then mB+mC=D, which property is shown?

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3 years ago

ludmilkaskok [199]

Given:

The point (9,-12) is on terminal side of angle theta in standard position.

To find:

The exact value of each of the six trigonometric functions of theta.

Solution:

The given point is (9,-12). Here, x-coordinate is positive and y-coordinate is negative. So, the point lies in 4th quadrant and only cos and sec are positive in 4th quadrant.

We know that,

$r=\sqrt{x^2+y^2}$

$r=\sqrt{9^2+(-12)^2}$

$r=\sqrt{81+144}$

$r=\sqrt{225}$

$r=15$

Now,

$\sin \theta=\dfrac{y}{r}=\dfrac{-12}{15}=-\dfrac{4}{5}$

$\cos \theta=\dfrac{x}{r}=\dfrac{9}{15}=\dfrac{3}{5}$

$\tan \theta=\dfrac{y}{x}=\dfrac{-12}{9}=-\dfrac{4}{3}$

$\cot \theta=\dfrac{1}{\tan \theta}=-\dfrac{3}{4}$

$\text{cosec} \theta=\dfrac{1}{\sin \theta}=-\dfrac{5}{4}$

$\sec \theta=\dfrac{1}{\cos \theta}=\dfrac{5}{3}$

Therefore, the values of six trigonometric functions of theta are $\sin \theta=-\dfrac{4}{5},\cos \theta=\dfrac{3}{5},\tan \theta=-\dfrac{4}{3},\cot \theta=-\dfrac{3}{4},\text{cosec} \theta=-\dfrac{5}{4},\sec \theta=\dfrac{5}{3}$ .

6 0

3 years ago

The value of 2(cos square 45 + tan square 60) is

Sergeu [11.5K]

Answer:

7

Step-by-step explanation:

Using the exact values for

cos45° = $\frac{1}{\sqrt{2} }$ and tan60° = $\sqrt{3}$ , then

2(cos²45° + tan²60°)

= 2( ( $\frac{1}{\sqrt{2} }$ )² + ( $\sqrt{3}$ )² )

= 2( $\frac{1}{2}$ + 3 ) ← distribute

= 1 + 6

= 7

7 0

2 years ago