PLEASE ANSWER CORRECTLY, I WILL GIVE BRAINLIEST <3 THANKS IN ADVANCE.

Mathematics

1 answer:

lilavasa [31]2 years ago

4 0

Answer:

B = 40° and D = 60°

Step-by-step explanation:

Just subtract the angles you already know from 90° (since perpendicular means right angle, which is 90°).

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1) construct 3 equations starting with x= -2

Kamila [148]

<h3>Hlo sis !!</h3><h3>ur answer is in above attatchment .</h3><h3>Hope this helps u sneha ✌✌</h3>

5 0

2 years ago

PLEASE HELP 15 POINTS<br><br><br> -8 - h = 2(-.5h + 4)

Vanyuwa [196]

Answer:

h= -6

Step-by-step explanation:

-8-h=2(-.5h+4)

-8-h=h+4

-2h=12

h=-6

3 0

3 years ago

Root 3 cosec140° - sec140°=4<br>prove that<br><br>

Lorico [155]

Answer:

Step-by-step explanation:

We are to show that $\sqrt{3} cosec140^{0} - sec140^{0} = 4\\$

<u>Proof:</u>

From trigonometry identity;

$cosec \theta = \frac{1}{sin\theta} \\sec\theta = \frac{1}{cos\theta}$

$\sqrt{3} cosec140^{0} - sec140^{0} \\= \frac{\sqrt{3} }{sin140} - \frac{1}{cos140} \\= \frac{\sqrt{3}cos140-sin140 }{sin140cos140} \\$

From trigonometry, 2sinAcosA = Sin2A

$= \frac{\sqrt{3}cos140-sin140 }{sin140cos140} \\\\= \frac{\sqrt{3}cos140-sin140 }{sin280/2}\\= \frac{4(\sqrt{3}/2cos140-1/2sin140) }{2sin280}\\\\= \frac{4(\sqrt{3}/2cos140-1/2sin140) }{sin280}\\since sin420 = \sqrt{3}/2 \ and \ cos420 = 1/2 \\ then\\\frac{4(sin420cos140-cos420sin140) }{sin280}$