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2 years ago

Can someone pls solve for y

Mathematics

1 answer:

Paha777 [63]2 years ago

8 0

Answer:

y =7.642881742

Step-by-step explanation:

tan A = opp side/adjacent side

tan 63 = 15/y

y tan 63 = 15

y = 15/ tan 63

y =7.642881742

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Root 3 cosec140° - sec140°=4 prove that

Lorico [155]

Answer:

Step-by-step explanation:

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

Proof:

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

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

Also note that sin(B-C) = sinBcosC - cosBsinC

sin420cos140 - cos420sin140 = sin(420-140)

The resulting equation becomes;

$\frac{4(sin(420-140)) }{sin280}$

= $\frac{4sin280}{sin280}\\ = 4 \ Proved!$

3 0

2 years ago

The approximate time between two high tides or two low tides is?

NeX [460]

Answer:

It's (very roughly) 12 hours from one high tide to the next, and of course the same for low tides. That's because the tides produced when the Moon is directly overhead or in the opposite direction are the same. Same with two low tides, since there are two tides a day (both low and high).

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2 years ago

Read 2 more answers

Wich one is bigger 3/3 or 7/8

MrRissso [65]

3/3=1 and 7/8 is < 1 so 3/3 is bigger

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2 years ago

How do you solve secx-2=0

siniylev [52]

It will be secx = 2
or, cosx = 1/2
or x = Π/3 , 5Π/3

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