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ch4aika [34]
3 years ago
5

How do i do question 3 a b c

Mathematics
1 answer:
viva [34]3 years ago
3 0

Answer:

a) 240°

b) 30°

c) 225°

Step-by-step explanation:

To solve these equations you have to use the inverse of the given trigonometric functions. The inverse of <em>sin</em> is <em>arcsin</em>, and the inverse of <em>tan </em>is <em>arctan. </em>Instead of giving an angle, what is its sine?, the question is: given a sine,  what is the angle?.

a)

sin(θ) = -√3/2

θ = arcsin(-√3/2)

θ = -60°

Given the periodicity of sine function, sin(-60°) is equivalent to sin(240°) (-60+180) and sin(300°) (-60+360).

b)

tan(θ) = 1/√3

θ = arctan(1/√3)

θ = 30°

c)

csc means cosecant, by definition:

csc(θ) = 1/sin(θ)

csc(θ) = -√2

1/sin(θ) = -√2

sin(θ) = -1/√2

θ = arcsin(-1/√2)

θ = -45° or 360-45 = 315° or 180+45 = 225°

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In the diagram below, P and Q are points on a circle with centre O.
Gnoma [55]

Answer:

<h2>∠PQT = 72°</h2>

Step-by-step explanation:

According to the diagram shown, ∠OPQ = ∠OQP = 18°. If PQT is a tangent to the circle, it can be inferred that line OQ is perpendicular to line QT. Ths shows that ∠OQT = 90°.

Also from the diagram, ∠OQP + ∠PQT = ∠OQT;

∠PQT  = ∠OQT -  ∠OQP

Given ∠OQP = 18° and ∠OQT = 90°

∠PQT = 90°-18°

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3 years ago
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Artyom0805 [142]
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Maybe the fifth difference is -7.
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<span>3, -6, 12, 4, 20, 13</span>
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How can I solve this proof?? PLZ HELP I DONT UNDERSTAND!! Thanks !!
love history [14]

Hello from MrBillDoesMath!

Answer:

See below


Discussion:

Triangles BAD and BCD are congruent by the SAS postulate.

          * Side AD = DC because BD is the perpendicular bisector  of AC.

          * Angle ADB and CDB are both right angles, and hence equal, because

            BD is the perpendicular bisector of AC)


Corresponding parts of congruent triangles are equal so angle BAD = angle BCD,  which was to be proven.


Regards,  

MrB

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