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3 years ago

URGENT PLEASE IF UR GOOD AT MATH

Mathematics

1 answer:

PIT_PIT [208]3 years ago

6 0

9514 1404 393

Answer:

a) θ = 14°

b) θ = 62°, A = 28°

Step-by-step explanation:

The mnemonic SOH CAH TOA is intended to remind you of the relation ...

Tan = Opposite/Adjacent

This is the useful relation in each case.

a) tan(θ) = 43/172

θ = arctan(43/172) ≈ 14°

b) tan(θ) = 32/17

θ = arctan(32/17) ≈ 62°

A = complement of θ = 90° -62° = 28°

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8d – 4d – 6d – 8 = 2d

Combine common terms:
-2d - 8 = 2d

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Swtich sides:
4d = -8

Divide both sides by 4:
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Answer: - 2

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Answer

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step-by-step explanation:

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3 years ago

Match each of the trigonometric expressions below with the equivalent non-trigonometric function from the following list. Enter

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Answer:

Match each of the trigonometric expressions below with theequivalent non-trigonometric function from the following list.Enter the appropiate letter(A,B, C, D or E)in each blank

A . tan(arcsin(x/8))

B . cos (arsin (x/8))

C. (1/2)sin (2arcsin (x/8))

D . sin ( arctan (x/8))

E. cos (arctan (x/8))

These are the spaces to fill out :

.. ..........x/64 (sqrt(64-x^2))

.............x/sqrt(64+x^2)

.............sqrt(64-x^2)/8

..............x/sqrt(64-x^2)

..............8/sqrt(64+x^2)

A. ........tan(arcsin(x/8)) =......x/sqrt(64-x^2)

B . cos (arsin (x/8)) ....sqrt(64-x^2)/8

Step-by-step explanation:

To solve this we have to find the missing sides to each of the triange discribed in prenthesis thus

A we have the sides of the triangle given by x, 8 and $\sqrt{8^{2} - x^{2} }$ or $\sqrt{64 - x^{2} }$

thus tan(arcsin(x/8)) = $\frac{x}{\sqrt{64 - x^{2} }}$ =

Therefore ........tan(arcsin(x/8)) =......x/sqrt(64-x^2)

Here we have cos = adjacent/hypotenuse where adjacent side is $\sqrt{64 - x^{2} }$ and hypothenuse = 8 we have $\sqrt{64 - x^{2} }$ /8

B . cos (arsin (x/8)) ....sqrt(64-x^2)/8

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3 years ago