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VLD [36.1K]
3 years ago
8

Solve for x someone help me I kinda suck at geometry

Mathematics
2 answers:
Lapatulllka [165]3 years ago
5 0

We have:

equilateral SIT triangle (hypothetical)

so angle S = angle I = angle T

Inferred: angle S = 180 degrees: 3 = 60 degrees

We have: angle S = (4x + 12) degrees = 60 degrees

                          4x = 60 degrees - 12 degrees

                          4x = 48 degrees

                             x = 48 degrees: 4

                             x = 12 degrees

So the value of x is 12 degrees

Setler [38]3 years ago
3 0

Answer:

48x

Step-by-step explanation:

Its the answer

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2 years ago
Prove that (Root of Sec A - 1 / Root of Sec A + 1) + (Root of Sec A + 1 / Root of Sec A - 1) = 2 cosec A
iVinArrow [24]

Answer:

The answer is below

Step-by-step explanation:

We need to prove that:

(Root of Sec A - 1 / Root of Sec A + 1) + (Root of Sec A + 1 / Root of Sec A - 1) = 2 cosec A.

Firstly, 1 / cos A = sec A, 1 / sin A = cosec A and tanA = sinA / cosA.

Also, 1 + tan²A = sec²A; sec²A - 1 = tan²A

\frac{\sqrt{secA-1} }{\sqrt{secA+1} } +\frac{\sqrt{secA+1} }{\sqrt{secA-1} } =\frac{(\sqrt{secA-1)}(\sqrt{secA-1})+(\sqrt{secA+1)}(\sqrt{secA+1}) }{(\sqrt{secA+1})(\sqrt{secA-1}) } \\\\=\frac{secA-1+(secA+1)}{\sqrt{sec^2A-secA+secA-1} } \\\\=\frac{2secA}{\sqrt{sec^2A-1} } \\\\=\frac{2secA}{\sqrt{tan^2A} } \\\\=\frac{2secA}{tanA} \\\\=\frac{2*\frac{1}{cosA} }{\frac{sinA}{cosA} }\\\\= 2*\frac{1}{cosA}*\frac{cosA}{sinA}\\\\=2*\frac{1}{sinAA}\\\\=2cosecA

7 0
3 years ago
a hexagon inscribed in a circle has three consecutive sides each of length 3 and three consecutive sides each of length 5. the c
allochka39001 [22]

According to given conditions, m+n is equal to 409.

Consider the diagram below.

In the hexagon ABCDEF, let

AB = BC = CD = 3;

DE = EF = FA = 5;

Arc BAF is equal to one-third of the circle's circumference.

Hence, ∠BCF = ∠BEF = 60°;

Similarly, ∠CBE = ∠CFE = 60°;

Let the point of intersection of BE and CF be P, BE and AD be Q and CF and AD be R.

∴ Δ EFP and Δ BCP are equilateral, and so Δ PQR is also equilateral.

Also, ∠ BAD and ∠ BED subtend the same arc and so do ∠ ABE and ∠ ADE.

∴ Δ ABQ is similar to Δ EDQ

\frac{AQ}{EQ}  = \frac{BQ}{DQ} = \frac{AB}{ED} = \frac{3}{5}

Also,

\frac{\frac{AD - PQ}{2} }{PQ + 5} = \frac{3}{5}  

and \frac{3 - PQ}{\frac{AD + PQ}{2} }  = \frac{3}{5}

On solving these simultaneous equations, we get AD = 360/49

∴ m + n = 409.

To learn more about similarity of triangles, refer to this link:

brainly.com/question/25882965

#SPJ4

4 0
1 year ago
PLEASE HELP!!!!!!!!!!!!!!!!!!!!!!!!
charle [14.2K]

Answer:

I think it is C.2920

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2 years ago
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What is the area of this pentagon?
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Answer:

56

Step-by-step explanation:

I thibk

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3 years ago
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