Triangle A B C is reflected across the line of reflection m to form triangle A prime B prime C prime. Triangle A prime B prime C

Answer:

The correct statements are;

1) ΔBCD is similar to ΔBSR

2) BR/RD = BS/SC

3) (BR)(SC) = (RD)(BS)

Step-by-step explanation:

1) Since RS is parallel to DC, we conclude that;

∠BDC = ∠BRS (Angles formed on the same side of the transversal)

Furthermore;

∠BCD = ∠BSR (Angles formed on the same side of the transversal)

∠CBD = ∠CBD (Reflexive property)

Thus;

ΔBCD ~ ΔBSR by the Angle-Angle-Angle (AAA) similarity criterion.

2) Given that  ΔBCD ~ ΔBSR, we obtain;

BC/BS = BD/BR → (BS + SC)/BS = (BR + RD)/BR = 1 + SC/BS = RD/BR + 1

1 + SC/BS = 1 + RD/BR thus, SC/BS = 1 + BR/RD - 1

SC/BS = RD/BR

By inverting both sides we find;

BR/RD = BS/SC

3) From BR/RD = BS/SC, we apply cross multiplication;

BR/RD = BS/SC leads to;

BR × SC = RD × BS → (BR)(SC) = (RD)(BS).

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Assuming arcs are measured in degrees, let S represent the following sum:

S = sin 1° + sin 2° + sin 3° +... + sin 359° + sin 360°


By rearranging, S can be reformulated as

S = [sin 1° + sin 359°] + [sin 2° + sin 358°] +... + [sin 179° + sin 181°] + sin 180° +
+ sin 360°

S = [sin 1° + sin(360° – 1°)] + [sin 2° + sin(360° – 2°)] +... + [sin 179° + sin(360° – 179)°]
+ sin 180° + sin 360°          (i)


However, for any real k,

sin(360° – k) = – sin k


Thus,

S = [sin 1° – sin 1°] + [sin 2° – sin 2°] +... + [sin 179° – sin 179°] + sin 180° + sin 360°

S results in 0 + 0 +... + 0 + 0 + 0        (... since sine of 180° and 360° are both equal to 0)

Therefore, S equals 0.


Each pair within the brackets negates itself, leading the sum to total zero.

∴   sin 1° + sin 2° + sin 3° +... + sin 359° + sin 360° equals 0.          ✔


I hope this clarifies things. =)


Tags:  sum summatory trigonometric trig function sine sin trigonometry

18

He can combine one hat with a scarf, resulting in 18 different combinations for Sam when multiplying 6 by 3.

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When rounding 1199.28856995 to the nearest tenth, the result is 119.3.

I hope this is helpful!