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3 years ago

Solve the formula thanks peeps

Mathematics

2 answers:

alukav5142 [94]3 years ago

5 0

P=2b+2c
p-2b=2b+2c-2b
p-2b=2c
divide both sides by 2
(p-2b)/2=2c/2
(p-2b)/2=c
p/2-2b/2=c
p/2-b=c so the correct answer is D

fenix001 [56]3 years ago

3 0

P = 2b + 2c...subtract 2b from both sides
P - 2b = 2c ...divide both sides by 2
(P - 2b) / 2 = c
P/2 - b = c <=== ur answer is D

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3 years ago

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3 years ago

Solve the triangle. Round your answers to the nearest tenth. A. m∠A=43, m∠B=55, a=16 B. m∠A=48, m∠B=50, a=23 C. m∠A=48, m∠B=50,

alexgriva [62]

Answer:

D. m∠A=43, m∠B=55, a=20

Step-by-step explanation:

Given:

∆ABC,

m<C = 82°

AB = c = 29

AC = b = 24

Required:

m<A, m<C, and a (BC)

SOLUTION:

Find m<B using the law of sines:

$\frac{sin(B)}{b} = \frac{sin(C)}{c}$

$\frac{sin(B)}{24} = \frac{sin(82)}{29}$

$sin(B)*29 = sin(82)*24$

$\frac{sin(B)*29}{29} = \frac{sin(82)*24}{29}$

$sin(B) = \frac{sin(82)*24}{29}$

$sin(B) = 0.8195$

$B = sin^{-1}(0.8195)$

$B = 55.0$

m<B = 55°

Find m<A:

m<A = 180 - (82 + 55) => sum of angles in a triangle.

= 180 - 137

m<A = 43°

Find a using the law of sines:

$\frac{a}{sin(A)} = \frac{b}{sin(B)}$

$\frac{a}{sin(43)43} = \frac{24}{sin(55)}$

Cross multiply

$a*sin(55) = 25*sin(43)$

$a = \frac{25*sin(43)}{sin(53)}$

$a = 20$ (approximated)

8 0

3 years ago

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Answer:

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7 0

2 years ago

If sinA=√3-1/2√2,then prove that cos2A=√3/2 prove that

Ivan

Answer:

$\boxed{\sf cos2A =\dfrac{\sqrt3}{2}}$

Step-by-step explanation:

Here we are given that the value of sinA is √3-1/2√2 , and we need to prove that the value of cos2A is √3/2 .

Given :-

• $\sf\implies sinA =\dfrac{\sqrt3-1}{2\sqrt2}$

To Prove :-

• $\sf\implies cos2A =\dfrac{\sqrt3}{2}$

Proof :-

We know that ,

$\sf\implies cos2A = 1 - 2sin^2A$

Therefore , here substituting the value of sinA , we have ,

$\sf\implies cos2A = 1 - 2\bigg( \dfrac{\sqrt3-1}{2\sqrt2}\bigg)^2$

Simplify the whole square ,

$\sf\implies cos2A = 1 -2\times \dfrac{ 3 +1-2\sqrt3}{8}$

Add the numbers in numerator ,

$\sf\implies cos2A = 1-2\times \dfrac{4-2\sqrt3}{8}$

Multiply it by 2 ,

$\sf\implies cos2A = 1 - \dfrac{ 4-2\sqrt3}{4}$

Take out 2 common from the numerator ,

$\sf\implies cos2A = 1-\dfrac{2(2-\sqrt3)}{4}$

Simplify ,

$\sf\implies cos2A = 1 -\dfrac{ 2-\sqrt3}{2}$

Subtract the numbers ,

$\sf\implies cos2A = \dfrac{ 2-2+\sqrt3}{2}$

Simplify,

$\sf\implies \boxed{\pink{\sf cos2A =\dfrac{\sqrt3}{2}} }$

Hence Proved !

8 0

3 years ago