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cupoosta [38]
3 years ago
13

Help I am very bad at math I need answers by tomorrow if you can help please do​

Mathematics
1 answer:
d1i1m1o1n [39]3 years ago
3 0

Answer:

Step-by-step explanation:

(a - b)(a +b) = a² - b²

1 - Sin² A = Cos² A

LHS = \frac{1}{1- Sin A} + \frac{1}{1 + Sin A}\\\\= \frac{1*(1 + Sin A)}{(1- Sin A)(1 + Sin A)} + \frac{1*(1- Sin A)}{(1 + Sin A)(1- Sin A)}\\\\= \frac{1 + Sin A+ 1 - Sin A}{1^{2}-  Sin^{2} A}\\\\= \frac{2}{1 - Sin^{2} A}\\\\= \frac{2}{Cos^{2} A}\\\\= 2 Sec^{2} A

2)  Sec² A - Tan² A = 1

LHS = \frac{1}{Sec A - Tan A}\\\\=\frac{1*(Sec A + Tan A)}{(Sec A -  Tan A)(Sec A + Tan A)}\\\\=\frac{Sec A + Tan A}{Sec^{2} A - Tan^{2} A}\\\\=\frac{Sec A + Tan A }{1}\\\\= Sec A + Tan A = RHS\\\\\\

3) LHS  = Cosec² A + Cot² A

             = Cosec² A +  Cosec² A - 1

            = 2Cosec² A - 1   = RHS

4) LHS = \frac{Sec A}{Cos A}- \frac{Tan A}{Cot A}\\\\          = Sec A*\frac{1}{Cos A}-Tan A*\frac{1}{Cot A}\\\\ = Sec A * Sec A - Tan A * Tan A\\\\= Sec^{2} A - Tan^{2} A \\\\= 1

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We know that
case 1)
Applying the law of sines
a/Sin A=b/Sin B
A=56°
a=12
b=14
so
a*Sin B=b*Sin A----> Sin B=b*Sin A/a---> Sin B=14*Sin 56°/12
Sin B=0.9672
B=arc sin (0.9672)------> B=75.29°-----> B=75.3°

find angle C
A+B+C=180°-----> C=180-(A+B)----> C=180-(56+75.3)----> C=48.7°

find c
a/Sin A=c/Sin C----> c=a*Sin C/Sin A----> c=12*Sin 48.7°/Sin 56°)
c=10.87-----> c=10.9

the answer Part 1)
the dimensions of the triangle N 1
are
a=12  A=56°
b=14  B=75.3°
c=10.9  C=48.7°

case 2) 
A=56°
a=12
b=14
B=180-75.3----> B=104.7°

find angle C
A+B+C=180°-----> C=180-(A+B)----> C=180-(56+104.7)----> C=19.3°

find c
a/Sin A=c/Sin C----> c=a*Sin C/Sin A----> c=12*Sin 19.3°/Sin 56°)
c=4.78-----> c=4.8

the answer Part 2)
the dimensions of the triangle N 2
are
a=12  A=56°
b=14  B=104.7°
c=4.8    C=19.3°
3 0
3 years ago
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