Answer:
A + B + C = π ...... (1)
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L.H.S.
= ( cos A + cos B ) + cos C
= { 2 · cos[ ( A+B) / 2 ] · cos [ ( A-B) / 2 ] } + cos C
= { 2 · cos [ (π/2) - (C/2) ] · cos [ (A-B) / 2 ] } + cos C
= { 2 · sin( C/2 ) · cos [ (A-B) / 2 ] } + { 1 - 2 · sin² ( C/2 ) }
= 1 + 2 sin ( C/2 )· { cos [ (A -B) / 2 ] - sin ( C/2 ) }
= 1 + 2 sin ( C/2 )· { cos [ (A-B) / 2 ] - sin [ (π/2) - ( (A+B)/2 ) ] }
= 1 + 2 sin ( C/2 )· { cos [ (A-B) / 2 ] - cos [ (A+B)/ 2 ] }
= 1 + 2 sin ( C/2 )· 2 sin ( A/2 )· sin( B/2 ) ... ... ... (2)
= 1 + 4 sin(A/2) sin(B/2) sin(C/2)
= R.H.S. ............................. Q.E.D.
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In step (2), we used the Factorization formula
cos x - cos y = 2 sin [ (x+y)/2 ] · sin [ (y-x)/2 ]
Step-by-step explanation:
Answer:
Step-by-step explanation:
D)hope it helpef
Answer:
65
Step-by-step explanation:
You can use f(15) = 40 to solve for C, then find f(0), the initial temperature.
40 = f(15)
40 = Ce^(-0.045·15) +14 = .50916C +14
26 = .50916C
26/.50916 = C ≈ 51.065
Then f(0) is ...
f(0) = 51.065·e^0 +14 = 65.065 ≈ 65
The initial temperature of the water was 65 degrees Fahrenheit.
A wall is in the shape of a trapezoid and it can be divided into a rectangle and a triangle. A triangle is with angles 45°- 45° - 90°. The hypotenuse of that triangle is 13√2 ft.Using the 45° - 45° - 90° theorem, sides of that triangle are in the proportion:x : x : x√2, and since that x√2 = 13√2 ( hypotenuse ), x = 13.Therefore h = 13 ft.We can check it: c² = 13² + 13²,c² = 169 + 169c² = 338c = √ 338 = 13√2Answer: h = 13 ft
