solution:
Z1 = 5(cos25˚+isin25˚)
Z2 = 2(cos80˚+isin80˚)
Z1.Z2 = 5(cos25˚+isin25˚). 2(cos80˚+isin80˚)
Z1.Z2 = 10{(cos25˚cos80˚ + isin25˚cos80˚+i^2sin25˚sin80˚) }
Z1.Z2 =10{(cos25˚cos80˚- sin25˚sin80˚+ i(cos25˚sin80˚+sin25˚cos80˚))}
(i^2 = -1)
Cos(A+B) = cosAcosB – sinAsinB
Sin(A+B) = sinAcosB + cosAsinB
Z1.Z2 = 10(cos(25˚+80˚) +isin(25˚+80˚)
Z1.Z2 = 10(cos105˚+ isin105˚)
Step-by-step explanation:
ABC is an isoceles triangle (both legs are equally long). and AB is its baseline.
OC is now a median for ABC splitting the angle at C and AB in half.
so, we have 2 right-angled triangles : OAC and OBC.
the half-angles at C are 42/2 = 21°.
the angles at A and B are 90°.
and the half-angles at O are 180 - 90 - 21 = 69°.
remember that the sum of all angles in a triangle must always be 180°.
AB are the heights of both of these triangles.
the single height is sin(69)×7 = 6.535062985... cm
and so,
AB = 2× height = 13.07012597... cm.
Your answer should be 101°. 52+53+154= 259. 360-259= 101. Hope that helps
Answer:
The decimal number is multiplied by 100
Step-by-step explanation:
Answer:
Step-by-step explanation:
From the picture attached,
a). Triangle in the figure is ΔBCF
b). Since, and are the parallel lines and m is a transversal line,
m∠FBC = m∠BFG [Alternate interior angles]
Since, and are the parallel lines and n is a transversal line,
m∠BCF = m∠CFE [Alternate interior angles]
By triangle sum theorem in ΔBCF
m∠FBC + m∠BCF + m∠BFC = 180°
From the properties given above,
m∠BFG + m∠CFE + m∠BFC = 180°
m∠GFE = 180°
Therefore, angle GFE is the straight angle that will be useful in proving that the sum of the measures of the interior angles of the triangle is 180°.