It'd be 9.096491 hope that helps.
Answer:
17.7 cm^2
Step-by-step explanation:
Use trig to find the height of the triangle. Then the area is bh/2.
Extend side BC to the right until it is vertically below point A. Draw a segment from point A vertically down until it intersects the extension of side BC. Call the point of intersection D. <D is a right angle.
Use triangle ABD to find the height, AD, of triangle ABC.
For <B of 37 deg, AD is the opposite leg. AB is the hypotenuse. The trig ratio that relates the opposite lefg to the hypotenuse is the sine.
sin B = opp/hyp
sin 37 deg = AD/13.1
AD = 13.1 * sin 37 deg
AD = 7.9
AD is the height of triangle ABC. BC is the base. We can find the area of triangle ABC.
area = bh/2
area = (4.5 cm)(7.9 cm)/2
area = 17.7 cm^2
Answer:
.05 im sure of it hope it helps.
Step-by-step explanation:
Answer:
angle t is 55 degrees
Step-by-step explanation:
Here's my work:
Isoscelese means two of the same length sides. A triangle is 180 degrees. 180-70=110
110/2=55 (two because those are the two angles associated with the equal sides)
A.)
<span>s= 30m
u = ? ( initial velocity of the object )
a = 9.81 m/s^2 ( accn of free fall )
t = 1.5 s
s = ut + 1/2 at^2
\[u = \frac{ S - 1/2 a t^2 }{ t }\]
\[u = \frac{ 30 - ( 0.5 \times 9.81 \times 1.5^2) }{ 1.5 } \]
\[u = 12.6 m/s\]
</span>
b.)
<span>s = ut + 1/2 a t^2
u = 0 ,
s = 1/2 a t^2
\[s = \frac{ 1 }{ 2 } \times a \times t ^{2}\]
\[s = \frac{ 1 }{ 2 } \times 9.81 \times \left( \frac{ 12.6 }{ 9.81 } \right)^{2}\]
\[s = 8.0917...\]
\[therfore total distance = 8.0917 + 30 = 38.0917.. = 38.1 m \] </span>
