NEED HELP ASAP!! This is a trigonometry question and I really need help, I do not understand it at all. thank you.
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Answer:
AD = 11.87
Step-by-step explanation:
So first we need to figure out the missing angles.
In ΔBCD, we have two angles given, 40° and 25°, since the angles of a triangle must equal 180°, we need to subtract the sum of these two angles from 180° to find the remaining angle: 180° - (40° + 25°) = 115°
Now we can find the remaining angles in ΔABD. To find ∠D, we can subtract the 115° on the other side from 180° because segment AC is a straight line, which is 180°. This makes that angle 65°. Following the same steps we did before, we can subtract the sum of the two angles in the triangle from 180° to find the remaining angle: 180° - (68° + 65°) = 47°
Now that we have found all the angles, we can start finding the lengths of the segments by using the identity [tex]\frac{a}{sin(A)}=\frac{b}{sin(B)}=\frac{c}{sin(C)} =\frac{d}{sin(D)}[/tex]
So we only need one segment length, BD, in order to find segment AD. To find segment BD, we can use [tex]\frac{b}{sin(B)}=\frac{c}{sin(C)}[/tex], in this case [tex]\frac{DC}{sin(B)}=\frac{BD}{sin(C)}[/tex]
Solving this equation for BD, we get [tex]\frac{DCsin(C)}{sin(B)}={BD}[/tex]
Plugging in the values we have we get [tex]BD =\frac{15sin(25)}{sin(40)} =9.86[/tex]
Now we can go over to ΔABD and use [tex]\frac{a}{sin(A)}=\frac{b}{sin(B)}[/tex], in this case [tex]\frac{BD}{sin(A)}=\frac{AD}{sin(B)}[/tex]
Solving this equation for AD, we get [tex]\frac{BDsin(B)}{sin(A)} =AD[/tex]
Plugging in the values we have we get [tex]AD=\frac{9.36sin(68)}{sin(47)} =11.87[/tex]
Segment AD = 11.87