## anonymous 2 years ago Solve the triangle. Round lengths to the nearest tenth and angle measures to the nearest degree. a = 7, b = 7, c = 5

1. anonymous

B. A = 69°, B = 69°, C = 42°??

2. anonymous

sounds good.

3. anonymous

wrong

4. anonymous

Try Using cosine law

5. anonymous

$7^{2} = 7^{2} + 5^{2} - 2(7)(5)cosA$

6. anonymous

my answers need to be in degrees

7. anonymous

you get $cosA = \frac{ -25 }{ -70 }$ than do arc cosine. you get angle 69.07

8. anonymous

i need degrees for a b n c ??

9. anonymous

you get angles 41.8 and 69.07. again

10. anonymous

A. A = 70°, B = 70°, C = 40° B. A = 69°, B = 69°, C = 42° C. A = 42°, B = 69°, C = 69° D. A = 69°, B = 42°, C = 69°

11. anonymous

|dw:1375984676609:dw| ?

12. anonymous

Try D.

13. anonymous

have to go.

14. anonymous

thanks

15. anonymous

deff not 69 deg

16. anonymous

Note that the triangle is isosceles:|dw:1375984026354:dw|

17. anonymous

ok ? still confused

18. anonymous

Now use the cosine law:$\bf a^2=b^2+c^2-2bc \cos(A) \implies 5^2=7^2+7^2-2(7)(7)\cos(A)$Note that 'A' is the angle opposite side BC. Now re-arrange and solve for cos(A):$\bf \implies \cos(A)=\frac{-73}{-98}=0.7445 \implies \cos^{-1}(0.7445)=A=41.884 \ degrees$

19. anonymous

Now because the triangle is isosceles, the remaining two angles B and C are equal so we can find each of them by subtracting A from 180 and dividing the result by 2:$\bf \angle B=\angle C=\frac{ 180 -41.884}{ 2 }=69.058 degrees$Rounding off our results to the nearest degree we get:$\bf \angle A =42 \ degrees$$\bf \angle B=\angle C=69 \ degrees$ @britbrat4290

20. anonymous