I need help on trigonometry

Question

In ?ABC, m∠A=52 °, c=11, and m∠B=19°. Find a to the nearest tenth

Answer 1

 

In all triangles, the sum of the internal angles is 180°.

The measure of 2 of the 3 angles is given ... ∠A = 52°, and ∠B = 19°.

Thus, ∠C = 180º - (52º + 19º) = 180° - 71° = 109°.

 

We can use the law of sines to find the length of side "a".

 

    a                  c

 ---------  = -------------

 sine A          sine C

 

     a                 11

----------- = --------------

sine 52°       sine 109º

 

Multiply both sides by sine 52°

 

a•(sine 52°)           11•(sine 52°)

------------------- = ----------------------

   sine 52°                  sine 109°

 

a = 11•(.986628)÷(.816743)

a = 9.17°

 

Rounded to the nearest tenth ... a = 9.2°

 

NOTE:  Make sure your calculator is in degree mode (not radian mode)

 

 

Answer 2

Hi Gwendolyn.

 

We can use two laws/properties about triangles here.

1) the sum of the angles of a triangle is 180^o

With this we have A = 52^o and B = 19^o.  There fore C = 180-(52+19) = 109^o

 

2) The second is the law of sines, which says  sinA/a = sinB/b = sinC/c

We have A, c and C so we can calculate SinA and SinC

Sin A = 0.788

Sin C = 0.9455

 

Then SinA/a = SinC/c

 

.788/a = .9455/11 and a = 9.168 or 9.2 rounded to the nearest tenth.