Question

8-10 deal with the ambiguous SSA case. for each find all possible solutions and sketch the triangle in each case.

Answer 1

Answer:

8.) C = 129.341°

9.) C = 90°

10.) C = 90°

Step-by-step explanation:

For part 8.

given A=50°, a=15, b=12

In any triangle, the ratio of side length of to the sine of its opposite angle  

is the same for all three sides:

$\frac{a}{SinA}=\frac{b}{SinB}=\frac{c}{SinC}$

Now put the value of  A=50°, a=15, b=12 in above formula

$\frac{15}{Sin50}=\frac{12}{SinB}$

Since, the value of Sin50°=0.766

$\frac{15}{0.766}=\frac{12}{SinB}$

$\frac{15}{0.766}=\frac{12}{SinB}$

$19.58=\frac{12}{SinB}$

simplify the above,

$SinB=\frac{12}{19.58}$

$SinB=0.6128$

take inv sin both the sides,

$B=0.659\,\text{degree}$

since some of angles of triangle is 180

so,

         A + B + C = 180°

50 + 0.659 + C = 180°

     50.659 + C = 180°

subtract 50.659 from both the sides,

                     C = 180°-50.659

                    C = 129.341°

For part 9

given A=50°, a=10, b=15

Since, $\frac{a}{SinA}=\frac{b}{SinB}=\frac{c}{SinC}$

Now put the value of  A=50°, a=10, b=15 in above formula

$\frac{10}{Sin50}=\frac{15}{SinB}$

Since, the value of Sin50°=0.766

$\frac{10}{0.766}=\frac{15}{SinB}$

$13.0548=\frac{15}{SinB}$

simplify the above,

$SinB=\frac{15}{13.0548}$

$SinB=1.149$

take inv sin both the sides,

$B=40\,\text{degree}$

since some of angles of triangle is 180

so,

         A + B + C = 180°

50° + 40° + C = 180°

        90° + C = 180°

subtract 90 from both the sides,

                    C = 180°-90°

                    C = 90°

             

For Part 10

given A=47°, a=1.5, b=2

Since, $\frac{a}{SinA}=\frac{b}{SinB}=\frac{c}{SinC}$

Now put the value of  A=47°, a=1.5, b=2 in above formula

$\frac{1.5}{Sin47}=\frac{2}{SinB}$

Since, the value of Sin47°=0.123

$\frac{1.5}{0.123}=\frac{2}{SinB}$

$12.195=\frac{2}{SinB}$

simplify the above,

$SinB=\frac{15}{12.195}$

$SinB=1.23$

take inv sin both the sides,

$B=43\,\text{degree}$

since some of angles of triangle is 180

so,

         A + B + C = 180°

    47° + 43° + C = 180°

             90° + C = 180°

subtract 90 from both the sides,

                    C = 180°-90°

                    C = 90°

Answer 2

For problem 8:

sin A/a sinB/b = sinC/c 

so sin50/15 = sinB/12 

0.766/15 =sinB/12 

0.051066 = sin B/12 

sin B = 0.6128 
B = 37.79 degrees 

sum of angles must equal 180 deg therefore C = 180-50-37.79 = 92.21 degrees and .766/15 = sin92.21/c 

.051066= 0.99925/c 
c = .99925/0.051066 = 19.567 

Problems 9 and 10 can also be done with the same method as problem 8.

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