# How do you solve the right triangle ABC if b=120m, A=12?

Mar 26, 2018

color(maroon)("Case 1 : "
color(indigo)(a = 24.95m,hatB = 90^@, hatC = 78^@, c = 117.38m, A_t = 1464.32 " sq m"

color(green)("Case 2 : "
color(green)(a = 25.51m, hatB = 78^@, hatC = 90^@, c = 122.68m, A_t = 1530.6 " sq m"

#### Explanation:

Case 1 :

$b = 120 m , \hat{B} = {90}^{\circ} , \hat{A} = {12}^{\circ}$

$\hat{C} = 180 - 90 - 12 = {78}^{\circ}$

Applying the Law of Sines,

a / sin A = B / sin B = c ? sin C

$a = \frac{120 \cdot \sin 12}{\sin} 90 = 24.95 \text{ m}$

$c = \frac{120 \cdot \sin 78}{\sin} 90 = 117.38 \text{ m}$

$\text{Area of rt triangle " A_t = (1/2) * a * c = (1/2) * 24.95 * 117.38 = color(indigo)(1464.32 " sq m}$

Case 2 :

$b = 120 m , \hat{A} = {12}^{\circ} , \hat{A} = {90}^{\circ}$

$\hat{B} = 180 - 12 - 90 = {78}^{\circ}$

color(green)("Applying the Law of Sines,"

$\frac{a}{\sin} A = \frac{b}{\sin} B = \frac{c}{\sin} C$

$a = \frac{b \cdot \sin A}{\sin} B = \frac{120 \cdot \sin 12}{\sin} 78 = 25.51 \text{ m}$

$\text{Applying Pythagoras theorem,}$

$c = \sqrt{{a}^{2} + {b}^{2}} = s q r \left({25.51}^{2} + {120}^{2}\right) = 122.68 \text{ m}$

$\text{Area of rt triangle "A_t = (1/2) * a * b = (1/2) * 25.51 * 120 = color(green)(1530.6 " sq m}$