Question

find all solutions for a triangle with c=70 c=24 and a=25

Answer 1

Answer:

see the attachments for the two solutions

Step-by-step explanation:

When the given angle is opposite the shorter of the given sides, there will generally be two solutions. The exception is the case where the triangle is a right triangle (the ratio of the given sides is equal to the sine of the given angle). If the given angle is opposite the longer of the given sides, there is only one solution.

When a side and its opposite angle are given, as here, the law of sines can be used to solve the triangle(s). When the given angle is included between two given sides, the law of cosines can be used to solve the (one) triangle.

___

Here, the law of sines can be used to solve the triangle:

A = arcsin(a/c·sin(C)) = arcsin(25/24·sin(70°)) = 78.19° or 101.81°

B = 180° -70° -A = 31.81° or 8.19°

b = 24·sin(B)/sin(70°) = 13.46 or 3.64

Answer 2

Answer:

Value of b is 13 , m∠B = 31° and m∠A = 79°.

Step-by-step explanation:

Given: m∠C = 70°  , c = 24  and a = 25

To find : All solutions of triangle that is m∠B , m∠A and b

We use law of sines which states that the ratio of the length of a side of triangle to the sine of the angle opposite that side is the same for all sides and angles in a given triangle.

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

Consider,

$\frac{a}{sin\,A}=\frac{c}{sin\,C}$

$\frac{25}{sin\,A}=\frac{24}{sin\,70}$

$sin\,A=\frac{25\times sin\,70}{24}$

$sin\,A=\frac{25\times0.94}{24}$

sin A = 0.98

sin A = sin 79

A = 79° (approx.)

∠A + ∠B + ∠C = 180°   ( Angle Sum Property of triangle )

79 + ∠B + 70 = 180

∠B = 180 - 149

m∠B = 31°

now Consider,

$\frac{a}{sin\,A}=\frac{b}{sin\,B}$

$\frac{25}{sin\,79}=\frac{b}{sin\,31}$

$b=\frac{25\times sin\,31}{sin\,79}$

$b=\frac{25\times0.52}{0.98}$

b = 13 (approx.)

Therefore, Value of b is 13 , m∠B = 31° and m∠A = 79°.