Question

PLEASE HELP Solve the right triangle.


Round your answers to the nearest tenth.

Answer 1

`\bold{\huge{\underline{ Solution }}}`

Let consider the given triangle be ABC

Here, It is given in the question that ,

• 
  `\sf{ {\angle} B = 90° }`
• 
  `\sf{ {\angle} C = 50° }`
• 
  `\sf{ AC = 14 }`

Therefore,

By using Angle sum property

• It states that the sum of all angles of triangles are equal to 180°

That is,

`\bold{\pink{ {\angle} A + {\angle}B + {\angle}C = 180{\degree}}}`

Subsitute the required values,

`\sf{ {\angle}A + 90{\degree} + 50{\degree}\: =\: 180{\degree} }`

`\sf{ {\angle}A + 140{\degree}\: = \:180{\degree} }`

`\sf{ {\angle}A\: = \: 180{\degree} - 140{\degree} }`

`\sf{ {\angle}A \: =\: 40{\degree} }`

Thus, The angle A is 40°

Now,

We have to find the side a and b

We know that,

`\bold{\red{ Sin{\theta} \:=\: }}{\bold{\red{(Perpendicular )/(Hypotenuse )}}}`

`\bold{\red{ Cos{\theta} \:=\: }}{\bold{\red{( Base )/(Hypotenuse )}}}`

For side A

`\sf{ Sin\: 40 {\degree} \:= \:}{\sf{( a)/( 14 )}}`

`\sf{ Sin(}{\sf{\frac{2{\pi}}{9}}}{\sf{) \:= \:}}{\sf{( a)/(14 )}}`

`\sf{Sin(}{\sf{\frac{2{*} 3.14 }{9}}}{\sf{) \:= \:}}{\sf{( a)/(14 )}}`

`\sf{sin(}{\sf{(6.28)/(9)}}{\sf{) \:=\: }}{\sf{( a)/(14 )}}`

`\sf{ a \:= \:14 {*} 0.64}`

`\sf{ a \: = \:14 {*} 0.64}`

`\bold{ a\: =\: 8.96\: \: or \:\:9\:\: (approx) }`

For Side B

`\sf{ Sin\: 50 {\degree} = }{\sf{( b )/( 14 )}}`

`\sf{Sin(}{\sf{\frac{5{\pi}}{18}}}{\sf{)\: =\: }}{\sf{( b)/(14 )}}`

`\sf{Sin(}{\sf{\frac{5{*} 3.14 }{18}}}{\sf{ ) \: = \:}}{\sf{( b)/(14 )}}`

`\sf{Sin(}{\sf{(15.7)/(18)}}{\sf{ )\: = \:}}{\sf{( b)/(14 )}}`

`\sf{ b\: = \:14 {*} 0.76}`

`\bold{ b\: = \: 10.64\:\: or \:\:10.7\:\: (approx) }`

Hence, The value of angle A , side a and b is 40° , 9 and 10.7 .

Answer 2

A = 40°

a = 9.0

b = 10.7

Explanation:

The sum of interior angles of a triangle is 180°

⇒ m∠A + 50° + 90° = 180°

⇒ m∠A + 140° = 180°

⇒ m∠A = 180° - 140°

⇒ m∠A = 40°

`\mathsf{\cos(\theta)=(adjacent\ side)/(hypotenuse)}`

Given:

• 
  `\theta` = 50°
• side adjacent to the angle = a
• hypotenuse = 14

`\implies \mathsf{\cos(50)=(a)/(14)}`

`\implies \mathsf{a=14\cos(50)}`

`\implies \mathsf{a=9.0\ (nearest\ tenth)}`

`\mathsf{\sin(\theta)=(opposite\ side)/(hypotenuse)}`

Given:

• 
  `\theta` = 50°
• side opposite to the angle = b
• hypotenuse = 14

`\implies \mathsf{\sin(50)=(b)/(14)}`

`\implies \mathsf{b=14\sin(50)}`

`\implies \mathsf{b=10.7\ (nearest\ tenth)}`