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3 years ago

In △ABC,c=71, m∠B=123°, and a=65. Find b. A. 101.5 B. 117.8 C. 123.0 D. 119.6

Mathematics

1 answer:

tia_tia [17]3 years ago

8 0

Answer:

Option D

Step-by-step explanation:

The questions which involve calculating the angles and the sides of a triangle either require the sine rule or the cosine rule. In this question, the two sides that are given are adjacent to each other the given angle is the included angle. This means that the angle B is formed by the intersection of the lines a and c. Therefore, cosine rule will be used to calculate the length of b. The cosine rule is:

b^2 = a^2 + c^2 - 2*a*c*cos(B).

The question specifies that c=71, B=123°, and a=65. Plugging in the values:

b^2 = 65^2 + 71^2 - 2(65)(71)*cos(123°).

Simplifying gives:

b^2 = 14293.0182932.

Taking square root on the both sides gives b = 119.6 (rounded to the one decimal place).

This means that the Option D is the correct choice!!!

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3 years ago

If p=(-3,-2) and q=(1,6) are the endpoints of the diameter of a circle find the equation of the circle

stira [4]

Answer:

The equation of the circle (x +1) )² +(y-(2))² = (2(√5))²

 or 

The equation of the circle x² + 2 x + y² - 4 y = 15

Step-by-step explanation:

Given points end Points are p(-3,-2) and q( 1,6)

The distance of two points formula

P Q = $\sqrt{x_{2} - x_{1})^{2} + ((y_{2} -y_{1})^{2} }$

P Q = $\sqrt{1 - (-3)^{2} + ((6 -(-2))^{2} }$

P Q = $\sqrt{16+64} = \sqrt{80}$

The diameter 'd' = 2 r

2 r = √80

= $\sqrt{16 X 5}$

= $4 \sqrt{5}$

r = 2√5

Mid-point of two end points

$(\frac{x_{1} + x_{2} }{2} , \frac{y_{1} +y_{2} }{2} ) = (\frac{-3+1}{2} ,\frac{-2 +6}{2} )$

= (-1 ,2)

Mid-point of two end points = center of the circle

(h,k) = (-1 , 2)

The equation of the circle

(x -h )² +(y-k)² = r²

(x -(-1) )² +(y-(2))² = (2(√5))²

x² + 2 x + 1 + y² - 4 y + 4 = 20

x² + 2 x + y² - 4 y = 20 -5

x² + 2 x + y² - 4 y = 15

Final answer:-

The equation of the circle (x +1) )² +(y-(2))² = (2(√5))²

 or 

The equation of the circle x² + 2 x + y² - 4 y = 15

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