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Alexandra [31]
3 years ago
7

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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8 0
3 years ago
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\small\begin{array}{ |c|c} \hline \bold{Area\ Between\ Curves} \\ \\ \textsf{Solving for the intersection of }\rm y = x^2 + 2\textsf{ and }\\ \rm y = 4, \\ \\ \qquad \begin{aligned} \rm y_1 &=\rm y_2 \\ \rm x^2 + 2 &=\rm 4 \\ \rm x^2 &= \rm 2 \\ \rm x &=\rm \pm \sqrt{2} \end{aligned} \\ \\ \textsf{We only need the first quadrant area bounded} \\ \textsf{by the given curves so the integral for the area} \\ \textsf{would then be} \\ \\ \boldsymbol{\displaystyle \rm A = \int_{\ a}^{\ b} {\left( \begin{array}{c}\text{upper} \\ \text{function}\end{array} \right) - \left( \begin{array}{c} \text{lower} \\ \text{function} \end{array} \right)\ dx}} \\ \\ \displaystyle \rm A = \int_{0}^{\sqrt{2}} \Big[4 - (x^2 + 2)\Big]\ dx \\ \\ \displaystyle \rm A = \int_{0}^{\sqrt{2}} (2 - x^2)\ dx \\ \\ \rm A = \left[2x - \dfrac{x^3}{3}\right]_{0}^{\sqrt{2}} \\ \\ \rm A = 2\sqrt{2} - \dfrac{\big(\sqrt{2}\big)^3}{3} \\ \\ \rm A = 2\sqrt{2} - \dfrac{2\sqrt{2}}{3} \\ \\\red{\boxed{\begin{array}{c} \rm A = \dfrac{4\sqrt{2}}{3}\textsf{ sq. units} \\ \textsf{or} \\ \rm A \approx 1.8856\textsf{ sq. units} \end{array}}} \\\\\hline\end{array}

#CarryOnLearning

#BrainlyForTrees

\qquad\qquad\qquad\qquad\qquad\qquad\tt{Monday\:at \: 04-04-2022} \\ \qquad\qquad\qquad\qquad\qquad\qquad\tt{12:10 \: pm}

5 0
2 years ago
HELP PLS HELP, DUE IN 10 MINUTES
amid [387]

Answer:

A.  26.10 cm

B. 118.95 cm

Step-by-step explanation:

ST = 41^2 - 40^2 =  c^2  = hypotenuse

ST = 1681 - 1600 = c^2

ST = c^2 = sq rt 681 =26.0959767014 = 26.1cm

Nearest 100th = 26.10

Length = 26.10 cm to nearest 100th

Perimeter of RSU we find (M) of SU first then add that to the other 3 lengths on the exterior of the triangle.

SU = 10^2 + 26.1^2 = c^2 = hypotenuse

SU = 100 + 681.21 = c^2

SU = c^2 = sqrt 781.21 = 27.9501341678 = 27.95cm

P TOTAL RSU  = SU + TR + RS + TU

= 27.95 + 40+ 41 + 10 = 118.95cm

5 0
3 years ago
Read 2 more answers
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