Mathematics, 27.07.2019 04:50 sirdre1982

Each base angle is an isosceles triangle measures 55 degrees 30 minutes. each of the congruent sides is 10 centimeters long. estimate the following problems to the nearest tenth. a. find the altitude of the triangle. b. what is the length of the base? c. fund the area of the triangle.

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Answer from: oktacos

To solve the problwm we must know about trignometric functions.

Trigonometric functions

$Sin \theta=\dfrac{Perpendicular}{Hypotenuse}$

$Cos \theta=\dfrac{Base}{Hypotenuse}$

$Tan \theta=\dfrac{Perpendicular}{Base}$

where perpendicular is the side of the triangle which is opposite to the angle, and the hypotenuse is the longest side of the triangle which is opposite to the 90° angle.

The area of the triangle is 46.672 cm².

Given to us

Base angle, ∠P = ∠R = 55° 30'Each Congruent Side, QP =QR = 10 cm

Assumption

Let the altitude(QS) be h, and the base of the triangle(PR) be b.

A.) The altitude of the triangle

The altitude of the triangle

In ΔPQS

$Sin(55^o\ 30') = \dfrac{Perpendicular}{Hypotenuse}\\\\Sin(55^o\ 30') = \dfrac{QS}{PQ}\\\\Sin(55^o\ 30') = \dfrac{h}{10}\\\\h = Sin(55^o\ 30') \times 10\\\\h = 8.24$

Hence, the altitude of the isosceles triangle is 8.24 cm.

B.) The length of the base

The length of the base

In ΔPQS

$Cos(55^o\ 30') = \dfrac{Base}{Hypotenuse}\\\\Cos(55^o\ 30') = \dfrac{PS}{PQ}\\\\Cos(55^o\ 30') = \dfrac{\dfrac{b}{2}}{10}\\\\\dfrac{b}{2} = cos(55^o\ 30') \times 10\\\\{b} = cos(55^o\ 30') \times 10\times 2\\\\b = 11.33$

Hence, the base of the isosceles triangle is 11.33 cm.

C.) The area of the triangle

The area of the triangle

$Area\ \triangle = \dfrac{1}{2}\times Height(altitude)\times base$

$Area\ \triangle PQR = \dfrac{1}{2}\times h\times b$

$= \dfrac{1}{2}\times 8.24\times 11.33\\\\=46.672\ cm^2$

Hence, the area of the triangle is 46.672 cm².

Learn more about Trignometric functions:

link

Answer from: adiboo2004

Answers:

55 deg 30 Minutes = 55.5 degrees

sin = opp/hyp

sin(55.5) = altitude/10

Altitude = 10*sin(55.5)

Altitude = 10 * sin(55.5) = 8.241261886

Rounded to nearest 10th

Altitude = 8.2 cm

b. for base use Low of Cosines

Let C = base

Side A and B = length 10 Cm

c^2 = 2*a^2 -2a^2*cos(<C)

<C = 180 - 2*55.5 = 180 -111 = 69 degrees

c = sqrt ( 2a^2 - 2a^2*cos(<C)

c = sqrt ( 2*100- 200*cos(69))

c = sqrt (200 - 200*cos(69) ) = 11.32812474

Answer base
c = 11.3 cms

c.
Area = (1/2)*b * h
Area = (1/2)*(answer a * answer b)
Area = (1/2)* 11.32812474* 8.241261886 = 46.67902132

Area = 46.7 (rounded)

Area = ( 1/2) * a*b*sin(<C) = (1/2)*10*10*sin(69) =
Area = 50* sin(69) = 46.67902132
Area = 46.7 (rounded)
Each base angle is an isosceles triangle measures 55 degrees 30 minutes. each of the congruent sides

Each base angle is an isosceles triangle measures 55 degrees 30 minutes. each of the congruent sides

Answer from: Quest

too hard the other guy did it

step-by-step explanation:

Answer from: Quest

answer: 30

step-by-step explanation: its 30 because im a basketball player and stephen curry hits most of his shots so if you put 30 you might make the shot goodluck

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