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Please, please help. Unit test is due. (K12 User)


​△ACB​ is bisected by CP←→. m∠PCB=65° and PB=11 ft.

What is the approximate area of each half of ​△ACB​ ?


28 ft²

56 ft²

67 ft²

143 ft²

Please please help Unit test is due K12 User ACB is bisected by CP mPCB65 and PB11 ft What is the approximate area of each half of ACB 28 ft 56 ft 67 ft 143 ft class=

Respuesta :

abidemiokin

The approximate area of each half of ​△ACB is 28 square feet

Area of a triangle

The formula for calculating the area of a triangle is expressed as:

A = 0.5absintheta

Given the following

PB =11ft

<PCB = 65degrees

Using the SOH CAH TOA identity

tan<PCB = 11/CP
CP = 11/tan65
CP = 5.13 ft

Determine the required area

Area = 0.5(11)(5.13)sin90

Area = 5.5 * 5.13

Area . = 28 square ft

Hence the approximate area of each half of ​△ACB is 28 square feet

Learn more on area of triangle here: https://brainly.com/question/136070

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