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Nguyễn Quân
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Nguyễn Lê Phước Thịnh
17 tháng 8 2023 lúc 19:32

3:

góc C=90-50=40 độ

Xét ΔABC vuông tại A có sin C=AB/BC

=>4/BC=sin40

=>\(BC\simeq6,22\left(cm\right)\)

\(AC=\sqrt{BC^2-AB^2}\simeq4,76\left(cm\right)\)

1:

góc C=90-60=30 độ

Xét ΔABC vuông tại A có

sin B=AC/BC

=>3/BC=sin60

=>\(BC=\dfrac{3}{sin60}=2\sqrt{3}\left(cm\right)\)

=>\(AB=\dfrac{2\sqrt{3}}{2}=\sqrt{3}\left(cm\right)\)

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Minh Lâm
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Cường Hoàng
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Bảo Duy
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Cee Hee
1 tháng 10 2023 lúc 19:33

Câu a) với b) tính cos, tan, sin là tính góc hay cạnh vậy cậu?

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Cee Hee
1 tháng 10 2023 lúc 20:24

a) Xét \(\Delta ABC\) vuông tại `A`

Ta có: \(BC^2=AB^2+AC^2\) (đl Pytago)

\(\Rightarrow5^2=4^2+AC^2\\ \Rightarrow AC^2=5^2-4^2\\ \Rightarrow AC^2=25-16=9\\ \Rightarrow AC=\sqrt{9}=3cm\) 

Vậy: \(AC=3cm\)

Ta có: \(CosC=\dfrac{AC}{BC}\left(tslg\right)\)

\(\Rightarrow CosC=\dfrac{3}{5}\\ \Rightarrow CosC\approx53^o\)

Vậy: Góc C khoảng \(53^o\)

Ta có: \(TanB=\dfrac{AC}{AB}\left(tslg\right)\)

\(\Rightarrow TanB=\dfrac{3}{4}\\ \Rightarrow TanB\approx37^o\)

Vậy: Góc B khoảng \(37^o\) 

_

b) Xét \(\Delta ABC\) vuông tại `A`

Ta có: \(BC^2=AB^2+AC^2\) (đl Pytago)

\(\Rightarrow10^2=5^2+AC^2\\ \Rightarrow AC^2=10^2-5^2\\\Rightarrow AC^2=100-25=75\\ \Rightarrow AC=\sqrt{75}=5\sqrt{3}cm\)

Vậy: \(AC=5\sqrt{3}cm\)

Ta có: \(SinC=\dfrac{AB}{BC}\left(tslg\right)\)

 \(\Rightarrow SinC=\dfrac{5}{10}\\ \Rightarrow30^o\)

Vậy: Góc C là \(30^o\)

Ta có: \(SinB=\dfrac{AC}{BC}\left(tslg\right)\)

\(\Rightarrow SinB=\dfrac{5\sqrt{3}}{10}\\ \Rightarrow SinB=60^o\)

Vậy: Góc B là \(60^o\).

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The Moon
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The Moon
20 tháng 8 2021 lúc 17:54

GẤP LẮM Ạ,NGAY BÂY GIỜ Ạ

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Hoàng Trang
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Cao Linh Chi
13 tháng 2 2016 lúc 11:30

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CUTE vô đối
7 tháng 3 2017 lúc 20:37

CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCGCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC

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cô nàng bạch dương
18 tháng 3 2017 lúc 12:04

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no name
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Hoang Anh
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scotty
8 tháng 2 2021 lúc 20:30

A B C 16 12 H

1) Có \(\Delta ABC\) vuông 

=> S\(\Delta ABC\) = \(\dfrac{AB.AC}{2}\) = \(\dfrac{16.12}{2}\) = 96 (cm2)

2) Có \(\Delta ABC\) vuông , theo định lý Pytago ta có :

 AB +  AC2 =  BC2

=> 162 + 122 = BC2

=> 400            = BC2

=> BC             = 20 (cm)

Ta có :  S\(\Delta ABC\)  =  S\(\Delta ABH\)  +  S\(\Delta ACH\)

=>  \(\dfrac{BH.AH}{2}+\dfrac{HC.AH}{2}=S\Delta ABC\)

=>  \(\dfrac{BH.AH+HC.AH}{2}=S\Delta ABC\)

=> \(\dfrac{AH.\left(BH+HC\right)}{2}=S\Delta ABC\)

=> \(\dfrac{AH.BC}{2}\)               =  96

=> AH                         =  96 .  \(\dfrac{2}{BC}\) = 96 .  \(\dfrac{2}{20}\) = 9.6 (cm)

3) Có \(\Delta ABH\) vuông , theo định lý Pytago ta có :

    BH2 = AB2 - AH2

=>BH= 162 - 9.62 = 163.84

=> BH = 12.8 (cm)

=> CH = BC - BH = 20 - 12.8 = 7.2 (cm)

 

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Đặng Thanh Hằng
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Võ Đăng Khoa
28 tháng 2 2015 lúc 20:26

khó vãi, giải cả bủi tấu mak 0 ra , mình sr nhá

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Đinh Phước Lợi
11 tháng 2 2018 lúc 14:47

https://docs.google.com/document/d/1Wuo1vFdubrUg8F8-Ng_f-K8sda_JE_rRM704rtBrI-Q/edit?usp=sharing

Ta có     H1+ H2+H3=180

E1+E2=180

mà E1=H1

nên E2=H2+H3

Tong 3 goc trong tam giác: E2+H2+A1=180

(H2+H3)+H2+A1=180

2.H2+H3+A1=180

SUY RA: H2=(180-90-A1):2        ***    H3=90 hihi

=45-A1/2

mà A1=90-2A2

thay vào *** ta có H2=45-(90-2.A2)/2=A2

vậy H2=A2 hay EH//AD

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Đinh Phước Lợi
11 tháng 2 2018 lúc 14:55

Ta có     H1+ H2+H3=180

E1+E2=180

mà E1=H1

nên E2=H2+H3

Tong 3 goc trong tam giác: E2+H2+A1=180

(H2+H3)+H2+A1=180

2.H2+H3+A1=180

SUY RA: H2=(180-90-A1):2        ***    H3=90 hihi

=45-A1/2

mà A1=90-2A2

thay vào *** ta có H2=45-(90-2.A2)/2=A2

vậy H2=A2 hay EH//AD


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Lê Nguyệt Tú
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