B

B

B

C

Chọn C

C

a) \(\approx\) 36,6

b) 130

c) 100

Câu hỏi là gì ạ?

a)A3=B1 vì // ( sole trong = nhau) nhớ tick nhé

b) vì // (trong cùng phía bù nhau)

nhớ tick nhé

kết quả nha

 1 Linh said that her uncle had worked in Hai Phong for 10 years

2 She told us that she was having dinner with Rachel the following day

3 The teacher said to the children that they must wash their hands

4 Susan said that  they hadn't left for London at 5 p.m the day before

5 Tom said to me that his brother would stay there the following Friday

6 He wanted to know  when the train left

7 Jack asked the old man how long he had worked in that company

8 My parents asked me what time your friend would arrive that weekend

9 They asked Mr. Quang if he had seen a UFO the evening before

10 The boy askes->asked Anna if he might use her phone

\(\text{#♪◈₰♣}\)

cho giá trị A1=5;B1=8.Hãy chọn kết quả của công thức=A1*2+B1*3
A.24
B.18
C.13
D.34

D.34

Bài 1: 

a) Áp dụng hệ thức lượng trong tam giác vuông vào ΔABC vuông tại A có AH là đường cao ứng với cạnh huyền BC, ta được:

\(AH^2=BH\cdot CH\)

\(\Leftrightarrow x=\dfrac{6^2}{12}=\dfrac{36}{12}=3\left(cm\right)\)

Áp dụng định lí Pytago vào ΔABH vuông tại H, ta được:

\(AB^2=AH^2+BH^2\)

\(\Leftrightarrow y^2=3^2+6^2=9+36=45\)

hay \(y=3\sqrt{5}\left(cm\right)\)

b) Ta có: \(\sin^2\widehat{A}+\cos^2\widehat{A}=1\)

\(\Leftrightarrow\cos^2\widehat{A}=\dfrac{3}{4}\)

hay \(\cos\widehat{A}=\dfrac{\sqrt{3}}{2}\)

Ta có: \(\tan\widehat{A}=\dfrac{\sin\widehat{A}}{\cos\widehat{A}}\)

\(=\dfrac{1}{2}:\dfrac{\sqrt{3}}{2}\)

\(=\dfrac{\sqrt{3}}{3}\)

Lời giải:

a. Theo định lý Pitago cho tam giác $ABH$:

$y^2=36+x^2(1)$

Theo hệ thức lượng trong tam giác vuông:

$y^2=x(x+12)(2)$

Từ $(1);(2)\Rightarrow x^2+36=x(x+12)$

$36=12x$

$x=3$ (cm)

$y^2=3(3+12)=45\Rightarrow y=3\sqrt{5}=6,708$ (cm)

b.

Giả sử $a = \widehat{B}$

$\cos a=\cos B=\frac{AB}{BC}=\frac{1}{2}$

$\Rightarrow AB=\frac{BC}{2}$

Theo Pitago: $AC=\sqrt{BC^2-AB^2}=\sqrt{BC^2-(\frac{BC}{2})^2}=\frac{\sqrt{3}}{2}BC$

$\sin a= \sin B=\frac{AC}{BC}=\frac{\sqrt{3}}{2}$

$\tan a=\tan B=\frac{AC}{AB}=\frac{\frac{BC}{2}}{\frac{\sqrt{3}}{2}BC}=\frac{\sqrt{3}}{3}$

Đáp án A