Question

Cho tam giác ABC, $\hat{B}={45}^o,$  $\hat{C}={30}^o,$ $BC=10cm$. Tính $AB$ và $AC$.

Answer 1

Trả lời:

Tam giác ABC có:

Sin B = AC/BC (hệ thức lượng)

=> AC = Sin B.BC = Sin 45⁰ . 10 = 5√2 (cm)

 Sin C = AB/BC

=> AB = Sin 30⁰ . 10 = 5 (cm)

Ta có tam giác ABC có: góc A + góc B + góc C = 180⁰ 

=> góc A = 180⁰ - 45⁰ - 30⁰ = 105⁰

Answer 2

Tam giác ABC có: Sin B = $\frac{AC}{BC}$ (hệ thức lượng) => AC = Sin B.BC = Sin 45⁰ . 10 = $5\sqrt{2}$ (cm)

 Sin C = $\frac{AB}{BC}$ (hệ thức lượng) => AB = Sin 30⁰ . 10 = 5 (cm)

Ta có tam giác ABC có: góc A + góc B + góc C = 180⁰ (định lý)

=> góc A = 180⁰ - 45⁰ - 30⁰ = 105⁰

Answer 3

Hạ $AH\perp BC$. Đặt $BH\; =\; x\; (0<x<10)\; =>\; AH=x\; ,\; HC=10-x$
 

AH=HC.tan 30 độ <=> x=(10-x) 1 phần canw3 <=>( căn 3-1)x=10
<=>(căn 3 +1)x=10 <=> x=10 phần căn 3 +1=5( căn 3-1)
=>AH=5( căn 3-1)
AC= AH phần sin c =AH phần sin30 độ =2.AH =10( canw3-1) cm 

Answer 4

Answer 5

Kẻ AH vuông góc với BC. Đặt BH=x (0<x<10)

\(\Rightarrow\) AH=x , HC=10-x

AH=HC.tan \(30^0\)\(\Leftrightarrow x=\left(10-x\right).\dfrac{1}{\sqrt{3}}\Leftrightarrow\left(\sqrt{3}+1\right).x=10\)

\(\Rightarrow AH=5.\left(\sqrt{3}-1\right)\)

AC=\(\dfrac{AH}{\sin C}=\dfrac{AH}{\sin30^0}=2.AH=10.\left(\sqrt{3}-1\right)cm\)

Áp dụng định lý Pytago vào\(\Delta ABC\) ta có:

AB=\(\sqrt{BC^2-AC^2}=\sqrt{10^2-[10.\left(\sqrt{3}-1\right)]^2}\approx6,813cm\)

Vậy AC=10.(\(\sqrt{3}\)-1); AB\(\approx\)6,813cm

Answer 6

 

 

Answer 7

 

Answer 8

 

Hạ $AH\perp BC$. Đặt $BH=x$ $\left(0<x<10\right)$ $\Rightarrow AH=x,$ $HC=10-x$.

AH=HC.\tan{{30}^\circ}\Leftrightarrow x=\left(10-x\right)\frac{1}{\sqrt3} \Leftrightarrow\left(\sqrt3+1\right)x=10AH=HC.tan30∘⇔x=(10−x)3​1​⇔(3​+1)x=10

\Leftrightarrow\left(\sqrt3+1\right)x=10\Leftrightarrow x=\frac{10}{\sqrt3+1}=5\left(\sqrt3-1\right).⇔(3​+1)x=10⇔x=3​+110​=5(3​−1).

Suy ra AH = 5(\sqrt{3}-1)AH=5(3​−1).

AC=\frac{AH}{\sin{C}}=\frac{AH}{\sin{{30}^o}}=2.AH=10\left(\sqrt3-1\right)\ cm.AC=sinCAH​=sin30oAH​=2.AH=10(3​−1) cm.

Answer 9

Hạ $AH\perp BC$. Đặt $BH=x$ $\left(0<x<10\right)$ $\Rightarrow AH=x,$ $HC=10-x$.

AH=HC.\tan{{30}^\circ}\Leftrightarrow x=\left(10-x\right)\frac{1}{\sqrt3} \Leftrightarrow\left(\sqrt3+1\right)x=10AH=HC.tan30∘⇔x=(10−x)3​1​⇔(3​+1)x=10

\Leftrightarrow\left(\sqrt3+1\right)x=10\Leftrightarrow x=\frac{10}{\sqrt3+1}=5\left(\sqrt3-1\right).⇔(3​+1)x=10⇔x=3​+110​=5(3​−1).

Suy ra AH = 5(\sqrt{3}-1)AH=5(3​−1).

AC=\frac{AH}{\sin{C}}=\frac{AH}{\sin{{30}^o}}=2.AH=10\left(\sqrt3-1\right)\ cm.AC=sinCAH​=sin30oAH​=2.AH=10(3​−1) cm.

Answer 10

 

 

Answer 11

Hạ $AH\perp BC$. Đặt $BH=x$ $\left(0<x<10\right)$ $\Rightarrow AH=x,$ $HC=10-x$.

AH=HC.\tan{{30}^\circ}\Leftrightarrow x=\left(10-x\right)\frac{1}{\sqrt3} \Leftrightarrow\left(\sqrt3+1\right)x=10AH=HC.tan30∘⇔x=(10−x)3​1​⇔(3​+1)x=10

\Leftrightarrow\left(\sqrt3+1\right)x=10\Leftrightarrow x=\frac{10}{\sqrt3+1}=5\left(\sqrt3-1\right).⇔(3​+1)x=10⇔x=3​+110​=5(3​−1).

Suy ra AH = 5(\sqrt{3}-1)AH=5(3​−1).

AC=\frac{AH}{\sin{C}}=\frac{AH}{\sin{{30}^o}}=2.AH=10\left(\sqrt3-1\right)\ cm.AC=sinCAH​=sin30oAH​=2.AH=10(3​−1) cm

Answer 12

 

Answer 13

 

Answer 14

Hạ $AH\perp BC$. Đặt $BH=x$ $\left(0<x<10\right)$ $\Rightarrow AH=x,$ $HC=10-x$.

$AH=HC.\tan {30}^{\circ }\iff x=\left(10-x\right)\frac{1}{\sqrt{3}}\iff \left(\sqrt{3}+1\right)x=10$

$\iff \left(\sqrt{3}+1\right)x=10\iff x=\frac{10}{\sqrt{3}+1}=5\left(\sqrt{3}-1\right).$

Suy ra $AH=5(\sqrt{3}-1)$.

$AC=\frac{AH}{\sin C}=\frac{AH}{\sin {30}^o}=2.AH=10\left(\sqrt{3}-1\right)cm.$

Answer 15

 

Answer 16

Hạ AH vuông vs BC. Đặt BH= x(0<x<10) ⇒AH=x, HC= 10-x

AH=HC.\(\tan30^o\)⇔ x= ( 10 - x ) . \(\dfrac{1}{\sqrt{3}}\)⇔(\(\sqrt{3}\)+1)x= 10 ⇔ x= \(\dfrac{10}{\sqrt{3}+1}\)= 5(\(\sqrt{3}\)- 1)

⇒AH= \(5\left(\sqrt{3}-1\right)\)

AC = \(\dfrac{AH}{\sin C}=\dfrac{AH}{\sin30^O}=2.AH=10\left(\sqrt{3}-1\right)\)cm

Answer 17

 

 

Answer 18

 

Answer 19

 

Answer 20

 

 

 

 

 

 

 

 

Answer 21

AC=5\(\sqrt{2}\)cm

 

AB=5cm

 

Answer 22

Hạ $AH\perp BC$. Đặt $BH=x$ $\left(0<x<10\right)$ $\Rightarrow AH=x,$ $HC=10-x$.

$AH=HC.\tan {30}^{\circ }\iff x=\left(10-x\right)\frac{1}{\sqrt{3}}\iff \left(\sqrt{3}+1\right)x=10$

$\iff \left(\sqrt{3}+1\right)x=10\iff x=\frac{10}{\sqrt{3}+1}=5\left(\sqrt{3}-1\right).$

Suy ra $AH=5(\sqrt{3}-1)$.

$AC=\frac{AH}{\sin C}=\frac{AH}{\sin {30}^o}=2.AH=10\left(\sqrt{3}-1\right)cm.$

Answer 23

Answer 24

Kẻ AH vuông góc với BC. Đặt BH=x

ð AH=x ; HC=10-x

AH=HC. Tan30

ó(căn 3+1)x=(10-x).1/căn3 ó(căn 3+1)x=10

óx= 5(căn 3-1)

ð AH=5(căn 3-1)

AC=AH/sinC=AH/sin30=2.AH=10(căn 3-1) cm

Answer 25

AH= 5( căn 3 -1)

AC = 10( căn 3 - 1)

Answer 26

 

 

Answer 27

 

 

Answer 28

Hạ $AH\perp BC$. Đặt $BH=x$ $\left(0<x<10\right)$ $\Rightarrow AH=x,$ $HC=10-x$.

$AH=HC.\tan {30}^{\circ }\iff x=\left(10-x\right)\frac{1}{\sqrt{3}}\iff \left(\sqrt{3}+1\right)x=10$

$\iff \left(\sqrt{3}+1\right)x=10\iff x=\frac{10}{\sqrt{3}+1}=5\left(\sqrt{3}-1\right).$

Suy ra $AH=5(\sqrt{3}-1)$.

$AC=\frac{AH}{\sin C}=\frac{AH}{\sin {30}^o}=2.AH=10\left(\sqrt{3}-1\right)cm.v$

Answer 29

Answer 30