MHB Solve Trigonometry Angles: Degrees, Minutes & Seconds

[Drain Brain]

please help me to solve these problems,

1. how many degrees, minutes, and seconds are respectively passed over in $11\frac{1}{9}$ minutes by the hour and minute hands of a watch?

2. The number of degrees in one acute angle of a right-angled triangle is equal to the number of grades in the other; express both angles in degrees.

 

[mathmari]

2. The number of degrees in one acute angle of a right-angled triangle is equal to the number of grades in the other; express both angles in degrees.

View attachment 2986

The sum of the angles of a triangle is equal to $180^{\circ}$.

Therefore:

$$\hat{A}+\hat{B}+\hat{C}=180^{\circ} \Rightarrow 90^{\circ}+\hat{B}+\hat{C}=180^{\circ} \Rightarrow \hat{B}+\hat{C}=90^{\circ}$$

"The number of degrees in one acute angle of a right-angled triangle is equal to the number of grades in the other":
The two acute angles of the above right-angles trianle are $\hat{B}$ and $\hat{C}$.

So, $\hat{B}=\hat{C}$.

Replacing this at the equation $\hat{B}+\hat{C}=90^{\circ}$, we get the following:

$$\hat{B}+\hat{C}=90^{\circ} \Rightarrow 2 \hat{B}=90^{\circ} \Rightarrow \hat{B}=\hat{C}=45^{\circ}$$

 

[MarkFL]

please help me to solve these problems,

1. how many degrees, minutes, and seconds are respectively passed over in $11\frac{1}{9}$ minutes by the hour and minute hands of a watch?

2. The number of degrees in one acute angle of a right-angled triangle is equal to the number of grades in the other; express both angles in degrees.

Can you show us what you have tried? This way our helpers know where you are stuck and can better help.

View attachment 2986

The sum of the angles of a triangle is equal to $180^{\circ}$.

Therefore:

$$\hat{A}+\hat{B}+\hat{C}=180^{\circ} \Rightarrow 90^{\circ}+\hat{B}+\hat{C}=180^{\circ} \Rightarrow \hat{B}+\hat{C}=90^{\circ}$$

"The number of degrees in one acute angle of a right-angled triangle is equal to the number of grades in the other":
The two acute angles of the above right-angles trianle are $\hat{B}$ and $\hat{C}$.

So, $\hat{B}=\hat{C}$.

Replacing this at the equation $\hat{B}+\hat{C}=90^{\circ}$, we get the following:

$$\hat{B}+\hat{C}=90^{\circ} \Rightarrow 2 \hat{B}=90^{\circ} \Rightarrow \hat{B}=\hat{C}=45^{\circ}$$

I believe the OP was saying that one angle is measured in gradians while the other it in degrees. 400 gradians is equivalent to 360 degrees.

 

[mathmari]

I believe the OP was saying that one angle is measured in gradians while the other it in degrees. 400 gradians is equivalent to 360 degrees.

Oh, I'm sorry.. (Lipssealed)(Wasntme)

The angle $A$ is $90^{\circ}$.
Let the acute angle $B$ be measured in gradians, so the angle $B$ is $x$ gradians which is equal to $\frac{360 x}{400}=0.9 x$ degrees.
The angle $C$ is $y$ degrees.

"The number of degrees in one acute angle of a right-angled triangle is equal to the number of grades in the other": $x=y$

The sum of the angles of the triangle is equal to $180$ degrees.

$$90^{\circ}+0.9 x+y=180^{\circ} \Rightarrow 0.9 x+x=90^{\circ} \Rightarrow 1.9x=90^{\circ} \Rightarrow x=\left (\frac{90}{1.9}\right )^{\circ}$$

Therefore, the angle $B$ is $ \left (0.9\cdot \frac{90}{1.9}\right )^{\circ}$ and the angle $C$ is $\left (\frac{90}{1.9}\right )^{\circ}$.