Let's begin noting that a triangle is isosceles if and only if two of its angles are congruent. We can thus find the angle <ABP, recalling that the sum of the interior angles of a triangle is equal to 180°.

$\angle \: ABP = \frac{180 - 80}{2} = 50 \degree$

Finally, let point K be the intersection between segments BC and PQ, and let's note that the triangle PQB is a right isosceles triangle, since all the angles in a square are equal to 90°, and the two triangles APB and BQC are congruent.

Therefore, the angle BKQ is equal to 180-50-45=85°.

Of course angle BKP=180-85=95°.

Hope this helps :)

6 0

3 years ago

Which equation can be solved to find one of the missing side lengths in the triangle?aB60512 unitsАCOS(60°) =22 믕cos(60°) = 12Su

Mekhanik [1.2K]

In a right rectangle, we have:

$\sin \alpha=\frac{opposite}{hypotenuse}$ $\cos \alpha=\frac{\text{adjacent}}{hypotenuse}$

For your exercice, hypotenuse=12

The exercise also inform the angle 60°, then:

$\sin \text{ 60}=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{b}{12}$ $\cos 60=\frac{adjacent}{\text{hypotenuse}}=\frac{a}{12}$

3 0

1 year ago

Is 4y=x a direct variation

worty [1.4K]

The direct variation equation is y =kx

Lets see:-

4y/4 = x/4
y = 1/4x

<span>Answer: Yes, this is a direct variation </span>

6 0

3 years ago

Read 2 more answers

Solve this equation using the radical equation method

Yuki888 [10]

The domain:
$3x+6\geq0\ \ |-6\\3x\geq-6\ \ \ |:3\\x\geq-2\\\\D:x\in\left< -2;\ \infty\right)$

$2x=\sqrt{3x+6}-1\ \ \ |+1\\\\2x+1=\sqrt{3x+6}\ \ \ \ |^2\\\\(2x+1)^2=\left(\sqrt{3x+6}\right)^2\ \ \ |use:\ (a+b)^2=a^2+2ab+b^2\\\\(2x)^2+2\cdot2x\cdot1+1^2=3x+6\\\\4x^2+4x+1=3x+6\ \ \ \ |-3x-6\\\\4x^2+x-5=0\\\\4x^2-4x+5x-5=0\\\\4x(x-1)+5(x-1)=0\\\\(x-1)(4x+5)=0\iff x-1=0\ \vee\ 4x+5=0\\\\x=1\ \vee\ 4x=-5\\\\x=1\in D\ \vee\ x=-1.25\in D$

$Answer:\ \boxed{x=-1.25\ \vee\ x=-1}$

6 0

3 years ago