Given:
The vertices of a triangle ABC are A(-3,-4), B(16,-2) and C(13,-10).
To show:
That the triangle ABC is a right angled triangle.
Solution:
Distance formula:
The vertices of a triangle ABC are A(-3,-4), B(16,-2) and C(13,-10).
Using the distance formula, we get
Similarly,
And,
Now, add the square of two smaller sides.
Since the sum of the square of two smaller sides is equal to the square of the largest side, therefore the given triangle is a right angle triangle by using Pythagoras theorem.
Hence proved.
Answer:
m∠P = 82°
m∠Q = 49°
m∠R = 49°
Step-by-step explanation:
<em>In the isosceles triangle, the base angles are equal in measures</em>
In Δ PQR
∵ PQ = PR
∴ Δ PQR is an isosceles triangle
∵ ∠Q and ∠R are the base angles
→ By using the fact above
∴ m∠Q = m∠R
∵ m∠Q = (3x + 25)°
∵ m∠R = (2x + 33)°
→ Equate them
∴ 3x + 25 = 2x + 33
→ Subtract 2x from both sides
∵ 3x - 2x + 25 = 2x - 2x + 33
∴ x + 25 = 33
→ Subtract 25 from both sides
∵ x + 25 - 25 = 33 - 25
∴ x = 8
→ Substitute the value of x in the measures of angles Q and R
∵ m∠Q = 3(8) + 25 = 24 + 25
∴ m∠Q = 49°
∵ m∠R = 2(8) + 33 = 16 + 33
∴ m∠R = 49°
∵ The sum of the measures of the interior angles of a Δ is 180°
∴ m∠P + m∠Q + m∠R = 180°
→ Substitute the measures of angles Q and R
∵ m∠P + 49 + 49 = 180
∴ m∠P + 98 = 180
→ Subtract 98 from both sides
∵ m∠P + 98 - 98 = 180 - 98
∴ m∠P = 82°
15% - 40. -25 = 1500 percent take out the zero because they have zero value and your left with 15 which is 15%
