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geometry cp quiz 4.1 form b name block 1. ∠r = ————— ∠q = ————— 2. ∠a =…

Question

geometry cp
quiz 4.1 form b
name
block

  1. ∠r = ————— ∠q = —————
  2. ∠a = ————— ab = —————
  3. ∠l = —————
  4. x = —————
  5. δrst is isosceles. write and solve an equation to find x. then find the measure of each angle.

x = —————
∠r = —————
∠s = —————
∠t = —————

  1. δabc is equilateral. write and solve an equation to find x. then find the measure of each angle.

x = —————
∠b = —————
∠a = —————
∠c = —————

Explanation:

Step1: Solve for \(\angle R\) in problem 1

In \(\triangle PQR\), since \(PQ = PR\), \(\angle R=\angle P = 74^{\circ}\).
Using the angle - sum property of a triangle (\(\angle P+\angle Q+\angle R = 180^{\circ}\)), we substitute \(\angle P=\angle R = 74^{\circ}\).
\(74^{\circ}+74^{\circ}+\angle Q=180^{\circ}\)
\(\angle Q=180^{\circ}-(74^{\circ}+74^{\circ}) = 32^{\circ}\)

Step2: Solve for \(\angle A\) and \(AB\) in problem 2

In \(\triangle ABC\), using the angle - sum property (\(\angle A+\angle B+\angle C = 180^{\circ}\)), with \(\angle B = \angle C=48^{\circ}\)
\(\angle A=180^{\circ}-(48^{\circ}+48^{\circ})=84^{\circ}\)
Since \(\angle B=\angle C\), \(AB = AC = 15\) in.

Step3: Solve for \(\angle L\) in problem 3

In \(\triangle JKL\), since \(JK = JL = KL\) (equilateral triangle), all angles are equal.
Using the angle - sum property (\(\angle J+\angle K+\angle L = 180^{\circ}\)), and \(\angle J=\angle K=\angle L\)
\(\angle L=\frac{180^{\circ}}{3}=60^{\circ}\)

Step4: Solve for \(x\) in problem 4

Assuming \(\triangle LMN\) is equilateral (since all angles are equal in an equilateral triangle, but if it's isosceles with two equal angles, and given the side length \(LM = 31\), if it's equilateral \(x = 31\)

Step5: Solve for \(x\), \(\angle R\), \(\angle S\), \(\angle T\) in problem 5

Since \(\triangle RST\) is isosceles with \(RS = RT\), \(\angle R+\angle S+\angle T = 180^{\circ}\), and \(\angle S=\angle T\)
\((15x - 31)+2(9x + 11)=180\)
\(15x-31 + 18x+22=180\)
\(33x-9 = 180\)
\(33x=189\)
\(x = 7\)
\(\angle R=15x-31=15\times7 - 31=74^{\circ}\)
\(\angle T=9x + 11=9\times7+11=74^{\circ}\)
\(\angle S=180-(74 + 74)=32^{\circ}\)

Step6: Solve for \(x\), \(\angle A\), \(\angle B\), \(\angle C\) in problem 6

Since \(\triangle ABC\) is equilateral, \(\angle A=\angle B=\angle C = 60^{\circ}\)
\(9x-3=60\)
\(9x=63\)
\(x = 7\)

Answer:

  1. \(\angle R = 74^{\circ}\), \(\angle Q = 32^{\circ}\)
  2. \(\angle A = 84^{\circ}\), \(AB = 15\) in
  3. \(\angle L = 60^{\circ}\)
  4. \(x = 31\) (assuming equilateral)
  5. \(x = 7\), \(\angle R = 74^{\circ}\), \(\angle S = 32^{\circ}\), \(\angle T = 74^{\circ}\)
  6. \(x = 7\), \(\angle A = 60^{\circ}\), \(\angle B = 60^{\circ}\), \(\angle C = 60^{\circ}\)