Find the third angle of a triangle (using a parallel line) when two of the angles are: Class 7

Find the third angle of a triangle (using a parallel line) when two of the angles are: Class 7

Question 1.

Find the third angle of a triangle (using a parallel line) when two of the angles are : Class 7

(a) 36°, 72°

(b) 150°, 15°

(c) 90°, 30°

(d) 75°, 45°

Solution:

Do as directed, following the same procedure as given in the textbook. We will get:

(a) 180° - (36° + 72°) = 180° - 108° = 72°

(b) 180° - (150° + 15°) = 180° - 165° = 15°

(c) 180° - (90° + 30°) = 180° - 120° = 60°

(d) 180° - (75° + 45°) = 180° - 120° = 60°

Question 2.

Can you construct a triangle all of whose angles are equal to 70°? If two of the angles are 70° what would the third angle be? If all the angles in a triangle have to be equal, then what must its measure be? Explore and find out. Class 7

Solution:

(i) No, because 70° + 70° + 70° = 210° > 180°.

(ii) Third angle 180° - (70° + 70°) = 180° - 140° = 40°.

(iii) If all angles of a triangle are equal, then each angle will be of [latex]\frac{180^{\circ}}{3}[/latex] = 60°.

Question 3.

Here is a triangle in which we know ∠B = ∠C and ∠A = 50°. Can you find ∠B and ∠C? Class 7

Solution:

∠B + ∠C = 180°- ∠A

= 180° - 50° = 130°

So, 2∠B = 130°

i.e., ∠B = [latex]\frac{130^{\circ}}{3}[/latex] = 65°,

and ∠C = 65°