Triangle Solver worksheet

1. Area: 16.28 square units

1. A triangle's angles always add up to 180°, so angle C is 180° minus the other two

   $C$$=180\text{°}-42\text{°}-70\text{°}$$=68\text{°}$

2. Find side b with the law of sines, which links each side to the angle across from it: b = a × sin B ÷ sin A

   $b$$=\frac{5\times \sin \left(70\text{°}\right)}{\sin \left(42\text{°}\right)}$$=7.021743$

3. Find side c with the law of sines, which links each side to the angle across from it: c = a × sin C ÷ sin A

   $c$$=\frac{5\times \sin \left(68\text{°}\right)}{\sin \left(42\text{°}\right)}$$=6.928273$

4. Find half the perimeter (s): add the three sides and divide by 2

   $s$$=\frac{5+7.021743+6.928273}{2}$$=9.475008$

5. Find the area with Heron's formula, which needs only the three sides, no height

   $\mathrm{Area}$$=\sqrt{9.475008\times \left(9.475008-5\right)\times \left(9.475008-7.021743\right)\times \left(9.475008-6.928273\right)}$$=\sqrt{264.911974}$$=16.276117$

6. Perimeter: add the three sides

   $P$$=5+7.021743+6.928273$$=18.950016$

2. Area: 8.68 square units

1. A triangle's angles always add up to 180°, so angle C is 180° minus the other two

   $C$$=180\text{°}-39\text{°}-50\text{°}$$=91\text{°}$

2. Find side a with the law of sines, which links each side to the angle across from it: a = c × sin A ÷ sin C

   $a$$=\frac{6\times \sin \left(39\text{°}\right)}{\sin \left(91\text{°}\right)}$$=3.776498$

3. Find side b with the law of sines, which links each side to the angle across from it: b = c × sin B ÷ sin C

   $b$$=\frac{6\times \sin \left(50\text{°}\right)}{\sin \left(91\text{°}\right)}$$=4.596967$

4. Find half the perimeter (s): add the three sides and divide by 2

   $s$$=\frac{3.776498+4.596967+6}{2}$$=7.186732$

5. Find the area with Heron's formula, which needs only the three sides, no height

   $\mathrm{Area}$$=\sqrt{7.186732\times \left(7.186732-3.776498\right)\times \left(7.186732-4.596967\right)\times \left(7.186732-6\right)}$$=\sqrt{75.323215}$$=8.678895$

6. Perimeter: add the three sides

   $P$$=3.776498+4.596967+6$$=14.373464$

3. Area: 16.64 square units

1. Use the law of cosines to find the side across from the angle you know. It gives c squared first

   $c^2$$=5^2+7^2-2\times 5\times 7\times \cos \left(72\text{°}\right)$$=52.36881$

2. Take the square root to get side c itself

   $c$$=\sqrt{52.36881}$$=7.23663$

3. Use the law of cosines to find the cosine of angle A from sides b, c and a

   $\mathrm{cos\; A}$$=\frac{7^2+{7.23663}^2-5^2}{2\times 7\times 7.23663}$$=0.75379219$

4. Turn that cosine back into an angle (inverse cosine, the cos⁻¹ key) to get angle A

   $A$$=\arccos \left(0.75379219\right)$$=41.080054\text{°}$

5. A triangle's angles always add up to 180°, so angle B is 180° minus the other two

   $B$$=180\text{°}-41.080054\text{°}-72\text{°}$$=66.919946\text{°}$

6. Find half the perimeter (s): add the three sides and divide by 2

   $s$$=\frac{5+7+7.23663}{2}$$=9.618315$

7. Find the area with Heron's formula, which needs only the three sides, no height

   $\mathrm{Area}$$=\sqrt{9.618315\times \left(9.618315-5\right)\times \left(9.618315-7\right)\times \left(9.618315-7.23663\right)}$$=\sqrt{277.005727}$$=16.643489$

8. Perimeter: add the three sides

   $P$$=5+7+7.23663$$=19.23663$

4. Area: 10.88 square units

1. Use the law of cosines to find the side across from the angle you know. It gives c squared first

   $c^2$$=4^2+6^2-2\times 4\times 6\times \cos \left(65\text{°}\right)$$=31.714323$

2. Take the square root to get side c itself

   $c$$=\sqrt{31.714323}$$=5.631547$

3. Use the law of cosines to find the cosine of angle A from sides b, c and a

   $\mathrm{cos\; A}$$=\frac{6^2+{5.631547}^2-4^2}{2\times 6\times 5.631547}$$=0.76524742$

4. Turn that cosine back into an angle (inverse cosine, the cos⁻¹ key) to get angle A

   $A$$=\arccos \left(0.76524742\right)$$=40.070991\text{°}$

5. A triangle's angles always add up to 180°, so angle B is 180° minus the other two

   $B$$=180\text{°}-40.070991\text{°}-65\text{°}$$=74.929009\text{°}$

6. Find half the perimeter (s): add the three sides and divide by 2

   $s$$=\frac{4+6+5.631547}{2}$$=7.815774$

7. Find the area with Heron's formula, which needs only the three sides, no height

   $\mathrm{Area}$$=\sqrt{7.815774\times \left(7.815774-4\right)\times \left(7.815774-6\right)\times \left(7.815774-5.631547\right)}$$=\sqrt{118.280708}$$=10.875693$

8. Perimeter: add the three sides

   $P$$=4+6+5.631547$$=15.631547$

5. Area: 11.29 square units

1. A triangle's angles always add up to 180°, so angle C is 180° minus the other two

   $C$$=180\text{°}-55\text{°}-50\text{°}$$=75\text{°}$

2. Find side b with the law of sines, which links each side to the angle across from it: b = a × sin B ÷ sin A

   $b$$=\frac{5\times \sin \left(50\text{°}\right)}{\sin \left(55\text{°}\right)}$$=4.675838$

3. Find side c with the law of sines, which links each side to the angle across from it: c = a × sin C ÷ sin A

   $c$$=\frac{5\times \sin \left(75\text{°}\right)}{\sin \left(55\text{°}\right)}$$=5.895889$

4. Find half the perimeter (s): add the three sides and divide by 2

   $s$$=\frac{5+4.675838+5.895889}{2}$$=7.785863$

5. Find the area with Heron's formula, which needs only the three sides, no height

   $\mathrm{Area}$$=\sqrt{7.785863\times \left(7.785863-5\right)\times \left(7.785863-4.675838\right)\times \left(7.785863-5.895889\right)}$$=\sqrt{127.49304}$$=11.291282$

6. Perimeter: add the three sides

   $P$$=5+4.675838+5.895889$$=15.571726$

6. Area: 24.25 square units

1. Use the law of cosines to find the side across from the angle you know. It gives c squared first

   $c^2$$=7^2+8^2-2\times 7\times 8\times \cos \left(60\text{°}\right)$$=57$

2. Take the square root to get side c itself

   $c$$=\sqrt{57}$$=7.549834$

3. Use the law of cosines to find the cosine of angle A from sides b, c and a

   $\mathrm{cos\; A}$$=\frac{8^2+{7.549834}^2-7^2}{2\times 8\times 7.549834}$$=0.59603956$

4. Turn that cosine back into an angle (inverse cosine, the cos⁻¹ key) to get angle A

   $A$$=\arccos \left(0.59603956\right)$$=53.413224\text{°}$

5. A triangle's angles always add up to 180°, so angle B is 180° minus the other two

   $B$$=180\text{°}-53.413224\text{°}-60\text{°}$$=66.586776\text{°}$

6. Find half the perimeter (s): add the three sides and divide by 2

   $s$$=\frac{7+8+7.549834}{2}$$=11.274917$

7. Find the area with Heron's formula, which needs only the three sides, no height

   $\mathrm{Area}$$=\sqrt{11.274917\times \left(11.274917-7\right)\times \left(11.274917-8\right)\times \left(11.274917-7.549834\right)}$$=\sqrt{588}$$=24.248711$

8. Perimeter: add the three sides

   $P$$=7+8+7.549834$$=22.549834$

7. Area: 19.27 square units

1. A triangle's angles always add up to 180°, so angle C is 180° minus the other two

   $C$$=180\text{°}-44\text{°}-58\text{°}$$=78\text{°}$

2. Find side a with the law of sines, which links each side to the angle across from it: a = c × sin A ÷ sin C

   $a$$=\frac{8\times \sin \left(44\text{°}\right)}{\sin \left(78\text{°}\right)}$$=5.68142$

3. Find side b with the law of sines, which links each side to the angle across from it: b = c × sin B ÷ sin C

   $b$$=\frac{8\times \sin \left(58\text{°}\right)}{\sin \left(78\text{°}\right)}$$=6.935952$

4. Find half the perimeter (s): add the three sides and divide by 2

   $s$$=\frac{5.68142+6.935952+8}{2}$$=10.308686$

5. Find the area with Heron's formula, which needs only the three sides, no height

   $\mathrm{Area}$$=\sqrt{10.308686\times \left(10.308686-5.68142\right)\times \left(10.308686-6.935952\right)\times \left(10.308686-8\right)}$$=\sqrt{371.428036}$$=19.272468$

6. Perimeter: add the three sides

   $P$$=5.68142+6.935952+8$$=20.617372$

8. Area: 12.5 square units

1. Use the law of cosines to find the cosine of angle A from sides b, c and a

   $\mathrm{cos\; A}$$=\frac{5^2+7^2-5^2}{2\times 5\times 7}$$=0.7$

2. Turn that cosine back into an angle (inverse cosine, the cos⁻¹ key) to get angle A

   $A$$=\arccos \left(0.7\right)$$=45.572996\text{°}$

3. Use the law of cosines to find the cosine of angle B from sides a, c and b

   $\mathrm{cos\; B}$$=\frac{5^2+7^2-5^2}{2\times 5\times 7}$$=0.7$

4. Turn that cosine back into an angle (inverse cosine, the cos⁻¹ key) to get angle B

   $B$$=\arccos \left(0.7\right)$$=45.572996\text{°}$

5. A triangle's angles always add up to 180°, so angle C is 180° minus the other two

   $C$$=180\text{°}-45.572996\text{°}-45.572996\text{°}$$=88.854008\text{°}$

6. Find half the perimeter (s): add the three sides and divide by 2

   $s$$=\frac{5+5+7}{2}$$=8.5$

7. Find the area with Heron's formula, which needs only the three sides, no height

   $\mathrm{Area}$$=\sqrt{8.5\times \left(8.5-5\right)\times \left(8.5-5\right)\times \left(8.5-7\right)}$$=\sqrt{156.1875}$$=12.4975$

8. Perimeter: add the three sides

   $P$$=5+5+7$$=17$

9. Area: 17.39 square units

1. Use the law of cosines to find the side across from the angle you know. It gives c squared first

   $c^2$$=6^2+6^2-2\times 6\times 6\times \cos \left(75\text{°}\right)$$=53.365029$

2. Take the square root to get side c itself

   $c$$=\sqrt{53.365029}$$=7.305137$

3. Use the law of cosines to find the cosine of angle A from sides b, c and a

   $\mathrm{cos\; A}$$=\frac{6^2+{7.305137}^2-6^2}{2\times 6\times 7.305137}$$=0.60876143$

4. Turn that cosine back into an angle (inverse cosine, the cos⁻¹ key) to get angle A

   $A$$=\arccos \left(0.60876143\right)$$=52.5\text{°}$

5. A triangle's angles always add up to 180°, so angle B is 180° minus the other two

   $B$$=180\text{°}-52.5\text{°}-75\text{°}$$=52.5\text{°}$

6. Find half the perimeter (s): add the three sides and divide by 2

   $s$$=\frac{6+6+7.305137}{2}$$=9.652569$

7. Find the area with Heron's formula, which needs only the three sides, no height

   $\mathrm{Area}$$=\sqrt{9.652569\times \left(9.652569-6\right)\times \left(9.652569-6\right)\times \left(9.652569-7.305137\right)}$$=\sqrt{302.296115}$$=17.386665$

8. Perimeter: add the three sides

   $P$$=6+6+7.305137$$=19.305137$

10. Area: 14.14 square units

1. Use the law of cosines to find the cosine of angle A from sides b, c and a

   $\mathrm{cos\; A}$$=\frac{5^2+9^2-6^2}{2\times 5\times 9}$$=0.77777778$

2. Turn that cosine back into an angle (inverse cosine, the cos⁻¹ key) to get angle A

   $A$$=\arccos \left(0.77777778\right)$$=38.942441\text{°}$

3. Use the law of cosines to find the cosine of angle B from sides a, c and b

   $\mathrm{cos\; B}$$=\frac{6^2+9^2-5^2}{2\times 6\times 9}$$=0.85185185$

4. Turn that cosine back into an angle (inverse cosine, the cos⁻¹ key) to get angle B

   $B$$=\arccos \left(0.85185185\right)$$=31.586338\text{°}$

5. A triangle's angles always add up to 180°, so angle C is 180° minus the other two

   $C$$=180\text{°}-38.942441\text{°}-31.586338\text{°}$$=109.471221\text{°}$

6. Find half the perimeter (s): add the three sides and divide by 2

   $s$$=\frac{6+5+9}{2}$$=10$

7. Find the area with Heron's formula, which needs only the three sides, no height

   $\mathrm{Area}$$=\sqrt{10\times \left(10-6\right)\times \left(10-5\right)\times \left(10-9\right)}$$=\sqrt{200}$$=14.142136$

8. Perimeter: add the three sides

   $P$$=6+5+9$$=20$