![Solution of Right Angled Triangle Solution of Right Angled Triangle](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjmxRKA7XOYqYxD9REckTyha_56KHG6trVJjzaxaqd0o50DeFd-XG3PmKvOPgQ77AymVNEVuEws9xWGkc-erMwn5_KJF0RWFG3X_2ioJzIbMp7jZQS_uui5Ydvyn4WFzTiLkZcLUjkXz_mo/s16000/Solution+of+Right+Angled+Triangle.png)
Solution of Right Angled Triangle
A triangle has three sides and three angles. Angles and sides
together are said to be the six parts of a triangle.
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In a right-angled triangle, one angle is 90°, so there are 5
parts unknown.
Under a given condition, the finding of the unknown parts of a right angled triangle is known as the solution of a right-angled triangle.
Let ABC be a right-angled triangle, right-angled at B. We denote its angles ∠CAB, ∠ABC, and ∠BCA by A, B, and C respectively.
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgYcwGB1UMbI4DalnDLqWHqhBq19DzsE8r68-Kq2uTC2HQec4BkQbQhU4ieEfvA6RpHgYlxFpJ_TabJb0rm-_IyTLjo3itAykUuwjLF-dqqTTqP2evCre9XcGFh25O5lBFj9gsj4N7yqVA0/s0/right+angled+triangle.png)
Here, B = 90°.
The sides opposite to angles A, B, and C are denoted by a, b and
c respectively. So, BC = a, CA = b and AB = c.
Then,
By Pythagoras relation h2 = p2 + b2,
we have
b2 = a2 + c2
Again, by the sum of two acute angles,
A + C = 90°
If we know one side and any one of the remaining four parts, we
can find the other parts of the right-angled triangle. This process of
calculating the unknown parts is known as the solution of a right-angled triangle.
Methods of Solving a Right Angled Triangle
- When two sides of a right-angled triangle are given
(i)
When a and c, are
given.
Then, b2 = a2 + c2.
∴ b = √(a2 + c2)
Again, sinA = a/b and so A is found.
At last, C = 90° - A.
(ii)
When, a and b are
given.
Then, c2 = b2 – a2.
∴ c = √(b2 – a2)
Again, sinA = a/b and it gives A.
At last, C = 90° - A.
- When one side and one acute angle are given
(i)
When A and b are
given.
Then, C = 90° - A
sinA = a/b
∴ a = b sinA
cosA = c/b
∴ c = b cosA
(ii)
When, A and a are
given.
Then, C = 90° - A
sinA = a/b
∴ b = a/sinA = a cosecA
cosA = c/b
∴ c = b cosA
Worked Out Examples
Example 1: Find the unknown angles and the sides of the following right angled
triangles:
Solution:
(i)
Figure: Right angled triangle ABC
![Right angled triangle ABC Figure: Right angled triangle ABC](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgCeMKA_iHAT7tfHibQCZA8OJeDHcURpfNN-D1Z9MqhDxP0AGFpudpV0rflusSJohAQQRzfC2Dy772kS8lZrTGZa9gMBUlpc5gCPLuC6n7eJ6femvD0eVQTaBhbBAjOcDMrksFRXZx0R-38/s16000/example+1%2528i%2529.png)
In DABC, ∠B = 90°, ∠C = 45° and BC = √2
∠A = 90° – ∠C = 90° – 45° = 45°
Now,
tan45° = AB/BC
or, 1
= AB/√2
or, AB
= √2
Again,
cos45° = BC/AC
or, 1/√2
= √2/AC
or, AC
= 2
∴ ∠A = 45°, AB = √2 and AC = 2
Ans.
(ii) Figure: Triangle ABC
![Triangle ABC Figure: Triangle ABC](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjSG2gVNNH93tqoC2uVhvrJ3A-BcgC6wgNqKsbeeutvodH0yzXBybRxYLiEdenGONtEWKnqBGaJU9utkJJ3wT9NL5IbCVkWdSEBqlL8uC0lzTR3Qg-5haxm-ieR23vZDR8bbr96B87nhm_U/s16000/example+1%2528ii%2529.png)
In triangle ABC, AB = √3, BC = 1 and AC = 2
Now,
AB2 + BC2 =
(√3)2 + (1)2
=
3 + 1
=
4
=
(2)2
=
AC2
Since, AB2 + BC2 = AC2
∴ ∠B = 90°
Now,
sinC = AB/AC
or,
sinC = √3/2
or,
sinC = sin60°
∴ C = 60°
And, ∠A = 90° –
C = 90° – 60° = 30°
∴ ∠B = 90°, ∠C = 60°, and ∠A = 30° Ans.
Example 2: Solve the right angled triangle ABC in which ∠C = 90°, ∠B = 30° and a = 4√3 cms.
![Right angled triangle ABC In right angled triangle ABC, ∠C = 90°, ∠B = 30° and a = BC = 4√3 cms.](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhgnTU8LuKahW239PTnlmmB8h_LNCRzTxf2P19G8X6p8OYgzGgpRaoJPf2nGlrEzwpvrymVcSIhhxc_z5hObmMrj0ZGH6PhaV8Q4rwSHSo6kUFnx1yFIXKbCslaYxJTxqU-n_xkZJO2iPFN/s16000/example+2.png)
Solution: Here,
In DABC,
∠C = 90°, ∠B = 30° and a = BC = 4√3
cms.
∠A = 90° – C = 90° – 30° = 60°
Now,
cos30° = BC/AB
or, √3/2 = 4√3/AB
or, AB√3 = 8√3
or, AB = 8√3/√3 = 8 cms.
Again,
sin30° = AC/AB
or, 1/2 = AC/8
or, 2AC = 8
or, AC = 8/2 = 4 cms.
∴ c = AB = 8cms, b = AC =
4cms, and ∠A = 60° Ans.
Example 3: In a right angled triangle ABC, if ∠B = 90°, a = 2√3 and b = 4, find the remaining angles and the sides.
![Right angled triangle ABC In triangle ABC, ∠B = 90°, a = BC = 2√3 and b = AC = 4.](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgtTlF8QrlowVuU2GQftUnzfttjnkB1Gj4Dv9qxsrjh6sZK2Lje5oYb4W8evRF021bZuf8-7CH6pG8A0mlTPpKCzhYnlOiHjPelHZE8TQNQIT9MGSQIoc7a5V9AAgYc82gxS74k3lO0Ewhw/s16000/example+3.png)
Solution: Here,
In triangle ABC,
∠B = 90°
a = BC = 2√3
b = AC = 4
Now,
cosC = BC/AC = 2√3/4 = √3/2 = cos30°
∴ C = 30°
And,
∠A = 90° – C = 90° – 30° = 60°
Again,
sin30° = AB/AC
or, 1/2 = AB/4
or, 2AB = 4
or, AB = 4/2 = 2
∴ ∠A = 60°, ∠C = 30°, and c = AB =
2 Ans.
Example 4: Solve the triangle ABC if ∠A = 90°, b = 16 and c = 16.
![Right angled triangle ABC In triangle ABC, ∠A = 90°, b = 16 and c = 16.](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj9v7tLldYSXMG-rZVm7eteeeSEHje1_AS3OuPb9Es2XzFZB0sFVy-YUPha3vb9ns55BzteDXggNPrWMrf2gykWJvf_ipynd-nftbdNAa0oLoYYXYft5bzVLTT0dsbswrbGi5pW8TDTAUvo/s16000/example+4.png)
Solution: Here,
In triangle ABC,
∠A = 90°
b = AC = 16
c = AB = 16
Since ∠A = 90°, so a is the
hypotenuse.
∴ a2 =
b2 + c2
= (16)2 + (16)2
= 2 × (16)2
or, a = 16√2
Now,
tanB = AC/AB = b/c = 16/16 = 1 = tan45°
∴ B = 45°
And,
∠C = 90° – B = 90° – 45° = 45°
∴ ∠B = 45°, ∠C = 45° and a = BC = 16√2
Ans.
Example 5: Solve the triangle ABC if a = 5√3, b = 5 and c = 10.
![Triangle ABC In triangle ABC, a = BC = 5√3, b = AC = 5 and c = AB = 10.](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgsJUhzJV3UzC-RXVbp32BBql_ezblZviVMJQpFHRjDnsaVk7nVxdI1a3GhDkDq-ze06sSHP-Z7zKo_tVq9lEf4CnYfpuFfE9l7EDHERDpkR6FoY982t7UoIQ06lunTlY8mGKPasg7MHzI3/s16000/example+5.png)
Solution: Here,
In triangle ABC,
a = BC = 5√3
b = AC = 5
c = AB = 10
Now,
AC2 + BC2 =
(5)2 + (5√3)2
=
25 + 75
=
100 = (10)2 = AB2
Since, AC2 + BC2
= AB2
∴ ∠C = 90°
Now,
cosA = AC/AB = 5/10 = ½ = cos60°
∴ A = 60°
And,
∠B = 90° – A = 90° – 60° = 30°
∴ ∠C = 90°, ∠B = 30°, and ∠A = 60° Ans.
Example 6: Solve a triangle ABC if ∠B = 60°, ∠A = 30° and a = 20.
![Triangle ABC In triangle ABC, ∠B = 60°, ∠A = 30° and a = BC = 20.](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg2J-IodSJIA6yFE5W8Qs2t4KeBkrskEggdzvX3_lBggUkvAEAmZmwzW7PB0pQWUbTZUGjJzd1-FRHcJRRnDLaFks2u6eRvTK2zO4GAlBbmXU10rheYY5GqWszQH-lnrA3SryJKfY67bXgk/s16000/example+6.png)
Solution: Here,
In triangle ABC,
∠B = 60°
∠A = 30°
a = BC = 20
Now,
∠C = 180° – (∠A + ∠B)
= 180° – (30° + 60°)
= 180° – 90°
= 90°
And,
sin30° = BC/AB
or, 1/2 = 20/AB
or, AB = 40
Again,
cos30° = AC/AB
or, √3/2 = AC/40
or, 2AC = 40√3
or, AC = 40√3/2 = 20√3
∴ ∠C = 90°, c = AB = 40 and b = AC = 20√3 Ans.
Example 7: In a right angled triangle the sides containing the right angle are 4√3cm and 12cm. Find the hypotenuse and acute angles.
![Right angled triangle ABC In triangle ABC, ∠B = 90°, c = AB = 4√3 cms and a = BC = 12 cms.](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgST_UlxiaendBiJbmQJvzcdDSVabJgkz6-KFZErc-PUahiCgXj8KVwDiGT6hlvulBjXgKWkH0smrVVnqnVlHZEjIUVTrK3nbeVgQT-Z5ih7ukzoPfQR7794OlYZIxJvj6I7dM5u7fX-jLw/s16000/example+7.png)
Solution: Here,
In triangle ABC,
∠B = 90°
c = AB = 4√3cm.
a = BC = 12cm.
Now,
tanC = AB/BC = 4√3/12 = 1/√3 = tan30°
∴ C = 30°
And,
∠A = 90° – C = 90° – 30° = 60°
Again,
sinC = AB/AC
or, sin30° = 4√3/AC
or, ½ = 4√3/AC
or, AC = 8√3 cm.
∴ ∠A = 60°, ∠C = 30°, and hypotenuse =
AC = 8√3cm. Ans.
Example 8: Solve the triangle ABC if a = 3, b = 4 and c = 5 [Take cos54° = 3/5]
![Triangle ABC In triangle ABC, a = BC = 3, b = AC = 4 and c = AB = 5.](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhkYkqSZUErAsYRJbLwd0VxWZt52Atd9fwRps9I4ULn8spS5HoFRCSzQHvQyr09IpOMPN31jFQJlfWu4sM8SNp0JAi6Wa1AiRT4LLnU5SD-oIEk0cUHpi_ZXHAvh2ZR9h2gHrPHuee1gyKe/s16000/example+8.png)
In triangle ABC,
a = BC = 3
b = AC = 4
c = AB = 5
Now,
a2 + b2 = 32 + 42 =
9 + 16 = 25 = 52 = c2
i.e. a2 + b2
= c2
∴ ∠C = 90°
Again,
cosB = a/c = 3/5 = cos54°
∴ B = 54°
And, ∠A = 90° – B
= 90° – 54°
=
36°
∴ ∠A = 36°, ∠B = 54° and ∠C = 90° Ans.
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