Question

Given a triangle PQR . Find all the angles of the triangle and show that the sum of three angles of PQR is 180°.
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Answer 1

Answer:

Angle Q = 90°

Angle R = 58°

Angle P = ?

So by the angle sum property we will find it

Sum of all three angles of triangle are 180°

So 90° + 58° + Angle P = 180°

148° + Angle P = 180°

Angle P = 180° - 148°

= 32°

Hope so the above answer will be helpful to u

Answer 2

Question:

Given a triangle PQR . Find all the angles of the triangle and show that the sum of three angles of PQR is 180°.

Given:

Angle Q Is 90°

Angle R Is 32°

To Find:

Angle P

Solution:

$\fbox{the \: sum \: of \: all \: angles \: of \: a \: triangle \: is \: 180\degree}$

$\bf{so</u><u>,</u><u>}$

$</u><u>\sf{\angle p + \angle q + \angle r \ \longmapsto 180 ^\circ}\\\sf{\angle p + 90^\circ + 58 ^\circ \longmapsto 180 ^\circ }\\\sf{ \angle p + 148 ^\circ \longmapsto180 ^\circ} \\ \sf{\angle p \ \longmapsto \ 180 ^\circ - 148 ^\circ} \\ \sf{ \angle p \ \longmapsto 32 ^\circ}</u><u>$

So,

$\sf{ \angle p \longmapsto 32 \degree}$

To Know More:

$\longrightarrow\sf{perimeter \: of \: triangle \: is \: 3 \times side} \\ \longrightarrow\sf {area \: of \: triangle \: is \: \frac{1}{2} \times base \times height} \\ \longrightarrow \sf{cone \: is \: a \: 3rd \: dimension \: shape \: of \: triangle}$